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MATH 1200: CALCULUS FOR SCIENTISTS I Pamini Thangarajah Mount Royal University Mount Royal University MATH 1200: Calculus for Scientists I Pamini Thangarajah This open text is disseminated via the Open Education Resource (OER) LibreTexts Project (https://LibreTexts.org) and like the hundreds of other open texts available within this powerful platform, it is licensed to be freely used, adapted, and distributed. This book is openly licensed which allows you to make changes, save, and print this book as long; the applicable license is indicated at the bottom of each page. Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new technologies to support learning. 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This text was compiled on 10/21/2020 TABLE OF CONTENTS This course provides an introduction to calculus with exposure to applications in sciences. The main concepts covered are limits, derivatives, and integrals. Derivatives of transcendental functions are used to solve optimization, linear approximation, and related rates problems. Studying calculus will expose students to a variety of important mathematical ideas and help students develop their critical thinking and problem solving skills. 0: PRE-CALCULUS REFRESHER 0.1: REAL NUMBERS 0.2: ANALYTIC GEOMETRY 0.3: TRIGONOMETRY 0.4: LOGARITHMS AND EXPONENTS 0E: EXERCISES COMMON MISTAKES PREFACE 1: LIMIT AND CONTINUITY OF FUNCTIONS 1.0: LIBRARY OF FUNCTIONS 1.0E: EXERCISES 1.1: INTRODUCTION TO CONCEPT OF A LIMIT 1.1E: EXERCISES 1.2: ONE SIDED LIMITS AND VERTICAL ASYMPTOTES 1.2E: EXERCISES 1.3: LIMIT CALCULATIONS FOR ALGEBRAIC EXPRESSIONS 1.3E: EXERCISES 1.4: LIMITS AT INFINITY AND HORIZONTAL ASYMPTOTES : END BEHAVIOR 1.4E: EXERCISES 1.5: FORMAL DEFINITION OF A LIMIT (OPTIONAL) 1.5E: EXERCISES 1.6: CONTINUITY AND THE INTERMEDIATE VALUE THEOREM 1.6E: EXCERCISES 1.7: LIMIT OF TRIGONOMETRIC FUNCTIONS 1.7E: EXERCISES 1.8: LIMITS AND CONTINUITY OF INVERSE TRIGONOMETRIC FUNCTIONS 1.8E: EXERCISES 1.9: LIMIT OF EXPONENTIAL FUNCTIONS AND LOGARITHMIC FUNCTIONS 1.9E: EXERCISES 1E: REVIEW EXERCISES 2: DERIVATIVES 2.0: TANGENT LINES AND RATES OF CHANGE 2.0E: EXERCISES 2.1: DERIVATIVE AS A FUNCTION 2.1E: EXERCISES 2.2: TECHNIQUES OF DIFFERENTIATION 2.2E: EXERCISES 2.3: DERIVATIVE AS A RATE OF CHANGE 1 10/21/2020 2.3E: EXERCISES 2.4: DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 2.4E: EXERCISES 2.5: CHAIN RULE 2.5E: EXERCISES 2.6: IMPLICIT DIFFERENTIATION 2.6E: EXERCISES 2.7: DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS 2.7E: EXERCISES 2.8: DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS 2.8E: EXERCISES 2.9: L'HÔPITAL'S RULE 2.9E: EXERCISES 2E: EXERCISES 3: APPLICATIONS OF DERIVATIVES 3.0 INTRODUCTION TO APPLICATIONS OF DERIVATIVE 3.1: RELATED RATES 3.1E: EXERCISES 3.2 LINEAR APPROXIMATIONS AND DIFFERENTIALS 3.2E: EXERCISES 3.3: EXTREMAS 3.3E: EXERCISES 3.4 THE MEAN VALUE THEOREM 3.4E: EXERCISES 3.5 DERIVATIVE TESTS 3.5E: EXERCISES 3.6: APPLIED OPTIMIZATION PROBLEMS 3.6E: EXERCISES 3.7: CURVE SKECTCHING 3.7E: EXERCISES 3.8: NEWTON'S METHOD 3.8E: EXERCISES 3.9: ANTI DERIVATIVES AND RECTILINEAR MOTION 3.9E: EXERCISES 3E: CHAPTER EXERCISES 4: INTEGRAL CALCULUS 4.0: ANTIDERVATIVES AND INDEFINITE INTEGRATION (REVISITED) 4.0E: EXERCISES 4.1: INTEGRATION BY SUBSTITUTION 4.1E: EXERCISES 4.2: DEFINITE INTEGRAL- AN INTRODUCTION 4.2E: EXERCISES 4.3: APPROXIMATING AREAS 4.3E: EXERCISES 4.4: THE DEFINITE INTEGRAL 2 10/21/2020 4.4E: EXERCISES 4.5: THE FUNDAMENTAL THEOREM OF CALCULUS 4.5E: EXERCISES 4.6: INTEGRATION FORMULAS AND THE NET CHANGE THEOREM 4.6E: EXERCISES 4.7: DEFINITE INTEGRALS BY SUBSTITUTION. 4.7E: EXERCISES 4.8: AREA BETWEEN TWO CURVES 4.8E: EXERCISES 4.9: APPLICATIONS OF DEFINITE INTEGRALS 4.9E: EXERCISES 4E: EXERCISES ADDITIONAL RESOURCES GLOSSARY KEY EQUATIONS NOTATIONS TABLE OF DERIVATIVES TABLE OF INTEGRALS TABLE OF LIMITS MOCK EXAMS (CELEBRATION OF LEARNING) PRECALCULUS REVIEW TEST 1 TEST 2 TEST 3 BACK MATTER INDEX GLOSSARY 3 10/21/2020 Preface Course Description: This course provides an introduction to calculus with exposure to applications in science, business, and economics. The main concepts covered are limits, derivatives, and integrals. Derivatives of exponential, logarithmic, trigonometric and inverse trigonometric functions are used to solve optimization, linear approximation, and related rates problems. Techniques of integration and applications are also introduced. Studying calculus will expose students to a variety of important mathematical ideas and help students develop their calculation, critical thinking, and problem-solving skills. Prerequisite(s): A grade of 60% or higher in Mathematics 30-1 or in MATH 0130, or a grade of C- or higher in any of MATH 1283 or MATH 1285 or equivalent. The following information gives a quick overview of further learning opportunities at Mount Royal University (MRU) as they relate to this course's scope: If you are interested in learning more in depth, you can choose to take these courses at MRU: Math 2200 Calculus for Scientists II & Math 3200 Mathematical Methods Course Learning Outcomes: Upon successful completion of this course, students will be able to: Calculate limits, derivatives, and integrals for a large variety of functions Use Calculus concepts in applications Communicate solutions clearly in the language of mathematics A Note on Formatting: Throughout this resource, practice exercises can be found at the end of each chapter. No answer key is provided, for some of the problems. This is to encourage students to experience mathematics as a synthetic and creative field and also to attend class to ask questions. The "Thinking Out Loud" sections are to prompt discussion - take these up with your classmates and see if you can justify your position using what you know. Acknowledgments: The creation of this resource would not have been possible without significant help from a variety of sources. They are, in no particular order, Provost's Teaching and Learning Innovation Grant, Mount Royal University Professor Delmar Larsen, LibreTexts, for his unconditional support The Academic Development Center, Mount Royal University The Media Production Group, Mount Royal University The Department of Mathematics and Computing, Mount Royal University The Faculty of Science & Technology, Mount Royal University Former students, who have taken this class in person, and who donated their class notes as reference material, namely, Ms. Merry Faye Graff, Ramandeep Kaur, and Pam Rosete. Christopher Gergess, Undergraduate Research Assistant at Mount Royal University Rosie Jones, Undergraduate Research Assistant at Mount Royal University Jillian Perillat, Undergraduate Research Assistant at Mount Royal University Maryanne Thangarajah, University of Waterloo, Undergraduate Research Assistant-Volunteer Patrick Thangarajah, University of British Columbia, Undergraduate Research Assistant-Volunteer Thank you all so very much for your help, insights, and resources. Pamini Thangarajah, PhD 10/21/2020 1 CC-BY-NC-SA https://math.libretexts.org/@go/page/4796 Calgary, Alberta September 2018 email: pthangarajah@mtroyal.ca 10/21/2020 2 CC-BY-NC-SA https://math.libretexts.org/@go/page/4796 CHAPTER OVERVIEW 0: PRE-CALCULUS REFRESHER 0.1: REAL NUMBERS 0.2: ANALYTIC GEOMETRY 0.3: TRIGONOMETRY 0.4: LOGARITHMS AND EXPONENTS 0E: EXERCISES COMMON MISTAKES 1 10/21/2020 0.1: Real numbers Real number arithmetic Rules 1. The distributive law: If a, b and c are real numbers, then a(b + c) = ab + ac (0.1.1) (b + c)a = ba + ca. (0.1.2) and 2. The commutative laws: If a and b are real numbers, then ab = ba (0.1.3) a + b = b + a. (0.1.4) and 3. The associative law: If a, b and c are real numbers, then a + (b + c) = (a + b) + c (0.1.5) a(bc) = (ab)c. (0.1.6) and 4. The existence of 0: The real number 0 exists so that, for any real number a , a + 0 = 0 + a = a. (0.1.7) 5. The existence of 1: The real number 1 exists so that, for any real number a , a ⋅ 1 = 1 ⋅ a = a. (0.1.8) 6. Subtraction: For each real number a, there exists a real number −a, so that a + (−a) = 0 = (−a) + a. 7. Division: For each nonzero real number a, there exists a real number 1 a( , a so that 1 ) =( a 1 (0.1.9) ) a = 1. (0.1.10) a The laws above form the foundation of arithmetic and algebra of real numbers. They are the laws that we have accepted and used with no reserve. They are mentioned here to encourage the reader to develop an appreciation for them and an awareness that they must be respected in all calculations involving real numbers. The partitioning of the real number system The collection of all real numbers contains a number of important sets. These are introduced next together with the appropriate standard notation. 1. The collection of counting numbers otherwise known as the collection of natural numbers is usually denoted by N. We write N = {1, 2, 3, 4, …}. (0.1.11) 2. The collection of the integers is usually denoted by Z and we write Pamini Thangarajah 10/21/2020 0.1.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10821 Z = {… , −3, −2, −1, 0, 1, 2, 3, 4, …}. (0.1.12) Notice that N is properly contained in Z, denoted by N ⫋ Z. (0.1.13) 3. The collection of all rational numbers (fractions) is usually denoted by Q and We write a Q= { : a and b are integers, b ≠ 0} . (0.1.14) b Notice that Z is properly contained in Q That is, Z ⫋ Q. (0.1.15) 4. The collection of all irrational numbers is denoted by I . This set contains all the real numbers which are not rational numbers. The rational numbers and irrational numbers have no elements in common. That means Q ∩ I = ∅. 5. The collection of all real numbers is denoted by numbers. We write R (0.1.16) . This set contains all of the rational numbers and all of the irrational Q ∪ I = R. (0.1.17) Definitions 1. A real number is called positive if it is greater than 0. 2. A real number is called non-negative if it is greater than or equal to 0. 3. An integer n is an even number if there is an integer m such that n = 2m . 4. An integer n is an odd number if there is an integer m such that n = 2m + 1 . Special properties of the real numbers It is worthwhile mentioning the following three properties: Property 0.1.1 The collection of real numbers has an order: Given any two distinct real numbers a and b , one of the following statements will be true: 1. a < b, 2. b < a. Example 0.1.1 : Consider the real numbers 3.111 and 3.11101,it is not difficult to see that 3.111 < 3.11101. (0.1.18) Property 0.1.1 The collection of the real numbers is complete: Given any two distinct real numbers, there will always be a third real number which will lie in between the two given. Example 0.1.2 : Given the real numbers two. Pamini Thangarajah 1.99999 and 1.999991, we can find the real number 10/21/2020 0.1.2 CC-BY-NC-SA 1.9999905 which certainly lies in between https://math.libretexts.org/@go/page/10821 Property0.1.1 The collection of the real numbers has neither the greatest element nor does it have a least element. Useful hints for calculations without your calculator Often in calculations, multiplication and division will be necessary, and although calculators are now available to help us through these chores, it is useful to develop a facility with these two operations. To this end, we conclude here some very useful facts. 1. An integer is an even number if and only if it is divisible by 2. 2. An integer is divisible by 3 if and only if the sum of its digits is divisible by 3. Example 0.1.3 : 3579 is divisible by 3 since the sum of its digits is 24 which is divisible by 3. 476911 is not divisible by 3 since the sum of its digits is 28 which is not divisible by 3. 3. An integer is divisible by 5 if and only if its last digit (the units digit) is either 0 or 5. Example 0.1.4 : 27795 3714 and 46790 are both divisible by 5. is not divisible by 5 since its last digit (the units digit) is 4. 4. Given any integer, if the difference between the sum of every other digit and the sum of the digits which remain is divisible by 11, then the integer is itself divisible by 11. Example 0.1.5 : 46791272 is divisible by 11, since and (6 + 9 + 2 + 2) = 19, (4 + 7 + 1 + 7) = 19 (4 + 7 + 1 + 7) − (6 + 9 + 2 + 2) = 0 and 46791272 is divisible by 11. 9123917 is divisible by 11, since and 1 + 3 + 1 = 5, 9 + 2 + 9 + 7 = 27 (9 + 2 + 9 + 7) − (1 + 3 + 1) = 22, which is divisible by 11. Intervals on the real line R = (−∞, ∞) in keeping with the fact that the collection of all real numbers has neither a least element nor a greatest element. The symbol ∞ does not represent a real number, but is used to denote the open endedness of the set. 1. For real numbers a and b , with a < b, {x ∈ R : a < x < b} which is denoted by (a, b). This is an open interval since the end-points of the interval are not contained in the interval. 2. For real numbers a and b , with a < b, Pamini Thangarajah {x ∈ R : a ≤ x ≤ b}} which is denoted by [a, b]. 10/21/2020 0.1.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10821 This is a closed interval since the end-points of the interval are contained in the interval. 3. For real numbers a and b , with a < b, {x ∈ R : a < x ≤ b}} which is denoted by (a, b]. and {x ∈ R : a ≤ x < b}} which is denoted by [a, b). These are half open or half closed intervals since only one of the end-points must be included in the interval. 4. For real numbers a and b , with a < b, {x ∈ R : x < b}} {x ∈ R : x > b}} which is denoted by (b, ∞). which is denoted by (−∞, b). This notation must not be confused with the notation adopted for representing a point in the plane. Both are used freely in the literature and usually the reader should find it easy to distinguish between the two. The partial order < on the set of all real numbers Partial ORDER If a, b and c are real numbers, then the following statements are true: 1. If a 0 then ac < bc. 4. If a bc. Example 0.1.6 : 2 <3 and 4 < 6. Example 0.1.7 : 2 <3 and −4 > −6. Pamini Thangarajah 10/21/2020 0.1.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10821 The absolute value Definition For any real number x the absolute value of x is denoted by |x| and defined by |x| = { x if x ≥ 0, −x if x < 0. (0.1.19) Consequently we see that | − 2| = |2| = 2. RULES For real numbers a, b and c, the following hold: 1. |a| ≥ 0, with equality when a = 0. 2. |a | = a = |a| . 3. |ab| = |a| ⋅ |b|. 4. This is a special case of above . If k is positive real number, then |ka| = k ⋅ |a|. 2 a∣ 5. ∣∣ b 2 2 ∣ |a| = provided b ≠ 0. , |b| 6. |a + b| ≤ |a| + |b|. A few examples to emphasize the importance of applying these rules as stated are given below. Example 0.1.8 : |6 − 2| = |6 + (−2)| = |4| = 4. (0.1.20) Yet |6| + | − 2| = 6 + 2 = 8. Clearly then |6 − 2| < |6| + | − 2|. In rule 4, it is necessary that our choice of k be positive number. For example if k = −2, |(−2)3| = | − 2| ⋅ |3| = 6. (0.1.21) |(−2)b| = | − 2| ⋅ |b| = 2|b|. (0.1.22) More generally, for any real number b, Laws of exponents Rules For real numbers a, b, m and n, with a ≠ 0 and b ≠ 0, the following rules hold: 1. a m n a m+n =a m 2. a m−n =a n a 3. a 0 =1 4. (ab) n n n =a b Pamini Thangarajah 10/21/2020 0.1.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10821 ) a = n b b 6. (a m 7. n n a 5. ( ) n mn =a 1 −n =a n a Square roots For any positive real number a, we can find another real number b, with the property that (b)(b) = a. (0.1.23) − In this case b is called the square root of a, and b = √− a means that b is the positive square root of a. RUles For positive real numbers a and b , − − − 1. √− a √a = a − − − − 2. √a√b = √ab 3. − − √a − − a =√ b √b − − 4. √a 2 = |a| −− − − − 2 a = k√a If k is a positive real number, then √k − 5. √− a =a 1/2 − 6. √− a =a 1/n n Caution − −−−−− − 2 2 + b ) ≠ a + b, for positive real numbers. Except in special cases, √(a − −−−−− − In particular, √(3 2 − − 2 + 4 ) = √25 = 5 ≠ 3 + 4. The quadratic formula The quadratic formula Given the quadratic equation ax 2 + bx + c = 0, a, b, and c being real numbers a ≠ 0 , then − −−−−−− − 2 −b ± √ (b − 4ac) x = , (0.1.24) 2a where b 2 − 4ac is called the discriminant. If the discriminant is 0 (that is b 2 Pamini Thangarajah − 4ac = 0 ) then the quadratic equation has exactly one solution (root) and that solution is 10/21/2020 0.1.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10821 −b x = . (0.1.25) 2a If the discriminant is positive (that is b 2 If the discriminant is negative (that is b 2 − 4ac > 0 − 4ac < 0 ) then the quadratic equation has two distinct solutions (roots). ) then the quadratic equation has no real number solutions. Example 0.1.9 : The quadratic 2 = (−20 ) equation 2 4x has − 20x + 25 = 0 exactly one solution since the discriminant − (4)(4)(25) = 400 − 400 = 0. In fact, x = −(−20) 5 = 2(4) . 2 Example 0.1.10 : Given that 3x 2 + 2x − 7 = 0. Here a = 3, b = 2, c = −7. 2 The discriminant = 2 − (4)(3)(−7) = 4 + 84 (0.1.26) = 88. Since the discriminant is positive, the equation has two distinct real number solutions. In fact, − − −2 ± √88 x = 2(3) − − 2(−1 ± √22) = (0.1.27) 2(3) − − −1 ± √22 = . 3 Example 0.1.11 : Given that 3x 2 − 2x + 7 = 0. Here a = 3, b = −2, c = 7. 2 The discriminant = (−2 ) − (4)(3)(7) = 4 − 84 (0.1.28) = −80 < 0. Since the discriminant is negative, the equation has no real number solutions. Pamini Thangarajah 10/21/2020 0.1.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10821 0.2: Analytic geometry The straight line Theorem 0.2.1 Given two points A = (x 1, 1. The slope of AB = y1 ) y2 − y1 and B = (x 2, y2 ), rise = . x2 − x1 run − −−−−−−−−−−−−−−−− − 2 2 + (y2 − y1 ) . 2 − x1 ) 2. The length of the line segment AB = √(x 3. The mid-point C , of the line AB has coordinates ( (x1 + x2 ) (y1 + y2 ) , ). 2 2 It is often useful to be able to describe the path or locus of a point as a relationship between the coordinates of an arbitary point on the path. Some of the standard paths are described below. It is worth noting here that the equation of a straight line is completely determined if we know the slope of the line and one point on the line. Equations of the straight line Equations of the straight line 1. The point slope form The equation of the straight line which has slope m and which passes through the point (x 1, y1 ) is, y − y1 = m(x − x1 ). (0.2.1) 2. The standard form The equation of the straight line which has slope m and makes an intercept of b units on the y−axis is y = mx + b. (0.2.2) Any straight line in the plane has an equation of the form ax + by + c = 0. Where a, b and c are real numbers. This straight line has slope −a b (0.2.3) and y−intercept −c , b provided b ≠ 0. Parallel straight lines and perpendicular straight lines Slopes 1. Two straight lines are parallel if and only if their slopes are equal. If L is a straight line with slope m and L is a straight line with slope m , and if neither line is parallel to either x−axis or the y−axis, then L is perpendicular to L if and only if m m = −1. 1 1 2 1 2 2 1 2 (Alternatively we note that the slope of L is the negative reciprocal of slope of L .) 1 2 2. The perpendicular distance of a point (h, k) from the straight line with equation ax + by + c = 0 is given by |ah + bk + c| d = Pamini Thangarajah − − − − − − √ a2 + b2 . 10/21/2020 0.2.1 CC-BY-NC-SA (0.2.4) https://math.libretexts.org/@go/page/10822 Though it is not essential to memorize this formula, it can sometimes be very useful. The circle Circle 1. The equation of a circle with the centre (h, k) and the radius r is 2 (x − h ) 2 + (y − k) 2 =r . (0.2.5) 2. The general form of the equation of a circle is 2 x this circle has centre (−g, −f ) and radius +y 2 + 2gx + 2f y + c = 0, (0.2.6) − −−−−−−− − 2 2 √g + f − c . The conic sections Conic Sections The general form of the equation of the conic section which has axes parrallel to the x− and y− axes is 2 ax + by 2 + cx + dy + e = 0, (0.2.7) provided that a and b are not both zero. The conic sections may be readily identified using the information given below: Equation ~0.2.7 represents 1. a circle provided that a = b . 2. a parabola with axis parallel to the x−axis if a = 0 , and b ≠ 0. 3. a parabola with axis parallel to the y−axis if b = 0 , and a ≠ 0. 4. an ellipse provided that a ≠ b and both a and b have the same sign. 5. a hyperbola provided that a and b differ in sign. 6. two straight lines exactly when bc − ad − 4abe = 0. 2 Pamini Thangarajah 2 10/21/2020 0.2.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10822 0.3: Trigonometry The right triangle Trigonometric functions Usually in a triangle when there is no chance of ambiguity, the sides opposite to each vertex have length denoted by the corresponding small letter. In the diagram, the triangle ABC has a right angle at B. length of the opposite side sin(A) = a = length of the hypotenuse length of the adjacent side cos(A) = . (0.3.2) b length of the opposite side tan(A) = (0.3.1) c = length of the hypotenuse . b length of the length of the adjacent side sin(A) a = = c . (0.3.3) cos(A) The remanining functions can be considered to be reciprocal functions. 1 csc(A) = , (0.3.4) sin(A) Pamini Thangarajah 10/21/2020 0.3.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10823 1 sec(A) = , (0.3.5) . (0.3.6) cos(A) 1 cot(A) = tan(A) Solving the general triangle Rules Consider the triangle ABC in the diagram: 1. The sine law: a b c = = sin(A) sin(B) (0.3.7) sin(C ) 2. The cosine law: 2 a 2 =b 2 +c − 2bc cos(A) (0.3.8) or equivalently, 2 b 2 +c 2 −a cos(A) = (0.3.9) 2bc 3. The area of the triangle ABC is bc sin(A). 4. If the perimeter of the triangle = 2s = a + b + c, then 1 2 − −−−−−−−−−−−−−−− − 2 sin(A) = √ s(s − a)(s − b)(s − c) (0.3.10) bc − −−−−−−−−−−−−−−− − and the area of the triangle is √s(s − a)(s − b)(s − c) . Unit Circle Definition Pamini Thangarajah 10/21/2020 0.3.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10823 Unit circle is circle with center at the origin (0, 0) and radius 1. Unit circle has the equation x + y = 1. 2 2 The radian measure of an angle is equal to the numerical value of the length of the arc of the unit circle from the point (1, 0) to the point (x, y). Thus, length of the arc AP = α radian units. Hence we obtain the conversion from degrees to radians and vice-versa: ∘ π radians = 180 (0.3.11) Definition of the trigonometric functions There are two ways to define trigonometric functions: Definition Using triangles or circles. 1. The definition using triangles is too restrictive in that it treats only angles between 0 degree and 90 degrees. 2. The definition using circles, on the other hand, leads to a much more general definition, since the angle can now take on any real value whatsoever. Pamini Thangarajah 10/21/2020 0.3.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10823 In the diagram above, A(1, 0) is the fixed point from which measurements begin. Angles and arc lengths are measured from A in a counter-clockwise direction. If P (x, y) is any point on the circumference of the unit circle, then if the arc length of radians, we define AP = α sin(α) = x and cos(α) = y. units, then ∠AOP = α (0.3.12) Further, we define sin(α) tan(α) = (0.3.13) cos(α) cos(α) cot(α) = , (0.3.14) , (0.3.15) . (0.3.16) sin(α) 1 sec(α) = cos(α) 1 csc(α) = sin(α) Some important relationships on unit circle Rules sin(π − α) = sin(α). ⎫ ⎪ cos(π − α) = − cos(α). ⎬ Quadrant II tan(π − α) = − tan(α). (0.3.17) ⎭ ⎪ sin(π + α) = − sin(α). ⎫ ⎪ cos(π + α) = − cos(α). ⎬ Quadrant III tan(π + α) = tan(α). sin(2π − α)) = sin(−α) = − sin(α). cos(2π − α) = cos(−α = cos(α). tan(2π − α) = tan(−α) = − tan(α). ⎫ ⎪ ⎬ Quadrant IV (0.3.19) ⎭ ⎪ sin(π/2 − α) = cos(α). Pamini Thangarajah (0.3.18) ⎭ ⎪ 10/21/2020 0.3.4 CC-BY-NC-SA (0.3.20) https://math.libretexts.org/@go/page/10823 cos(π/2 − α) = sin(α). (0.3.21) Basic identities Basic identities 2 2 sin (A) + cos (A) = 1. (0.3.22) 2 2 2 2 1 + tan (A) = sec (A). (0.3.23) 1 + cot (A) = csc (A). (0.3.24) cos(A + B) = cos(A) cos(B) − sin(A) sin(B). (0.3.25) cos(A − B) = cos(A) cos(B) + sin(A) sin(B). (0.3.26) sin(A + B) = sin(A) cos(B) + cos(A) sin(B). (0.3.27) sin(A − B) = sin(A) cos(B) − cos(A) sin(B). (0.3.28) sin(2A) = 2 sin(A) cos(A). 2 (0.3.29) 2 cos(2A) = cos (A) − sin (A) (0.3.30) 2 = 2 cos (A) − 1 (0.3.31) 2 = 1 − 2 sin (A). (0.3.32) tan(A) + tan(B) tan(A + B) = . (0.3.33) . (0.3.34) 1 − tan(A) tan(B) tan(A) − tan(B) tan(A − B) = 1 + tan(A) tan(B) 2 tan(A) tan(2A) = 2 . (0.3.35) 1 − tan (A) An app for trigonometric functions Pamini Thangarajah 10/21/2020 0.3.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10823 y Values (x,y) = (1.000, 0.000) angle = 0.0° cosθ = x = 1.000 1 x degrees radians cosθ Trig Tour Pamini Thangarajah 10/21/2020 0.3.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10823 0.4: Logarithms and Exponents Definition The function y = a is called an exponential function of base a and with exponent x provided that a > 0, a ≠ 1. x The function y = log a (x) is called logarithimic function with base a, provided that a > 0 and a ≠ 1, and is defined by y y = log (x) if and only if x = a a provided that a > 0 and a ≠ 1. (0.4.1) Rules for logarithms: 1. log (x) 2. log (xy) = loga (x) + loga (y) 3. log ( a a a is defined only when x > 0 . x y 4. n log a ) = loga (x) − loga (y) 5. log (1) = 0 6. log (a) = 1 a a n (x) = loga (x ) Rules for Exponents: The transcendental number e is one of the most common bases employed. This notation was introduced in honor of the Swiss mathematician Euler (1707-1783). The definition of e which follows is certainly beyond the scope of these notes since it uses the concept of the limit. However, it is included here for completeness and as partial preparation for first calculus course. Definition n 1 e = lim (1 + n→∞ ) (0.4.2) n and it has been found that e ∼ 2.71828 … (0.4.3) It has been found that scientific calculations involving exponential and logarithmic functions are simpler, if the base involved is e. Natural Exponents and Natural Logarithms Pamini Thangarajah 10/21/2020 0.4.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10824 0E: Exercises The real number system Exercise 0E. 1 Answer True or False for each of the statements below. If the statement is false, try to provide an example to demonstrate this. 1. The sum of any two integers is an integer. 2. The sum of any natural numbers is a natural number. 3. The difference between any two natural numbers is a natural number. 4. The sum of any two rational number is a rational number. 5. The sum of any two irrational number is an irrational number. 6. If the sum of the digits of an integer is divisible by 3, then the integer is divisible by 3. 7. If an integer has units digit 0 or 5 ( ends in either 0 or 5 ), then the integer is divisible by 5. 8. If n is an integer then a. 2n − 1 and 2n + 1 are both odd integers. b. 2n is an even integer. c. 3n is an odd integer. 9. For any real number x, |x| ≤ 0. 10. For any real number x, |x| ≥ 0. 11. For any real number x, and y , |x| + |y| = |x + y|. 12. For any real number x, and y , |x| + |y| ≥ |x + y|. 13. For any real number x, it follows that √− x is again real number. − − a. √x 2 b. = x. − − √x2 = −x if x < 0. Answer 1. True. 2. True. 3. False. 5 − 7 = −2 is not a natural number. 4. True. (Note: The rational numbers include the integers.) – – 5. False. √2 + −√2 = 0 . 6. True. 7. True. 8. a. True. b. True. c. False. 3 × 2 = 6 is even. 9. False. |3| = 3 > 0 . 10. True. 11. False. |2| + | − 2| = 2 + 2 = 4 but |2 − 2| = 0 . 12. False. Use same example as above. Note: The correct statement is |x + y| ≤ |x| + |y| . − − − − − 13. a. False. √(−2) 2 – = √4 = 2 ≠ −2 . b. True. Pamini Thangarajah 10/21/2020 0E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 Exercise 0E. 2 For each of the following terms, identify (a) its coefficient and (b) its degree in the variables which occur. 1. −2x 5 2. 7x 3 y 3. 5 4. −4x 2 3 y z 4 Answer 1. Coefficient: −2 ; degree 5. 2. Coefficient: 7; degree 4. 3. Coefficient: 5; degree 0. 4. Coefficient: −4 ; degree 9. Exercise 0E. 3 Classify each of the expressions below as polynomial expression, rational expression or neither. If the expression is a polynomial, determine the coefficients of each of its terms. 1. 2x 4 3 2 − 3x 2. 3x 7 + 7x 3 − 13 x − 14x − 19 3/2 −x − 21 1 3. − 5x + 31 2 x 4. −2 8x − 1 2 5. 5x + 3x − 1 7 6. 2x − 7. 1 x −1 3x + 1 − 3 √x 8. 17 9. 0 10. 11. −4 2 2x 2 x −x +1 − 5x + 1 2 1 − 2x − 3x 12. (x − 3)(x + 5)(2x − 1) 1 13. x −3 2− 14. 15. 1 x − − − − − − − √(3x + 1) x2 − x − 2 1 −x 3 + 5x 16. |x| ⎧ ⎪ 2x−1 , x >0 x − 1, x ≤0 x 17. ⎨ ⎩ ⎪ 18. 3x + 2 1 x Pamini Thangarajah 10/21/2020 0E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 2 7x 19. 3 √x , x ≠ 0. 2/3 x 20. x − 2√− x −1 Answer 1. Polynomial; degree 4; coefficients 2, −3 , 7, −14, 19. 2. Neither. 3. Neither. 4. Neither. 5. Polynomial; degree 2; coefficient 5 7 , 3 7 , −1 7 . 6. Neither. 7. Neither. 8. Polynomial; degree 0; coefficient 17. 9. Polynomial; degree 0; coefficient 0. 10. Rational Expression. 11. Rational Expression. 12. Neither. 13. Neither. 14. Neither. 15. Rational Expression. 16. Neither. 17. Neither. 18. Neither. 19. Neither. 20. Neither. Exercise 0E. 4 Identify each of the following numbers as natural numbers, integers, whole numbers, rational numbers, or irrational numbers. −3 −4 , −15, 89, 4 −23 – − − , −√9, √13, 3 – , −√2 (0E.1) 10 Answer −3 4 : rational number. : integer, rational number. −15 : natural number, whole number, integer, rational number. 89 −4 −23 : rational number. – −√9 = −3 : integer, rational number. − − √13 : irrational number. 3 10 : rational number. – −√2 : irrational number. Exercise 0E. 5 Pamini Thangarajah 10/21/2020 0E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 A collection of real numbers is represented geometrically in each diagram below. In each case, give the interval representation. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. Pamini Thangarajah 10/21/2020 0E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 15. Answer 1. [4, ∞) 2. (−∞, −7) 3. (−∞, 1] 4. (−2, 7] 5. (−∞, 8) ∪ [15, ∞) 6. (−∞, 7) ∪ (7, ∞) 7. (−1, 2) 8. [−1, 2] 9. [−10, −3) ∪ [1, 5] 10. [1, 2) ∪ (2, 3] 11. (−∞, −1) ∪ (−1, 3) ∪ [5, ∞) 12. 1, 3, 5 13. (−∞, −1) ∪ (−1, 2) ∪ (2, ∞) 14. (−2, 1] ∪ (10, ∞) 15. [1, 5) ∪ (5, ∞) Exercise 0E. 6 Use long division (not calculators) to obtain a decimal representation of the following fractions. 1. 2. 3. 4. 5. 6. 7. 3 8 2 7 2 3 1 7 2 9 3 7 3 4 8. 4 5 9. 4 7 10. 11. 12. 5 7 6 7 5 9 Pamini Thangarajah 10/21/2020 0E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 8 13. 9 7 14. 9 Answer 1. 2. 3 = 0.375 8 2 ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ = 0. 285714 7 3. 4. 5. 6. 7. 8. 9. 2 ¯ ¯ ¯ = 0. 6 3 1 ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ = 0. 142857 7 2 ¯ ¯ ¯ = 0. 2 9 3 ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ = 0. 428571 7 3 = 0.75 4 4 = 0.2 5 4 ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ = 0. 571428 7 5 10. ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ = 0. 714285 7 6 11. ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ = 0. 857142 7 5 12. ¯ ¯ ¯ = 0. 5 9 8 13. ¯ ¯ ¯ = 0. 8 9 7 14. ¯ ¯ ¯ = 0. 7 9 Exercise 0E. 7 Express each number given below as a product of its prime factors (in exponential form): 1. 81 2. 24 3. 1080 Answer 1. 3 4 2. 2 ⋅3 3. 2 ⋅3 2 3 3 ⋅5 Exercise 0E. 8 Evaluate each of the following using exponential laws: 1. 2 6 2. (−3) 5 3. 1 −3 7 Pamini Thangarajah 10/21/2020 0E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 4. 9 −3 5 2 5. −2 2 −2 6. 3 −1 3 4 −2 7. ( ) −3 3 2x 8. ( ) 5y 9. (2 3 2 ) 10. (5 −2 −1 ) 10 5 11. 7 5 12. ( 13. ( −2 3 ) 4 0 51 ) 67 Answer 1. 2 6 = 64 2. (−3) 5 = −243 1 3. 3 =7 −3 = 343 7 1 4. 9 −3 = 1 = 3 729 9 5 2 5. 5−(−2) =2 −2 7 =2 = 128 2 −2 6. 3 −2−(−1) −1 =3 =3 1 = −1 3 3 7. ( 4 −2 4 (−2) ) 8. ( = 4 3 3 8x = = (5y)3 5y 2 3⋅2 ) 10. (5 3 (2x) ) 3 81 3 2x 9. (2 16 = 3 =2 −2 125y 3 6 =2 −1 = 64 (−2)⋅(−1) ) =5 2 =5 = 25 10 11. 5 10−7 7 3 =5 =5 = 125 5 12. ( 13. ( −2 3 ) 2 4 = 4 51 2 3 16 = 9 0 ) =1 67 Exercise 0E. 9 Express each of the following in radical form: 1. x 1/3 2. 1 1/2 x Pamini Thangarajah 10/21/2020 0E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 3. x −2/5 1 4. −3/4 x 5. x −1/9 Answer 1. x 1/3 3 − = √x 1 2. − √x 1 = − = √x 1/2 x 3. x −2/5 1 = 2/5 x 4. − 5 − √x3 1 = 1 x 3/4 =x − 5 − √x2 = x − − 4 3 = √x x−3/4 5. x −1/9 1 − − 9 √x8 1 = = 1/9 x = 9 − √x x Exercise 0E. 10 Write the following radical expressions in exponent form: 1. √− x 4 − − 2. √x 3 3. 4. 5. 6. 7 1 − 4 − √x3 − √x 3 − √x 3 − 5 − √x √x , x >0 − √x 1 − − − √x−2 3 Answer 1. √− x =x 1/4 4 − − 2. √x 3 7 7/3 =x 1 3. −3/4 − 4 − √x3 =x − √x 4. 3 − √x 1/2−1/3 =x − 5− 3 √x √x 5. − √x 1/6 =x (1/3+1/5)−1/2 =x 1/30 =x 1 2/3 − − =x 3 − √x−2 6. Exercise 0E. 11 Determine a real number value of each of the following if one exists. If one does not exist, explain why. − − − − 1. √−64 3 − − − 2. √2 3 3. 15 1 − − √26 Pamini Thangarajah 10/21/2020 0E.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 −−− − 4. √−400 − − − − − 5. √(−2) 3 6. 3 − − − − √−16 4 Answer − − − − 1. √−64 = −4 3 − − − 2. √2 3 15 15/3 =2 1 3. 5 =2 1 − − √26 = = 32 1 = 6/2 3 1 = 8 2 2 4. −−− − √−400 5. − − − − − 3 √(−2) = −2 is undefined because the square root of a negative number is undefined. 3 − − − − 6. √−16 is undefined because the fourth root of a negative number is undefined. 4 Exercise 0E. 12 Simplify each of the following: − − − − − − − − − − − − 1. −3√98 + 5√12 − 3√24 + √27 − 2√54 + 4√50 – – – – 2. (2√5 − 3√2)(2√5 + 3√2) – – 3. (2 − √3)(7 + 2√3) 4. – – 2 – – – – (3 √5 − 2 √7) − (2 √5 + √7)(√5 − √7) 5 5. √3 −10 3 √3 6. − − − √128 – √8 − − – 7. √72 √8 Answer 1. − − − − − − − − − − − − −3 √98 + 5 √12 − 3 √24 + √27 − 2 √54 + 4 √50 – – – = −21 √2 + 10 √3 − 6 √6 – – – +3 √3 − 6 √6 + 20 √2 – – – = −√2 + 13 √3 − 12 √6 – – – – – 2. (2√5 − 3√2)(2√5 + 3√2) = (2√5) – – – 2 – 2 − (3 √2) = 4 ⋅ 5 − 9 ⋅ 2 = 2 – – 3. (2 − √3)(7 + 2√3) = 14 + 4√3 − 7√3 − 2(3) = 8 − 3√3 4. – – 2 – – – – (3 √5 − 2 √7) − (2 √5 + √7)(√5 − √7) = – – 9 ⋅ 5 − 2(3 √5)(2 √7) + 4 ⋅ 7 − − − − −2 ⋅ 5 + 2 √35 − √35 + 7 − − = 70 − 11 √35 5 5. – √3 – 5 ⋅ 3 √3 = −10 – √3(−10) 3 =− 2 – 3 √3 6. − − − √128 – √8 − − – 8 √2 = – 2 √2 – =4 – – 7. √72 ⋅ √8 = 3√8 ⋅ √8 = 3 ⋅ 8 = 24 Exercise 0E. 13 Pamini Thangarajah 10/21/2020 0E.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 Use order of operations to evaluate each of the following: 1. −31 + 35/7 − (2)(24) + 70 2 2. 2 3 − (−4 ) 7 + 2 24 1 −5 Answer 1. −4 7 2. 12 Basic algebra Exercise 0E. 1 Simplify each of the following expressions: 1. 5x 3 2. −3 y 2 3 − 21xy + 17 x y 5 xz + 2 3. (3x 3 4. 2x 2 2 y y 2 2 xz 3 7 + 2 xz − 2 3 − 7x y + 5y −2 ((y −2 + 24 x 2 − 19xy + 6 19 xz 3 2 + 4xz − 19 3 − 2) + (−5 x y 2 y − 11x) − (2x y 2 + 4 x y + 11) −2 −2 − 4x y − 11x + 1)) 5. 3xy(5x − 7y) 6. 3x 2 4 −1 y (2 x −2 + 9y 7. (3x − 2y)(4x 2 + 4xy) 2 −y ) 8. (x + y − 2z)(x − y + 2z) 9. ( 3 1 5 − x )( y 2 − x 2 ) y 3 − 10. x y −12 18 + x y Answer 1. 22x 3 y 2 2. 6xz − 3. −2x 3 4. 2x 2 2 6. 6x y 4 7. 12x 3 2 xz 2 − 19 3 y 2 2 − 3x y + 5y + 56 y 5. 15x 8. x − 40xy + 6 14 3 3 − 4x y − 21x y 2 + 27 x y − 3x y −y 2 2 3 +9 2 − 2x y 2 2 2 3 + 12 x y 2 5 − 8x y + 2y + 4yz − 4 z 3 2 9. 3 ( 1 − x 5 )( y 2 − x 3y − x ) = ( y 5y + 2x )( xy ) xy (5y + 2x)(3 − x) = 2 (xy) 10. Pamini Thangarajah 10/21/2020 0E.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 2 3 2y − 3x − x y xy = 12 − 18 −12y + 18x + x y xy (2y − 3x)(xy) = (−12y + 18x)(xy) 2y − 3x = −6(2y − 3x) 1 = − , 2y ≠ 3x, xy ≠ 0 6 Factoring polynomials with some applications Exercise 0E. 1 Factor each of the expressions given. It may be necessary to apply the Factorization theorem. 1. 15x 3 y 2 − 45x y 2. 12xz − 4x 2 3. 2x − 50 4. 7y − 56 2 3 5. (3x − 2) 2 z 2 − x − 56 7. x − 8x 4 2 8. x 4 − 10 x 9. x 10. z −9 2 6 +9 6 + 8y 2 + 36x − (4y + 3 ) 6. x 2 2 4 + 64 11. 6x 2 12. x 3 + 19x + 15 2 +x − 4x − 4 13. yz − 2ty + z 2 − 2tz 14. 2ac − 4ad + 3bc − 6bd 15. 4x 3 + 500 y 16. (x − 2y ) 2 3 − 2(x − 2y) − 8 17. 48 − 6t 3 18. x − 256 19. y − 4xy − 45 x 8 2 2 20. 2x 2 − 11x − 21 21. 24x − 98x − 45 22. 24x − 13x − 45 2 2 23. x 6 24. x 5 3 + 19 x 3 2 − 9x + 8x 25. 3(x − 2) 2 26. x +x 27. x − 12 x 4 4 2 − 216 − 72 5 (x + 4 ) 4 3 + 5(x + 4 ) (x − 2 ) +1 2 + 16 Answer 1. 15x 3 y 2 − 45x y 2. 12xz − 4x 2 Pamini Thangarajah z 2 4 2 2 = 15x y (x 2 2 − y ) = 15x y (x − y)(x + y) + 36x = 4x(3z − x z 2 + 9) 10/21/2020 0E.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 3. 2x − 50 = 2(x 4. 7y − 56 = 7(y 2 2 3 3 − 25) = 2(x − 5)(x + 5) − 8) = 7(y − 2)(y 2 + 2y + 4) 5. 2 (3x − 2 ) 6. x − x − 56 = (x − 8)(x + 7) 7. x − 8x 2 4 8. x 4 9. x 6 10. z 2 2 + 8y 2 11. 6x ((3x − 2) − (4y + 3))((3x − 2) + (4y + 3)) = (3x − 4y − 5)(3x + 4y + 1) 2 − 9)(x + 1) = (x − 3)(x + 3)(x 2 2 2 + 9 = (x 2 3 = (x ) + 64 2 = − 9 = (x − 10 x 6 2 2 − (4y + 3 ) − 9)(x 3 + (2y ) + 1) − 1) = (x − 3)(x + 3)(x − 1)(x + 1) 2 = (x 2 4 2 + 2 y )(x − 2x y 2 4 +y ) does not factor. + 19x + 15 = (3x + 5)(2x + 3) 12. 3 2 x +x 2 − 4x − 4 = x (x + 1) − 4(x + 1) 2 = (x + 1)(x = 13. yz − 2ty + z 2 − 4) (x + 1)(x − 2)(x + 2) − 2tz = (y + z)(z − 2t) 14. 2ac − 4ad + 3bc − 6bd = (c − 2d)(2a + 3b) 15. 4x 3 + 500 y 16. (x − 2y ) 2 17. 48 − 6t 3 3 2 = 4(x + 5y)(x 2 − 5xy + 25 y ) − 2(x − 2y) − 8 = ((x − 2y) − 4)((x − 2y) + 2) 3 2 = 6(8 − t ) = 6(2 − t)(4 + 2t + t ) 18. 8 x 4 − 256 = (x = = 19. y 2 2 − 4xy − 45 x 20. 2x 2 2 (x 4 − 16)(x 2 − 4)(x + 16) 4 + 4)(x 2 + 16) 4 (x − 2)(x + 2)(x + 4)(x + 16) = (y − 9x)(y + 5x) − 11x − 21 = (2x + 3)(x − 7) 21. 24x − 98x − 45 = (2x − 9)(12x + 5) 22. 24x − 13x − 45 = (3x − 5)(8x + 9) 2 2 23. 6 x 3 + 19 x 3 − 216 = (x 3 − 8)(x 2 = (x − 2)(x + 27) 2 + 2x + 4)(x + 3)(x − 3x + 9) 24. 5 x 3 − 9x 2 + 8x − 72 = 3 2 x (x 3 2 = (x = 2 − 9) + 8(x + 8)(x 2 (x − 2)(x − 9) − 9) + 2x + 4)(x − 3)(x + 3) 25. 2 5 3(x − 2 ) (x + 4 ) 4 3 + 5(x + 4 ) (x − 2 ) = = 2 4 (x − 2 ) (x + 4 ) (3(x + 4) + 5(x − 2)) 2 4 2(x − 2 ) (x + 4 ) (4x + 1) 26. 4 x 2 +x +1 = 4 2 x + 2x 2 = (x = Pamini Thangarajah 2 (x 2 + 1) 2 +1 −x 2 −x 2 + x + 1)(x − x + 1) 10/21/2020 0E.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 27. 4 x 2 − 12 x + 16 = 4 2 x − 8x 2 = (x = 2 (x 2 − 4) 2 + 16 − 4 x 2 − 4x 2 − 2x − 4)(x + 2x − 4) Exercise 0E. 2 Evaluate each of the following: 1. 2 4 − 14 4. 5. 10 −2 2. 6 ⋅ ( 3. 7 + 5 ) 3 17 1 5 − + 36 24 2 18 5 ÷ 19 38 2 1 −5 + 9 2 17 6. ( 26 ) ( 52 ) 34 −24 7. 7 −8 49 Answer −1 1. 70 2. −4 17 3. 24 4 4. 5 −77 5. 18 1 6. 4 7. 21 Exercise 0E. 3 Simplify each of the expressions below: 2 x 1. −4 x 3 x 2 x 2 x − 6x + 9 2x 2x 2 − 27 ) ( 2 4. x −4 − 2x − 8 2. ( 3. 1 − 2 2 (x 2 −8 x ) 3 (x − 3)2 2 − 2x − 35 − 2x ÷ 2x − 6 + 3x + 1 − 7x + 6 + 3x + 9)(x − 6) − 13x − 7 2 2x − 4x − 6 x − x2 − 9 Pamini Thangarajah 10/21/2020 0E.13 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 −1 3 − 5. x x −1 2x − 5 2 x + 3x − 10 1 3 x + 6. x −1 2 − x +3 (x + 2) 4 2 + x −5 x +2 2(x + h) − 1 2x − 1 − 7. 3(x + h) + 1 3x + 1 h Answer 1. 2 x 2 x −4 (x − 2)(x + 2) 1 − − 2x − 8 1 = − x −4 x −4 (x − 4)(x + 2) x −2 +1 = x −4 x −1 = , x ≠ −2 x −4 2. 3 x 2 x 2 − 27 x ⋅ 2 − 6x + 9 (x 2 (x − 3)(x − 7x + 6 = + 3x + 9)(x − 6)(x − 1) 2 + 3x + 9)(x − 6) 2 (x − 3 ) (x + 3x + 9)(x − 6) x −1 = ,x ≠6 x −3 3. 2 2x 2 2x 2 −8 x 2 − x − 35 − 2x ÷ 2x − 6 + 3x + 1 2x 2 − 13x − 7 2 2x −8 = − 4x − 6 (2x + 1)(x + 1) (x − 7)(x + 5) − 2(x − 3)(x + 1) × 2(x − 3) 2 2x (2x + 1)(x − 7) (x + 5)(x + 1) −8 = − 2x + 1 (2x + 1)(x + 1) 2 (2 x 2 − 8) − (x + 5)(x + 1 ) = (2x + 1)(x + 1) 2 2x 3 − 8 − (x 2 + 7x + 11x − 5) = (2x + 1)(x + 1) 3 −x 2 − 5x − 11x − 3 = (2x + 1)(x + 1) This cannot be reduced any further as 2x + 1 and x + 1 are not factors of the numerator. We can use the remainder theorem to check: Let p(x) = −x − 5x − 11x − 3 . p(−1/2) = 19/8 and p(−1) = 4 . Since neither of them are zero, then x = −1/2, −1 are not zeros of p(x) and therefore 2x + 1 and x + 1 are not factors of p(x) . 3 4. 3 2 (x − 3)2 = x2 − 9 2 3(x + 3) − x(x − 3) x − 9 + 6x − x = (x − 3 )2 (x + 3) (x − 3 )2 (x + 3) 5. Pamini Thangarajah 10/21/2020 0E.14 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 1 (x − 1) − (3x) 3 − x x(x − 1) x −1 = 2x − 5 2 x 2x − 5 + 3x − 10 (x + 5)(x − 2) −(2x + 1)(x + 5)(x − 2) = x(x − 1)(2x − 5) 6. 1 3 + x −1 2 2 2 (x + 2 ) (x + 3) + 3(x − 1)(x + 3) − x(x − 1)(x + 2 ) x − (x + 2) 2 x +3 (x − 1)(x + 2 ) (x + 3) = 4 2 4(x + 2) + 2(x − 5) + x −5 x +2 (x − 5)(x + 2) 4 3 −x 2 − 2x + 10 x = (x + 2)(x − 5) + 26x + 3 ⋅ 2 6x − 2 (x − 1)(x + 2 ) (x + 3) 4 (−x 3 − 2x 2 + 10 x + 26x + 3)(x − 5) = , x ≠ −2 (x − 1)(x + 2)(x + 3)(6x − 2) 7. 2(x + h) − 1 2x − 1 − 3x + 1 3(x + h) + 1 (2x + 2h − 1)(3x + 1) − (2x − 1)(3x + 3h + 1) 1 = h ( ) h (3x + 3h + 1)(3x + 1) 2 (6 x 2 + 6xh − 3x + 2x + 2h − 1) − (6 x + 6xh + 2x − 3x − 3h − 1) = h(3x + 3h + 1)(3x + 1) 5h = h(3x + 3h + 1)(3x + 1) 5 = ,h ≠0 (3x + 3h + 1)(3x + 1) Exercise 0E. 4 Simplify each of the expressions below: 2 1. x −4 x − 2x − 8 3 x 2. ( 2 x x −4 2 − 27 x ) ( − 6x + 9 2 2x 3. 1 − 2 (x 2 −8 x 2x ) 2 − 2x − 35 2x ÷ + 3x + 1 2x − 6 3 − 7x + 6 + 3x + 9)(x − 6) − 2 4. 2 − 13x − 7 2 2x − 4x − 6 x 2 − (x − 3) 2 x −1 −9 3 − 5. x x −1 2x − 5 2 x + 3x − 10 1 3 + 6. x −1 x (x + 2)2 4 − x +3 2 + x −5 x +2 2(x + h) − 1 2x − 1 − 7. 3(x + h) + 1 3x + 1 h Answer Under construction. Pamini Thangarajah 10/21/2020 0E.15 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 Exercise 0E. 5 Express your answer in its simplest form. 1. − − − − − − − √(2x + 3) − 1 2 x −x −2 1 2. − − − − − − −1 √(x + 1) 2 x 3. + 2x − − − − − − − − − − − − 2 √(2 x + 6x − 7) − 1 2 3x 4. + 2x − 5 − − − − − − − − − − − − − − − √(3x + 13) − √(17 − x) 2 1 −x 1 5. 1 − − – √x √2 2 x + 2x − 8 1 1 − 6. 7. x +1 − − − − − − − − 2 √(4 x + 3) − 2 − −−−−−−−−− − − − − − − − − √(2x + 2h + 3) − √(2x + 3) h 1 8. x 1 − − − − − − − 2 √(4 − h) h Answer 1. −−−− − √ 2x + 3 − 1 2 x −−−− − √ 2x + 3 − 1 = −x −2 −−−− − √ 2x + 3 + 1 ⋅ (x − 2)(x + 1) −−−− − √ 2x + 3 + 1 −−−− − 2 2 (√ 2x + 3 ) − (1 ) = −−−− − (x − 2)(x + 1)(√ 2x + 3 + 1) 2x + 2 = −−−− − (x − 2)(x + 1)(√ 2x + 3 + 1) 2 = −−−− − (x − 2)(√ 2x + 3 + 1) 2. 1 − − − − − √x + 1 −1 − − − − − √ x + 1 − (x + 1) = x2 + 2x − − − − − √ x + 1 x(x + 2) − − − − − √ x + 1 + 1) ⋅ − − − − − √ x + 1 + 1) − − − − − 2 2 (√ x + 1 ) − (1 ) = − − − − − − − − − − x(x + 2)√ x + 1 (√ x + 1 + 1)) x = − − − − − − − − − − x(x + 2)√ x + 1 (√ x + 1 + 1) −1 = − − − − − − − − − − (x + 2)√ x + 1 (√ x + 1 + 1) ,x ≠0 3. Pamini Thangarajah 10/21/2020 0E.16 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 − −−−−−−−− − √ 2 x2 + 6x − 7 − 1 2 3x − −−−−−−−− − √ 2 x2 + 6x − 7 − 1 = 2 + 2x − 5 3x ⋅ − −−−−−−−− − √ 2 x2 + 6x − 7 + 1 + 2x − 5 2 2x = − −−−−−−−− − √ 2 x2 + 6x − 7 + 1 + 6x − 7 − 1 − −−−−−−−− − 2 (3x + 5)(x − 1)(√ 2 x + 6x − 7 + 1) 2(x + 4)(x − 1) = − −−−−−−−− − 2 (3x + 5)(x − 1)(√ 2 x + 6x − 7 + 1) 2(x + 4 = ,x ≠1 − −−−−−−−− − 2 (3x + 5)(√ 2 x + 6x − 7 + 1) 4. − −−−− − −−−− − √ 3x + 13 − √ 17 − x − −−−− − −−−− − √ 3x + 13 − √ 17 − x = 2 − −−−− − −−−− − √ 3x + 13 + √ 17 − x ⋅ 2 1 −x 1 −x − −−−− − −−−− − √ 3x + 13 + √ 17 − x (3x + 13) − (17 − x) = − −−−− − −−−− − (1 − x)(1 + x)(√ 3x + 13 + √ 17 − x ) 4(x − 1) = − −−−− − −−−− − (1 − x)(1 + x)(√ 3x + 13 + √ 17 − x ) = − 4 − −−−− − −−−− − ,x ≠1 (1 + x)(√ 3x + 13 + √ 17 − x ) 5. 1 − √x 2 x − √x 1 − – √2 – √2 − x = + 2x − 8 2 (x + 4)(x − 2) – − 2 √x − √2x = 2x(x + 4)(x − 2) – − 2 √x − √2x = – − 2 √x + √2x ⋅ – − 2 √x + √2x 2x(x + 4)(x − 2) 2 4x − 2x = – − 2x(x + 4)(x − 2)(2 √x + √2x) 1 = − – − (x + 4)(2 √x + √2x) , x ≠ 0, 2 6. 1 1 + x +1 2x + 1 x − −−−− − √ 4 x2 + 3 − 2 = − −−−− − 2 x(x + 1)(√ 4 x + 3 − 2) − −−−− − √ 4 x2 + 3 + 2 2x + 1 = − −−−− − 2 x(x + 1)(√ 4 x + 3 − 2) ⋅ − −−−− − √ 4 x2 + 3 + 2 − −−−− − (2x + 1)(√ 4 x2 + 3 + 2) = 2 x(x + 1)(4 x + 3 − 4) − −−−− − 2 (2x + 1)(√ 4 x + 3 + 2) = x(x + 1)(2x − 1)(2x + 1) − −−−− − √ 4 x2 + 3 + 2 = , x ≠ −1/2 x(x + 1)(2x − 1) 7. Pamini Thangarajah 10/21/2020 0E.17 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 − − − − − − − − − − −−−− − √ 2x + 2h + 3 − √ 2x + 3 − − − − − − − − − − −−−− − √ 2x + 2h + 3 − √ 2x + 3 = − − − − − − − − − − −−−− − √ 2x + 2h + 3 + √ 2x + 3 ⋅ h h − − − − − − − − − − −−−− − √ 2x + 2h + 3 + √ 2x + 3 (2x + 2h + 3) − (2x + 3) = − − − − − − − − − − −−−− − h(√ 2x + 2h + 3 + √ 2x + 3 ) 2h = = − − − − − − − − − − −−−− − h(√ 2x + 2h + 3 + √ 2x + 3 ) 2 − − − − − − − − − − −−−− − ,h ≠0 √ 2x + 2h + 3 + √ 2x + 3 8. 1 1 − − − − − √4 − h − − − − − − 2 − √4 − h 2 = h − − − − − 2h √ 4 − h − − − − − 2 − √4 − h = − − − − − 2h √ 4 − h − − − − − 2 + √4 − h ⋅ − − − − − 2 + √4 − h 4 − (4 − h) = − − − − − − − − − − 2h √ 4 − h (2 + √ 4 − h ) 1 = − − − − − − − − − − 2 √ 4 − h (2 + √ 4 − h ) ,h ≠0 Solving equations Exercise 0E. 1 Solve the following equations: 1. 3x − 17 = 0 2. −14z + 21 = 0 3. −1 7 x = 2 6 4. 3x − 18 =0 7 5. −5x = 0 6. −2 7 x = 3 5 Answer 1. x = 2. z = 3. x = 4. x = 17 3 3 2 −7 3 6 7 5. x = 0 6. x = −21 10 Exercise 0E. 2 Solve each of the folwing quadratic equations by (a)factoring, (b)using the method of completing the square, (c) using the quadratic formula. 1. x − x − 20 = 0 2. x − 12x + 36 = 0 2 2 Pamini Thangarajah 10/21/2020 0E.18 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 3. 4x 2 + 3x = 0 4. 10x 2 +x −3 = 0 5. 4x − 4x + 1 = 0 6. 2x − 5x − 3 = 0 2 2 Answer 1. x = 5, −4 2. x = 6 −3 3. x = 0, 4. x = 5. x = 4 1 −3 , 2 5 1 2 −1 6. x = 3, 2 Exercise 0E. 3 Use any method to solve the equations given below. 1. 3t 2 2. x 2 − 14t = 0 +1 = 0 3. x 2 +x +1 = 0 4. 2x 2 +x +5 = 0 5. 36x 2 6. 6x 2 − 25 = 0 − 31x + 18 = 0 Answer 1. t = 0, 14/3 2. There are no solutions: x 2 = −1 only has complex solutions. 3. There are no solutions: Using the quadratic formula, we get x = 4. There are no solutions: Using the quadratic formula, we get x = 5. x = ± 6. x = − − − −1 ± √−3 2 , which is a complex number. − − − − −1 ± √−39 4 , which is a complex number. 5 6 9 2 , 2 3 Exercise 0E. 4 Solve each of the following equations given for x. − − − − − − − 1. √(3x + 1) 2. =2 − − − − − − − − − − − − − − √(2x − 3) − √(4x + 1) = −2 − − − − − − 5x − 2 3. √ 4. 2 = 7x + 4 3 − − − − − − − √(3x − 5) − 4x + 9 = 0 − − − − − − − 5. √(2x − 7) 6. x 3 = x −5 −8 = 0 7. 8x 3 + 27 = 0 Pamini Thangarajah 10/21/2020 0E.19 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 8. 6x 4 2 − 5x 9. x 2/3 −6 = 0 1/3 − 5x −6 = 0 10. x − √− x − 12 = 0 11. 4x 2/5 12. x 3/2 13. t 4 14. y 1/5 − 4x 3/4 −x 2 − 9t 4 − 2y +1 = 0 −6 = 0 +8 = 0 2 −8 = 0 15. x − 10 x 16. x − 26 x 3 − 27 = 0 4 2 6 +9 = 0 2 17. x − √− x −2 = 0 18. x − 17 x 19. t −t 3 2 2/3 1/3 + 30x = 0 −6 = 0 20. (x − 1)(x + 5)(2x + 3) = 0 21. 22. x + 10 2 2x 2 2 x 2 x 24. 25. =0 3x − 1 + 7x − 15 x 23. 2x − x + 13 =0 +1 +1 1 = 4 + 31 2 9 − x −2 =0 x +2 − − − − − − − − √(3x + 15) − x = 5 − − − − − − − − 26. √(2x + 14) − 5x = −1 − − − − − − 27. √(x + 8) + 3 = 0 − − − − − − 28. √(x + 4) − − − − − − − = √(2x + 1) + 1 29. |4x − 3| = 5 30. |x − 1| = |7 − x| 31. |2x + 3| − 3x + 5 = 0 Answer 1. 2. −−−− − −−−− − √ 2x − 3 − √ 4x + 1 = −2 −−−− − −−−− − √ 2x − 3 = √ 4x + 1 − 2 2x − 3 = −−−− − (4x + 1) − 4 √ 4x + 1 + 4 −−−− − 2x − 3 = 4x + 5 − 4 √ 4x + 1 −−−− − 4 √ 4x + 1 = 2x + 8 2 16(4x + 1) = 4 x 2 x + 32x + 64 − 8x + 12 = 0 x = 6, 2 Check answers: – − − x = 6 : √9 − √25 = −2 √ – – x = 2 : √1 − √9 = −2 √ 3. Pamini Thangarajah 10/21/2020 0E.20 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 − − − − − − 5x − 2 √ 2 = 7x + 4 3 5x − 2 4 = 7x + 4 9 9(5x − 2) = 4(7x + 4) 17x = 34 x = 2 Check answer: −− − 8 √ − − 4 =√ 2 = 18 √ 9 3 4. −−−− − √ 3x − 5 − 4x + 9 = 0 −−−− − √ 3x − 5 = 4x − 9 2 3x − 5 = 16 x 2 16 x − 72x + 81 − 75x + 86 = 0 (16x − 43)(x − 2) = 0 x = 2, 43/16 Check answer: – x = 2 : √1 − 8 + 9 = 2 → Not a solution − − − − − x = 43/16 : √49/16 − 43/4 + 9 = −36/4 + 9 = 0 √ 5. −−−− − √ 2x − 7 = x − 5 2 2x − 7 = x 2 x − 10x + 25 − 12x + 32 = (x − 8)(x − 4) = 0 x = 8, 4 Check answers: – x = 8 : LHS = √9 = 3, RHS = 8 − 5 = 3 √ – x = 4 : LHS = √1 = 1, RHS = 4 − 5 = −1 → 6. x 3 2 − 8 = (x − 2)(x x =2 7. 8x 3 Not a solution. + 2x + 4) = 0 . 2 + 27 = (2x + 3)(2 x − 6x + 9) = 0 x = −3/2 8. 6x 4 2 − 5x 2 2 − 6 = (3 x + 2)(2 x 2 − 3) = (3 x – – – – + 2)(√2x − √3)(√2x + √3) = 0 − − 3 x = ±√ 2 9. Let u = x 1/3 2 u . − 5u − 6 = (u − 6)(u + 1) = 0 Therefore, x 1/3 10. Let = u = −1, 6 → x = −1, 216 − u = √x . . . Therefore, √− x = u = 4, −3 → x = 2 . 2 u − u − 12 = (u − 4)(u + 3) = 0 11. Let u = x 1/5 2 4u 2 Therefore, x 1/5 12. Let u = x u =0 5 = u = 1/2 → x = (1/2 ) 3/4 2 . − 4u + 1 = (2u − 1 ) . Then u 2 3/4 = (x − u − 6 = (u − 3)(u + 2) = 0 2 ) = 1/32 (3/4)∗2 =x 3/4 Pamini Thangarajah 3/2 =x . . Therefore, x = u = 3, −2 → x = 3 . Note: x = −2 is not a solution since x 3/4 . 4/3 3/4 ≥0 . 10/21/2020 0E.21 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 13. t 4 2 − 9t 2 2 + 8 = (t − 8)(t – – − 1) = (t − 2 √2)(t + 2 √2)(t − 1)(t + 1) = 0 – t = ±2 √2, ±1 14. y 4 − 2y 2 − 8 = (y 2 − 4)(y 2 + 2) = (y − 2)(y + 2)(y 2 + 2) = 0 y = ±2 15. x 4 2 − 10 x 2 + 9 = (x 2 − 9)(x − 1) = (x − 3)(x + 3)(x − 1)(x + 1) = 0 x = ±1, ±3 16. x 6 3 − 26 x 3 − 27(x 3 − 27)(x 2 + 1) = (x − 3)(x 2 + 3x + 9)(x + 1)(x + x + 1) = 0 x = 3, −1 17. Let u = √− x. 2 u − − u − 2 = (u − 2)(u + 1) = 0 → √x = u = 2, −1 → x = 4 18. x 3 2 − 17 x 19. Let u = t 1/3 2 u 2 + 30x = x(x − 17x + 30) = x(x − 15)(x − 2) = 0 → x = 0, 15, 2 . 1/3 − u − 6 = (u − 3)(u + 2) = 0 → t = u = 3, −2 → t = 27, −8 20. x = 1, −5, −3/2 21. Note: x ≠ −13, 1/3 x + 10 2x − x + 13 = 0 3x − 1 x + 10 2x = x + 13 3x − 1 (x + 10)(3x − 1) = 2x(x + 13) 2 2 3x + 29x − 10 = 2 x 2 x + 26x + 3x − 10 = 0 (x + 5)(x − 2) = 0 x = −5, 2 22. 2 2x + 7x − 15 2 x 2 2x = 0 +1 + 7x − 15 = 0 (2x − 3)(x + 5) = 0 x = 3/2, −5 23. 2 x +1 1 = 2 x 2 3x 4 + 31 2 4(x 2 + 1) = x + 31 − 27 = 0 3(x − 3)(x + 3) = 0 x = ±3 24. Note: x ≠ 2, −2 2 9 − x −2 = 0 x +2 2 9 = x −2 x +2 2(x + 2) = 9(x − 2) x = 22/7 25. Pamini Thangarajah 10/21/2020 0E.22 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 − −−−− − √ 3x + 15 − x = 5 − −−−− − √ 3x + 15 = x + 5 2 3x + 15 = x 2 x + 10x + 25 + 7x + 10 = 0 (x + 5)(x + 2) = 0 x = −5, −2 Check answers: – x = −5 : √0 + 5 = 5 √ – x = −2 : √9 + 2 = 3 + 2 = 5 √ 26. − −−−− − √ 2x + 14 − 5x = −1 − −−−− − √ 2x + 14 = 5x − 1 2 2x + 14 = 25 x 2 25 x − 10x + 1 − 12x − 13 = 0 (25x + 13)(x − 1) = 0 x = −13/25, 1 Check answers: − − − − − − x = −13/25 : √324/25 + 13/5 = 18/25 + 13/5 = 31/5 ≠ −1 . Therefore, x = −13/25 is not a solution. − − x = 1 : √16 − 5 = 4 − 5 = −1 √ . 27. − − − − − √x + 8 + 3 = 0 − − − − − √ x + 8 = −3 − − − − − No solutions because √x + 8 > 0. 28. − − − − − −−−− − √ x + 4 = √ 2x + 1 + 1 x +4 = −−−− − (2x + 1) + 2 √ 2x + 1 + 1 −−−− − 2 √ 2x + 1 = 2 − x 2 4(2x + 1) = 4 − 4x + x 2 x − 12x = 0 x(x − 12) = 0 x = 0, 12 Check answers: – – x = 0 : LHS = √4 = 2; RHS = √1 + 1 = 2 √ − − − − x = 12 : LHS = √16 = 4; RHS = √25 + 1 = 6 → LHS ≠ RHS, x = 12 29. Note: Use |a| 2 2 =a 2 = |a | is not a solution. . |4x − 3| = 5 2 (4x − 3) 2 16 x = 25 − 24x − 16 = 0 2 2x − 3x − 2 = 0 (2x + 1)(x − 2) = 0 x = −1/2, 2 30. |x − 1| = |7 − x| 2 x 2 − 2x + 1 = 49 − 14x + x 12x − 48 = 0 x = 4 31. Pamini Thangarajah 10/21/2020 0E.23 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 |2x + 3| − 3x + 5 = 0 |2x + 3| = 2 4x 2 5x 3x − 5 (Note: 3x − 5 ≥ 0 → x ≥ 5/3) 2 + 12x + 9 = 9 x − 30x + 25 − 42x + 16 = 0 (x − 8)(5x − 2) = 0 x = 8, 2/5 x = 2/5 < 5/3 . Therefore, it is not a solution. The solution set is x = 8 . Exercise 0E. 5 Solve each of the following systems of equations given below. 1. 3x − 4y = −5 2x + 5y = 12 2. 2x + y =1 x + 2y = −4 3. 1 2 x− y =1 2 3 −2x + y = 2 4. 3 x+ 1 y = 4 2 4x − y =1 5. xy = 1 2 x +y 2 =4 Leave your answer in radical form. 6. 2 x 2 x −y +y 2 2 =9 = 21 Leave your answer in radical form 7. 2x + y = 3 2 (x − 3y ) − (x − 3y) − 8 =0 Answer 1. Solve for y (or x , it doesn't matter) in one equation and substitute it into the other equation. This creates an equation with one unknown. Solve for the unknown. If we solve for y in the first equation, we get =y = ( into the second equation 2x + 5 ( x =1 1 ) (3x + 5) 4 1 ) (3x + 5)) = 12 4 . Substitute y . Simplify to (23/4)x = 23/4. Therefore, and y = 1/4(3 ⋅ 1 + 5) = 8/4 = 2 . 2. Solve for y in equation 1 and substitute it into equation 2. Solve for x . Pamini Thangarajah 10/21/2020 0E.24 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 y = 1 − 2x x + 2(1 − 2x) = −4 −3x + 2 = −4 x = 2 y = −3 3. Solve for y in equation 2 and substitute it into equation 1. Solve for x . y = 2 + 2x 1 2 x− 2 (2 + 2x) = 1 3 14 x = − 5 18 y = − 5 4. Solve for x in equation 1 and substitute it into equation 2. Solve for y. 1 x = 3 − 2 1 4( 3 − 2 y 4 y) − y = 1 4 1 y = 4 5 x = 16 5. Solve for y in equation 1 and substitute it into equation 2. Solve for x . 1 y = x 2 x 2 1 +( ) = 4 x 4 x +1 = 4 2 x 4 x 4 x 2 − 4x 2 + 1 = 4x +1 = 0 2 x 2 x Case 1: x 2 2 2 – = 2 ± √3 – = 2 + √3 − − − − − − – x = ±√2 + √3 , y = ± Case 2: x − − 4 ± √12 = 1 − − − − − − – √2 + √3 – = 2 − √3 − − − − − − – x = ±√2 − √3 , y = ± 1 − − − − − − – ±√2 − √3 6. Equation 2 - Equation 1 yields 2y = 12. – Solve for y: y = ±√6. − − Since x = 21 − y = 21 − 6 = 15 , x = ±√15 . 2 2 2 7. Solve for y in the first equation: y = 3 − 2x . Substitute y into the second equation: (x − 3(3 − 2x)) − 2(x − 3(3 − 2x)) − 8 = (7x − 9) Let u = 7x − 9 : u − 2u − 8 = (u − 4)(u + 2) = 0 . Therefore, 7x − 9 = u = 4, −2 . \\ In conclusion, x = 1, y = 1 or x = 13/7, y = −5/7. 2 2 − 2(7x − 9) − 8 = 0 . 2 Inequalities Exercise 0E. 1 Answer true or false in each of the following statements. If the conclusion of the statement is not correct, make the correct conclusion. Pamini Thangarajah 10/21/2020 0E.25 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 1. a < b and c < b implies that a < c. 2. a < c and c < 0 implies that a < 0. 3. a < b implies that 1 1 < . a b 4. a < b implies that −a < −b 5. a < b implies that a 2 2 <b 6. If a −b . 5. False. No correct conclusion. 6. True. Exercise 0E. 2 For each of the following, write without absolute value: 1. | − 29| 2. |45| 3. |2x + 5| 2 2 4. (5x − 1) 2 5. − |1 − 5x | |x| − x if x < 0 x 6. |x − 2| − x + 2 if x > 2 x −2 Answer 1. 29 2. 45 2 3. |2y + 5| 4. (5x − 1) 2 2 = (2y + 5 ) − |1 − 5x | |x| − x 5. 2 2 + 20y + 25 2 = (5x − 1 ) 2 − (1 − 5x ) =0 −x − x = x = −2, x ≠ 0 x |x − 2| − x + 2 6. = 4y x −2 −x +2 = x −2 =0 x −2 Exercise 0E. 3 Solve each of the following inequalities: 1. x 2 − 3x ≤ 4 2. 2x 2 − 5x + 2 > 0 3. 4 − 3x ≥ 0 4. 4x 2 − 28x + 49 ≤ 0 5. (x − 3)(x 2 6. x 2 + 1) < 0 + 24 < 0 Pamini Thangarajah 10/21/2020 0E.26 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 2 7. x − x − 20 ≤0 3 x (x + 1) 2 8. 2x −x −1 2 x <0 − 16 2 x 9. +1 <0 2 (x − 1 ) (2 − x) 2 10. 2 (2x − 1 ) (x + 1)(3x − 2) 2 (x − 2)(x ≤0 − x + 4) 11. |2x + 5| > 2 12. |1 − x| ≤ |x + 1| 13. |2x − 1| ≤ |5x − 1| 14. |2 + x| >1 |2 − x| 15. |4x − 1| < x 16. |2x − 3| ≥ x Answer 1. Isolate all of the terms on one side of the inequality: x 2 − 3x − 4 ≤ 0 . Factor: (x − 4)(x + 1) ≤ 0 . Make a sign chart: Solution: x ∈ [−1, 4]. 2. Factor: (2x − 1)(x − 2) . Make a sign chart: Solution: x ∈ (−∞, 1/2) ∪ (2, ∞) . 3. Since 4 − 3x is linear, solve for x : x ≤ 4/3 4. Factor: (4x − 7) . 2 Anything squared is always positive. Therefore, 4x 2 − 28x + 49 is never negative but it is zero. Solution: x = 7/4 5. Make a sign chart: Solution: x ∈ (−∞, 3) . 6. x 2 + 24 is always positive. Therefore, there is no solution. 7. Note: x ≠ 0, −1 Factor: (x − 5)(x + 4) 3 . x (x + 1) Make a sign chart: Pamini Thangarajah 10/21/2020 0E.27 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 Solution: x ∈ [−4, −1) ∪ (0, 5] . 8. Note: x ≠ ±4 Factor: (2x + 1)(x − 1) . (x − 4)(x + 4) Make a sign chart: Solution: x ∈ (−4, −1/2) ∪ (1, 4) . 9. Note: x ≠ 1, 2 Make a sign chart: Solution: x ∈ (2, ∞) . 10. Note: x ≠ 2 and x 2 is not factorable. −x +4 Make a sign chart: Solution: x ∈ [2/3, 2). 11. Case 1: If 2x + 5 ≥ 0 , then |2x + 5| = 2x + 5 2x + 5 > 2 → x > −3/2 Case 2: If 2x + 5 < 0 , then |2x + 5| = −(2x + 5) −2x − 5 > 2 → x < −7/2 Solution: x ∈ (−∞, −7/2) ∪ (−3/2, ∞) 12. |1 − x| ≤ |x + 1| 2 1 − 2x + x 2 ≤ x + 2x + 1 0 ≤ 4x 0 ≤ x 13. |2x − 1| ≤ |5x − 1| 2 4x 2 − 4x + 1 ≤ 25 x 2 0 ≤ 16 x Pamini Thangarajah 10/21/2020 0E.28 CC-BY-NC-SA − 10x + 1 + 6x = 2x(8x + 3) https://math.libretexts.org/@go/page/10827 Make a sign chart to get the solution x ∈ (−∞, −3/8] ∪ [0, ∞) . 14. Note: x ≠ 2 and (2 − x ) 2 >0 , therefore we can multiply both sides of the inequality by it. ∣ 2 +x ∣ ∣ ∣ > 1 Square both sides ∣ 2 −x ∣ 2 4 + 4x + x 2 > 1 Since (2 − x ) is positive we can cross multiply 2 4 − 4x + x 2 4 + 4x + x 2 > 4 − 4x + x 8x > 0 Solution: x > 0 15. Note: x ≥ 0 0 ≤ |4x − 1| < x 2 16 x 2 15 x 2 − 8x + 1 < x − 8x + 1 < 0 (3x − 1)(5x − 1) < 0 Make a sign chart to get the solution x ∈ (1/5, 1/3). 16. |2x − 3| ≥ x 2 4x 2 3x 2 − 12x + 9 ≥ x − 12x + 9 ≥ 0 3(x − 1)(x − 3) ≥ 0 Make a sign chart to get the solution x ∈ (−∞, 1) ∪ (3, ∞) . Analytic geometry Exercise 0E. 1 In each cases, Given points P and Q, determine a)the slope of the line segment P Q, b) the midpoint, M , of the line segment P Q and c) the lengh of the line segment P Q. 1. P = (−2, 9), Q = (2, 6) 2. P = (0, 0), Q = (−6, 8) 3. P = (5, −3), Q = (7, −3) 4. P = (7, −2), Q = (7, −9) 5. P = (11, 0), Q = (0, −60) 6. P = (6, −5), Q = (−6, 0) 7. P = (13, −2), Q = (5, 13) Answer 3 15 4 2 1. a) − , b) (0, , c) 5 ) 4 2. a) − , b) (−3, 4), c) 10 3 3. a) 0, b) (6, −3), c) 2 11 4. a) undefined, b) (7, − 5. a) 60 11 , b) ( Pamini Thangarajah 11 , c) 7 ) 2 , c) 61 , −30) 2 10/21/2020 0E.29 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 6. a) − 7. a) − 5 12 15 8 , b) (0, − , b) (9, 5 , c) 13 ) 2 11 , c) 17 ) 2 Exercise 0E. 2 1. Find the equation of the straight line passing through (1, −2) and which has a slope of −2. 2. Find the equation of the straight line passing through (2, −3) that is a) parallel to the line 2x − 5y + 3 = 0 . b) perpendicular to the line 2x − 5y + 3 = 0. Answer 1. y = −2x 2. a) y = b) y = − 2 19 x− 5 5 5 x +2 2 Exercise 0E. 3 1. Find the equation of the straight line passing through the midpoint of the line segment P = (−1, 3), and Q = (5, −7). PQ and perpendicular to PQ given that 2. Find the equation of the straight line which contains both of the points A = (4, −2), and B = (7, −2). 3. Find the equation of the straight line which contains the point (−1, 5) and is parallel to the x−axis. 4. Find the equation of the straight line which contains the point (10, −1) and is parallel to the y−axis. Answer 1. y = 3 16 x− 5 5 2. y = −2 3. y = 5 4. x = 10 Exercise 0E. 4 Which of the following pairs of lines is parallel?, perpendicular?, neither parallel nor perpendicular? 1. 4x + 3y − 2 =0 2x = 7y − 1 2. 1 3 x− 2 y +1 = 0 4 9y − 6x + 3 = 0 3. x + 11 = 0 2y − 7 = 0 4. 3x + 7y + 1 = 0 35x − 15y − 2 = 0 Pamini Thangarajah 10/21/2020 0E.30 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 Answer 1. Neither parallel nor perpendicular. 2. Parallel. 3. Perpendicular. 4. Perpendicular. Exercise 0E. 5 Sketch the graph of each of the following straight lines: 1. 3x − 6y − 2 = 0 2. 7x + 5y − 35 = 0 3. 2x − y = 0 4. 3x + 4y = 0 5. 2y − 11 = 0 6. 3x + 2 = 0 Answer 1. 2. Pamini Thangarajah 10/21/2020 0E.31 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 3. 4. 5. Pamini Thangarajah 10/21/2020 0E.32 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 6. Exercise 0E. 6 Determine the point of intersection of each of the following pairs of lines, if they exists. 1. 2x + 5y = −11 and 7x − 4y = 26 2. 2x + −3y = 1 and 1 3 x− 2 −1 y = 4 4 Answer 1. x = 2, y = −3 2. The lines are parallel. Therefore, there is no point of intersection. Exercise 0E. 7 1. Given that A = (3, 4), B = (−2, 3), and C which two of the sides have the same length.) 2. Given that A = (5, 3), Pythagoras'.) B = (−2, 2), 3. Show that the points P and C = (2, −1), = (2, −1), = (1, −1), Q = (0, 2) show that the triangle ABC is isoceles. (An isoceles triangle is a triangle in show that the triangle ABC has a right angle at C . (Hint: Recall Theorem of and R = (−1, 5) are collinear (lie on the same line). 4. Use the distance formula to show that the triangle ABC is an equilateral triangle, given that C = (3, 2/3). (An equilateral triangle is triangle in which all sides are of equal length). 5. If ABC D is parallelogram with A = (7, −3), B = (8, 3), and C = (3, 5), A = (1, 0), B = (5, 0), and determine the coordinates of the vertex D. Answer ¯ ¯¯¯¯¯¯ ¯ −−−−−−−−−− − − − 2 2 + (−1 ) = √26 ¯ ¯¯¯¯¯¯ ¯ 1. AB = (−5, −1) and |AB| = √(−5) ¯ ¯¯¯¯¯¯ ¯ AC = (−1, −5) and −−−−−−−−−− − − − | AC | = √(−1 )2 + (−5 )2 = √26 ¯ ¯¯¯¯¯¯ ¯ Since two of the lines of the triangle have equal length, the three points form an isoceles triangle. ¯ ¯¯¯¯¯¯ ¯ −−−−−−−−−− − – 2 2 + (−1 ) = 5 √2 ¯ ¯¯¯¯¯¯ ¯ 2. AB = (−7, −1) and |AB| = √(−7) ¯ ¯¯¯¯¯¯ ¯ AC = (−3, −4) ¯ ¯¯¯¯¯¯ ¯ BC = (4, −3) and −−−−−−−−−− − 2 2 | AC | = √(−3 ) + (−4 ) =5 ¯ ¯¯¯¯¯¯ ¯ − −−−−−−−− − 2 2 + (−3 ) =5 ¯ ¯¯¯¯¯¯ ¯ and |BC | = √(4) Pythagoras' theorem is only true for right angled triangles: ¯ ¯¯¯¯¯¯ ¯ 2 | AC | ¯ ¯¯¯¯¯¯ ¯ 2 + | BC | 2 = (5 ) 2 + (5 ) = 50 − − – ¯ ¯¯¯¯¯¯ ¯ . If the triangle is right angled, we would expect that √50 = 5√2 would equal |AB| . It does! 3. Find the equation of the line containing the points P and Q . Slope = =−3 . Line: y = −3x + 2 . Check if R lies on the line: If x = −1 , y = −3(−1) + 2 = 5 . Therefore, the point R = (−1, 5) is on the line and the three points are collinear. Pamini Thangarajah 10/21/2020 0E.33 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 ¯ ¯¯¯¯¯¯ ¯ −− − ¯ ¯¯¯¯¯¯ ¯ 4. AB = (4, 0) and |AB| = √(4) – AC = (2, 2 √3 ¯ ¯¯¯¯¯¯ ¯ and – BC = (2, 2 √3) ¯ ¯¯¯¯¯¯ ¯ ¯ ¯¯¯¯¯¯ ¯ 2 =4 − −−−−−−−−− − ¯ ¯¯¯¯¯¯ ¯ – 2 2 | AB| = √(2 ) + (2 √3) =4 −−−−−−−−−−− − – 2 2 + (2 √3) =4 ¯ ¯¯¯¯¯¯ ¯ and |AB| = √(−2) ¯ ¯¯¯¯¯¯ ¯ ¯ ¯¯¯¯¯¯ ¯ Since |AB| = |AC | = |BC | , it is an equilateral triangle. 5. Let D = (h, k) so that ABC D is a parallelogram. k+3 m¯¯¯¯¯¯¯¯ = m ¯¯¯¯¯¯¯¯ ⇒ AD BC 5 −3 = h −7 3 −8 −5(k + 3) = 2(h − 7) 2h + 5k = −1 k−5 m¯¯¯¯¯¯¯¯ = m¯¯¯¯¯¯¯¯ ⇒ AB CD (1) 6 = h −3 8 −7 k − 5 = 6(h − 3) 6h − k = 13 (2) To determine h and k solve (1) and (2) simultaneously to obtain h = 2 and k = −1 . Thus, D = (2, −1) . Exercise 0E. 8 Find an equation of the circle with center and radius as indicated: 1. Center (4, 3), radius = 9. – 2. Center (−1, 2), radius = √5. 3. Center (2, 0), radius = 1. 4. Center (0, 0), radius = 1. 5. Center (−7, −3), radius = − − √17 . 2 6. Center (0, −4), radius = 3. Answer 1. (x − 4) 2 2. (x + 1) 2 3. (x − 2) 2 4. x 2 +y 2 5. (x + 7) 6. x 2 + (y − 2 ) 2 +y = 81 =5 =1 =1 2 2 2 + (y − 3 ) 2 + (y + 3 ) 2 + (y + 4 ) = 17/4 =9 Exercise 0E. 9 Find the center and radius of the following circles: 1. (x − 1) 2 2. (x + 2) 2 3. x 2 +y 2 4. (x − 7) 2 + (y + 3 ) 2 + (y + 1 ) 2 +y 5. x + (y + 2 ) 6. x +y 7. x +y 2 2 2 8. x 2 +y 9. 4x 2 2 2 2 = 24 =3 2 2 = 16 = 64 =1 − 2x + 2y + 1 = 0 + 8x − 6y + 17 = 0 − 4y + 1 = 0 + 4y 2 − 4x + 16y + 13 = 0 Pamini Thangarajah 10/21/2020 0E.34 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 10. 9x 2 + 9y 2 + 18x + 8 = 0 Answer 1. Centre at (1, −3), radius = 4. – 2. Centre at (−2, −1) , radius = 2√6. – 3. Centre at (0, 0), radius = √3. 4. Centre at (7, 0), radius = 8. 5. Centre at (0, −2), radius = 1. 6. Complete the square to find the centre and radius. 2 (x − 1 ) 2 + (y + 1 ) =1 Centre at (1, −1), radius = 1. 7. (x + 4) + (y − 3) = 8 – Centre at (−4, 3), radius = 2√2. 2 2 8. x + (y − 2) = 3 – Centre at (0, 2), radius = √3. 2 2 9. (x − 1/2) + (y + 2) = 1 Centre at (1/2, −2), radius = 1. 2 10. (x + 1) 2 +y 2 2 = 1/9 Centre at (−1, 0), radius = 1 3 . Exercise 0E. 10 1. Find an equation of the circle centered at (1, 5) passing through the point (7, −1). 2. Determine the equation of the circle with diameter P Q, where P 3. Determine the equation of the circle with center C = (−1, 1) = (−3, 5) and tangent to = (3, −5) and tangent to and Q = (2, 0). a) the x−axis b) the y−axis 4. Determine the equation of the circle with center C a) the x−axis b) the y−axis Answer 1. We know that the equation will be (x + 1) + (y − 5) Substitute (7, −1) into the equation and solve for r . 2 2 8 2 + (−6 ) 2 2 = 64 + 36 = 100 = r Therefore, the equation of the circle is (x + 1) 2 2 + (y − 5 ) 2. The radius is half the distance between the two points: 2 (x − 1/2 ) 2 b) (x + 3) 2 2 + (y − 5 ) 4. a) (x − 3) 2 b) (x − 3) 2 2 + (y − 5 ) 2 2 = 100 (0E.2) . . 1 1 ,− 2 2 + (y + 1/2 ) ) 2 . = 5/2 (0E.3) = 25 =9 + (y + 5 ) + (y + 5 ) → r = 10 − − √10/2 The centre will be the midpoint between the two points: ( 3. a) (x + 3) , where r is the radius. 2 =r = 25 =9 Exercise 0E. 11 Pamini Thangarajah 10/21/2020 0E.35 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 1. Find an equation of the circle with center at (2, −3) and radius of length 4 units. 2. Find an equation of the circle which has center at the point (4, −7) and which contains the point (2, 1). 3. Find an equation of the circle which has diameter AB, where A = (2, 3) and B = (6, −1). 4. Find an equation of the circle which is tangent to the x−axis at the point (−2, 0), and is tangent to the y−axis at the point (0, 2). Answer 1. (x − 2) 2 2. (x − 4) 2 3. (x − 4) 2 4. (x + 2) 2 2 + (y + 3 ) 2 + (y + 7 ) 2 + (y − 1 ) 2 + (y − 2 ) =4 = 68 = 32 =4 Exercise 0E. 12 Classify each equation given as being an equation of a parabola, an ellipse, a circle, a hyperbola or none of these. 1. 4x 2 2. 4x 2 3. x 2 + 5y +y 4. 4x 2 2 6. 4x 7. 2x 2 + 8x + 4y + 3 = 0 2 − 8x + 15y = 1 + 12x − 8y = 0 +y −3 = 0 5. 4x + y 2 2 −y 2 +y − 2y + 4 = 0 2 − 3y + 3y − 4 = 0 2 − 6x + 12y = 1 8. 3x + 2y − 5 = 0 9. x 2 + 5y 10. x 3 +y 3 3 − 3x + 2y − 1 = 0 =1 11. (x − 1) 2 12. (x − 1) 2 2 + (y + 3 ) 2 − (y + 3 ) = 22 = 22 Answer The general form of a conic is Ax + By + C xy + Dx + Ey + F = 0 . If the conic is not degenerate, then we can use the general form to classify the conic. If A = B , then the conic is a circle. If A ≠ B , then the conic is an ellipse if A and B are both positive, the conic is a hyperbola if conic is a parabola if A or B is zero. 2 2 A and B are opposite signs, and the 1. Hyperbola. 2. Ellipse. 3. Circle. 4. Parabola. 5. Parabola. 6. Parabola. 7. Hyperbola. 8. Degenerate conic. 9. Not a conic. 10. Not a conic. 11. Circle. 12. Hyperbola. Pamini Thangarajah 10/21/2020 0E.36 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 Trignometry Exercise 0E. 1 1. In each of the triangle, determine the values of sin(α), cos(α), tan(α), sec(α), csc(α), and cot(α). a) b) 2. Use the information provided in the diagram to determine cot(t), sin(t), sec(t). Answer 1. a) The unknown side is 6. . sin α = 3/5, cos α = 4/5, tan α = 3/4, sec α = 5/4, csc α = 5/3, cot α = 4/3 b) The unknown side is 15. . sin α = 15/17, cos α = 8/17, tan α = 15/8, sec α = 17/8, csc α = 17/15, cot α = 8/15 2. The unknown side is 60. . cot t = 60/11, sin t = 11/61, sec t = 61/60 Pamini Thangarajah 10/21/2020 0E.37 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 Exercise 0E. 2 1. If cos(α) = , and if α + β = 90 3 ∘ 5 2. If sec(α) = 2, and if α + β = 90 ∘ , , find cot(β). find sin(α), csc(β) and tan(β). Answer 1. Since α + β = 90 , β is the third angle of the triangle. cot β = 4/3. ∘ – 2. sin α = √3, csc β = 2, tan β = – √3 3 Exercise 0E. 3 Convert each of the angles given in degree measure to radian measure: 1. 225 ∘ 2. 450 ∘ 3. −210 ∘ 4. 175 ∘ 5. 1 ∘ Answer 1. 2. 5π 4 5π 2 3. − 4. 5. 7π 6 35π 36 π 180 Exercise 0E. 4 Convert each of the angles given in radian measure to degree measure: 1. 2 2. 3. 4. 5. 6. π 5 −π 12 7π 24 11π 6 15π 8 Answer ∘ 1. 360 π 2. 36 ∘ 3. −15 ∘ 4. 52.5 ∘ 5. 330 ∘ Pamini Thangarajah 10/21/2020 0E.38 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 6. 337.5 ∘ Exercise 0E. 5 Find the reference angle for each of the following angles. 1. 60 ∘ 2. 225 ∘ 3. 120 ∘ 4. −120 ∘ 5. 6. 7. 8. π 4 4π 3 7π 6 5π 6 9. 11π 6 10. 3000 ∘ Answer 1. 60 ∘ 2. 45 ∘ 3. 60 ∘ 4. 60 ∘ 5. π 4 6. π 3 7. 8. 9. π 6 π 6 π 6 10. 60 ∘ Exercise 0E. 6 Complete the table below: Pamini Thangarajah 10/21/2020 0E.39 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 Answer Exercise 0E. 7 Pamini Thangarajah 10/21/2020 0E.40 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 Determine the exact values of each of the following: 3π 1. sin( ) 2 11π 2. cos( ) 6 3. sec(420 ∘ ) 4. tan(225 ∘ 5. tan( ) 29π ) 3 7π 6. sin( ) 6 7. csc(−120 ∘ ) 11π 8. sin( ) 3 Answer 1. −1 2. – √3 2 3. 2 4. 1 – 5. −√3 6. − 7. − 8. − – 2 √3 3 – 2 √3 3 – √3 2 Exercise 0E. 8 −3 1. If sin(α) = , 5 and α is an angle in the third quadrant, then find a) cos(α) b) tan(α) 17 2. If sec(α) = and \(270^{\circ}<\alpha < 360^{\circ} \(, then find , 5 a) cos(α) b) sin(α) c) tan(α) 3.If cos(t) = −12 , and tan(t) > 0, find 13 a) sin(t) b) tan(t) 4.If cot(β) = 4 , and sin(β) > 0, find 3 a) cos(β) b) csc(β) Pamini Thangarajah 10/21/2020 0E.41 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 1 –, √3 5. If tan(t) = and cos(t) < 0, find a) sin(t) b) cos(t) 6. If csc(t) = 2, and cos(t) > 0, find a) sin(t) b) tan(t) c) cos(t) Answer 1. a) cos α = −4/5 , b) tan α = 3/4. 2. a) cos α = 15/17, b) sin α = −8/17 , c) tan α = −8/15. 3. a) sin t = −5/13, b) tan t = 5/12. 4. a) cos β = 4/5, b) csc β = 5/3 . – 5. a) sin t = −1/2 , b) cos t = −√3/2 . – – 6. a) sint = 1/2, c) tan t = −√3/3 , c) cos t = −√3/2 . Exercise 0E. 9 Use the sum of difference formulas to determine the exact values of each of the following: 1. sec(15 ∘ 2. sin(75 ∘ ) ) 3. tan(75 ∘ 4. csc( ) 5π ) 12 5. cos(105 ∘ 6. sin(165 ∘ 7. sin( ) ) π ) 12 Answer 1. ∘ ∘ cos(15 ) = cos(45 ∘ − 30 ) ∘ ∘ ∘ ∘ = cos(45 ) cos(30 ) + sin(45 ) sin(30 ) – √2 = ( – √3 )( 2 – √2 ) +( 2 1 )( 2 ) 2 – – √2(√3 + 1) = 4 Therefore, sec(15 ∘ 2. sin(75 ∘ 4 ) = ∘ ) = sin(45 – – √2(√3 + 1) . – – √2(√3 + 1) ∘ + 30 ) = 4 3. tan(75 ∘ 4. sin( – √6 + 2 ) = – – 2 √3 − √2 5π 3π ) = sin( 12 Therefore, csc( Pamini Thangarajah 12 5π π ) = sin( 12 – – √2(√3 + 1) – – √2(√3 + 1) π + 4 4 ) = 12 2π + ) = 6 4 . 10/21/2020 0E.42 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 5. cos(105 ∘ – – √2(√3 − 1) ) = 4 6. ∘ ∘ sin(165 ) = sin(180 ∘ − 15 ) ∘ ∘ circ = sin(180 ) cos(15 ) − cos(180 ∘ ) sin(15 ) ∘ = sin(15 ) – – √2(√3 − 1) = 4 7. sin( – – √2(√3 − 1) π ) = 12 4 Exercise 0E. 10 1. If sin(α) = 2. If sec(α) = −3 and α is in the third quadrant, then determine cos(2α), sin(2α), and tan(2α). , 5 17 and 270 ∘ , 5 1 3. If tan(β) = – √3 , 6. If tan(t) = −4 π < t < π, 2 find sin(2t), csc(2t), sec(2t) and tan(2t). , and α is in the third quadrant, then find cos(α/2), sin(α/2), and tan(α/2). , and cos(t) < 0, find sin(t/2), cos(t/2) and tan(t/2). 5 1 – √3 , then find tan(2α). and cos(β) < 0, find sin(2β) and cos(2β). 4. If csc(t) = 2, and if 5. If sin(α) = ∘ < α < 360 3π 7. If csc(t) = −2, and if < t < 2π, 2 find cos(t/2), and tan(t/2). 8. If cot(α) = −1, and if sin(α) > 0 , then find cos(α/2), sin(α/2), and tan(α/2). 9. If sin(α) = 11 61 , and α is in the first quadrant, then find csc(α/2), and sec(α/2). Answer 1. cos(2α) = cos 2 2 α − sin α = 7/25 sin(2α) = 2 sin α cos α = 24/25 2 tan α tan(2α) = = 24/7 2 1 − tan α 2. tan(2α) = −240/161 – 3. sin(2β) = √3/2 cos(2β) = 1/2 – 4. sin(2t) = −√3/2 cos(2t) = 1/2 sec(2t) = 2 – tan(2t) = −√3 − − − − − − − − 1 + cos α 5. cos(α/2) = ±√ – 2 √5 =− 2 5 − − − − − − − − – 1 − cos α 4 √5 sin(α/2) = ±√ =− 2 5 , `-' because the angle is in the third quadrant. 1 − cos α tan(α/2) = = −2 sin α 6. − − − − − − − – 2 + √3 sin(t/2) = −√ 4 − − − − − − − – 2 − √3 cos(t/2) = −√ 4 – tan(t/2) = −2 − √3 Pamini Thangarajah 10/21/2020 0E.43 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 − − − − − − − – 2 + √3 cos(t/2) = √ 4 7. – tan(t/2) = −2 + √3 − − − − − − − – 2 + √2 sin(α/2) = √ 4 8. − − − − − − − – 2 − √2 cos(α/2) = −√ 4 – tan(α/2) = 1 + √2 − − − 9. csc(α/2) = √122 − − − − 122 sec(α/2) = √ 121 Exercise 0E. 11 For each of the following, determine the amplitude, phase shift and period. Use this information to sketch the graph. 1. y = 4 sin(2x + π/3) 1 2. y = cos(5x − π/4) 2 3. y = −2 cos(x/3) 4. y = 2 tan(3x + π/2) 5. y = −3 sin(4x) 4 6. y = cos(2x) 3 Answer 1. Amplitude: 4; Phase shift: 2. Amplitude: 1 2 ; Phase shift: π 3 ; Period: π . π 4 2π ; Period: 5 . 3. Amplitude: 2; Phase shift: none; Period: 6π. π 4. Amplitude: none; Phase shift: 2 ; Period: 5. Amplitude: 3; Phase shift: none; Period: 6. Amplitude: 4 3 ; Phase shift: none; Period: π 3 π 4 . . π 2 . Exercise 0E. 12 Solve each of the following trigonometric equations in the interval indicated. 1. 2 cos(x) + 1 = 0, ∘ ∘ [ 0 , 360 ) 2. 2 tan(x) sin(x) + 2 sin(x) − tan(x) − 1 = 0, 3. csc 4. sin 2 2 ∘ ∘ ∘ ∘ (t) − 4 = 0, [ 0 , 180 ] (t) − 3 = 0, [ 0 , 360 ) 5. 2 cos (x) − 3 cos(x) + 1 = 0, 6. 2 sin (x) + 3 sin(x) + 1 = 0, 2 2 [0, 2π) ∘ ∘ [ 0 , 360 ) 7. (2 sin(x) − 1)(tan(x) + 1) = 0, ∘ ∘ [ 0 , 360 ) – – 8. sin(x)(2 cos(x) + √3)(tan(x) − √3) = 0, 9. sin(2x) = 0, [0, π] ∘ ∘ [ 0 , 360 ) cos(2x) = 0, [0, 2π) Answer Pamini Thangarajah 10/21/2020 0E.44 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 1. x = 120 ∘ ∘ , 240 2. Factor by grouping to get tan x + 1 = 0 or 2 sin x − 1 = 0 . tan x + 1 = 0 yields the solution x = 3π/4 and 2 sin x − 1 = 0 yields the solutions x = π/6, 5π/6. 3. Need to solve csc t = ±2 . The solutions are t = 60 , 150 . ∘ ∘ – 4. Need to solve sin t = ±√3/2 . The solutions are t = 60 , 120 , 240 ∘ ∘ ∘ ∘ , 300 . 5. Need to solve cos x = 1, 1/2. The solutions are x = 0, π/3, 5π/3. 6. Need to solve sin x = −1, −1/2 . The solutions are x = 270 , 210 , 330 . ∘ ∘ ∘ 7. Need to solve sin x = 1/2 and tan x = −1 . The solutions are x = 30 , 150 , 135 , 315 . ∘ ∘ ∘ ∘ – – sin x = 0, cos x = −√3/2, tan x = √3 8. Need to solve The solutions are x = 0 ∘ ∘ ∘ ∘ ∘ ∘ , 180 , 150 , 210 , 60 , 240 . . 9. The solutions are x = 0, π/2, π, 3π/2, π/4, 3π/4, 5π/4, 7π/4 . Exercise 0E. 13 Given that tan(α/2) = t, express each of the following in terms of t : 1. sin(α/2) 2. cos(α/2) 3. tan(α/2) 4. sin(α) 5. cos(α) 6. tan(α) Answer Construct a right-angled triangle with α/2 as one of the angles. − − − − − Then the opposite side it t and the adjacent side is 1. Using Pythagoras' theorem, we obtain a hypotenuse of √t + 1 . 2 Therefore, sin(α/2) = t − − − − − √t2 + 1 and cos(α/2) = 1 − − − − − √t2 + 1 Using sin(α) = 2 sin(α/2) cos(α/2), we obtain sin(α) = . 2t 2 t 2 Using cos(α) = 2 cos 2 x −1 , we obtain cos(α) = −t + 2t − 1 2 t 2t Therefore, tan(α) = 2 −t . +1 . +1 . + 2t − 1 Note: This solution assumes 0 ≤ t < π/2 . If t was in another quadrant, how does this effect the solution? Exercise 0E. 14 In each of the following, find other information about the triangle ABC , using the information provided: 1. A = 130 , B = 20 , b = 10 2. B = 120 , C = 30 , a = 16 ∘ ∘ ∘ ∘ 3. C = 30 , a = 24, b = 34 4. C = 108 , a = 10, b = 15 ∘ 5. a = 12, ∘ b = 17, c = 19 Pamini Thangarajah 10/21/2020 0E.45 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 6. A = 40 , a = 7, c = 10 7. A = 40 , a = 12, c = 10 ∘ ∘ Answer 1. C ∘ = 108 ∘ ∘ − 130 ∘ − 20 = 30 ∘ b sin(130 ) a = = 22.398 ∘ sin(20 ) ∘ b sin(30 ) c = = 14.619 ∘ sin(20 ) 2. A = 180 ∘ ∘ ∘ − 120 ∘ − 30 = 30 ∘ a sin(120 ) b = = 27.713 ∘ sin 30 ) ∘ a sin(30 ) c = = 16 ∘ sin(30 ) −−−−−−−−−−−−−−−− − ∘ 2 2 + b − 2ab cos(30 ) = 17.851 3. c = √a 2 2 c +b 2 −a ∘ A = arccos( ) = 107.76 2cb 2 2 c +a 2 −b ∘ B = arccos( ) = 42.24 2ca − −−−−−−−−−−−−−−−− − ∘ ∘ 2 2 + b − 2ab cos(108 ) = 20.438 4. c = √a 2 2 c +b 2 −a ∘ A = arccos( ) = 27.73 2cb 2 2 c +a 2 −b ∘ B = arccos( ) = 44.27 2ca 2 c 5. A = arccos( 2 +b 2 −a ∘ ) = 38.44 2cb 2 2 c +a 2 −b ∘ B = arccos( ) = 61.73 2ca 2 a 2 +b 2 −c ∘ C = arccos( ) = 79.84 2ab ∘ 6. C c sin(40 ) = arcsin( ∘ ) = 66.67 a ∘ B = 180 ∘ ∘ − 40 − 66.67 ∘ = 73.33 ∘ a sin(73.33 ) b = = 10.432 ∘ sin(40 ) ∘ 7. C c sin(40 ) = arcsin( ∘ ) = 32.39 a ∘ B = 180 ∘ − 40 ∘ − 32.39 ∘ = 107.61 ∘ a sin(107.61 ) b = ∘ = 17.794 sin(40 ) Exercise 0E. 15 1. A ladder 10 meters long is leaning against the side of a building. It makes an angle of 75 with the ground. How far up the side of the building does the ladder reach? ∘ 2. A man 6 ft tall on the ground 12 ft away from a lamp-post. The light come from the lamp casts a shadow which is 8 ft long. Determine the height of the lamp-post. Pamini Thangarajah 10/21/2020 0E.46 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 3. From the top of a light house, 40 metres above sea-level, the angle of depression to a boat is observed to be distance between the boat and the light house. ∘ 25 . Find the horizontal Answer 1. The ladder reaches up 10 sin(75 ∘ ) m up the wall. 2. Using similar triangles, the lamppost is 15 ft high. 3. The horizontal distance is 40 tan(25 ∘ ) m. Logarithms and Exponents Exercise 0E. 1 Which of the following functions are exponential functions? If the function is exponential, determine its base. 1. f (x) = 6 x 2. f (x) = x 6 3. f (x) = 2 −3x − − − − 4. f (x) = √(5 3x ) Answer 1. The function is exponential and its base is 6. 2. The function is not exponential. 3. The function is exponential and its base is 2. 4. The function is exponential and its base is 5. Exercise 0E. 2 Sketch the graphs of each of the following: 1. y = 3 x 2. y = 3 −2x 3. y = 5 x 4. y = 2 −7 x−4 +1 Answer 1. Pamini Thangarajah 10/21/2020 0E.47 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 2. 3. 4. Exercise 0E. 3 Solve each of the following exponential equations: 1. 3 (2x−1) 2. 8 −2x 1 = 64 3. 81 x 4. 7 2−x = 21 x−2 5. 2 7/3 = 27 7x 1 = 343 1 = – √2 Pamini Thangarajah 10/21/2020 0E.48 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 6. 10 2x−1 7. 3 x =5 = 17 8. 4 x−2 9. 2 x 1−x =2 = 21 10. (2 3x 11. 3 x 12. 3 x 5x − 8)(2 −x +3 − 4) = 0 =2 10 −x +3 = 3 Answer 1. Write each side of the equation as a same base (in this case 3). 2x−1 3(7/3) 3 =3 7 =3 (0E.4) x =4 2. Since we have 8 −2x 3. Since we have 3 4x 6−3x =3 4. Since we have 7 5. Since we have 2 7x , then x = 1 −2 =8 x−2 , then x = 6/7. , then x = −1 . −3 =7 −1/2 =2 (0E.5) , then x = −1/14. 6. Since 10 and 5 cannot be neatly written in terms of the same base, we will use logarithms to solve. 2x−1 10 2x−1 log(10 = 5 ) = log(5) 2x − 1 = log(5) x = (log(5) + 1)/2 7. x = log 3 (17) 8. Since 2 2x+4 9. x = log 2 10. Since 2 , then x = 1 . 1−x =2 (21) 3x =8 or 2 5x =4 , then x = 1 or 11. Multiply both sides by 3 to obtain 3 x Therefore, let 3 x =1 2x 2 5 . x +1 = 2 ⋅ 3 . Bring all the terms to one side and recognize that this is a hidden quadratic. and x = 0 . 12. x = −1, 1 Exercise 0E. 4 Express each of the following in logarithmic form: 1. 2 x =y 2. 5 −3 1 = 125 3. 4 x+y 4. 29 0 = 15 =1 5. x = 2y 6. 4 =4 3 1 Answer 1. x = log 2 (y) Pamini Thangarajah 10/21/2020 0E.49 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 2. −3 = log 5 1 ( ) 125 3. x + y = log 4 4. 0 = log (15) (1) 29 5. 3 = log (2y) x 6. 1 = log 4 (4) Exercise 0E. 5 Express each of the following in exponential form: 1. x = log 2 (y) 2. y = log (4) x 1 3. log ( 4. log (7) = 1 5. log (1) = 0 5 7 3 ) = −3 125 6. log(y) = 2 7. log(.001) = −3 8. log 1/3 9. log 3 (81) = −4 (y + z) = x + y 10. log(x − 5) = 2 Answer 1. 2 x =y 2. 4 = x y 3. 1 −3 =5 125 4. 7 = 7 1 5. 1 = 3 0 6. y = 10 2 7. 0.001 = 10 −3 8. 81 = (1/3) −4 9. y + z = 3 x+y 10. x − 5 = 10 2 Exercise 0E. 6 Sketch each of the following graphs and compare them to the graphs of the exponential function given. 1. y = log compare with y = 4 . 2. y = log(x) compare with y = 10 . 4 (x) x x Answer 1. y = log 4 (x) x y =4 Pamini Thangarajah 10/21/2020 0E.50 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 2. y = log(x) x y = 10 Exercise 0E. 7 Express as a single logarithm: 1 1. 2 log 2 2. (x) − 5 log (x − 2) + 2 2 log (x 2 2 + 1) 1 2 log(2x − 1) − 52 log(x) − 7 log(1 + x) + 3 log(5x + 12) 3 Answer − −−− − 2 x √x2 + 1 1. log ( 2 2. log( ) 5 (x − 2) − −−−−−−−−−−−−− − 3 √(2x − 1)(5x + 12)2 2 7 ) x (1 + x ) Exercise 0E. 8 Express as a sum or difference of logarithms: 1. log 2 2. log − − − − − − − 2 2 x √(x + 1) 5 (x − 2) − − − − − − − 3 −−−−−−− − 3 2 √(2x − 1) √(5x + 12) 2 7 x (1 + x ) Pamini Thangarajah 10/21/2020 0E.51 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 2 x(1 − x) 3. log 5 4 (2 − x) 2 2/7 x (1 + x ) 4. log 1/2 3 4(x − 2) Answer 1 1. 2 log 2 2 (x) + log2 (x 2 1 2. + 1) − 5 log2 (x − 2) 2 log(2x − 1) + 3 log(5x + 12) − 2 log(x) − 7 log(1 + x) 3 3. log x + 2 log5 (1 − x) − 4 log5 (2 − x) 5 2 4. 2 log 1/2 x+ log1/2 (1 + x) − 2 − 3 log1/2 (x − 2) 7 Exercise 0E. 9 Simplify each of the following: 1. log 32 2 2. log(0.001) 3. log x 4. log 4 − log 3 x 2 4 16 Answer 1. log 2 5 (32) = log2 (2 ) = 5 2. log(0.001) = log(10 −3 ) = −3 3. log (x ) = 3 4. log (4) − log4 (16) = log2 (2 ) − log4 (4 ) = 0 3 x 2 2 2 Exercise 0E. 10 Solve each of the following equations: 1. log x = −2 3 2. log 1000 = x 3. log 16 = 4 x 4. log(0, 1) = 2x 5. log 2 (3x + 1) − log2 (x − 8) = −1 6. log 4 = −1 x 7. log x − log(2x − 1) = 0 8. log x − log5 (x + 7) = log5 2 − log5 (x − 3) 5 9. 2 log 2 1 x − log2 (x − 3 ) = log2 3 − 1 10. log x + log3 √2 (x − 3) = 2 11. log x + log3 √2 (x + 3) = 2 3 √2 3 √2 Answer 1. x = 3 −2 1 = 9 2. x = −3 3. x = 2 , since 2 4 Pamini Thangarajah = 16 10/21/2020 0E.52 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 4. x = −1 2 5. log2 (3x + 1) − log2 (x − 8) = −1 3x + 1 log2 ( ) = −1 x −8 3x + 1 −1 = 2 x −8 6x + 2 = x − 8 x = −2 But x = −2 is extraneous because if we x = −1 in the original equation, we get the logarithm of a negative number which is undefined. Therefore, the equation has no solution. 6. x = 1 4 7. log x − log(2x − 1) = 0 log x = log(2x − 1) x = 2x − 1 x = 1 8. log5 (x) − log5 (x + 7) = log5 (2) − log5 (x − 3) x 2 log ( 5 ) = ( x +7 ) x −3 x 2 = x +7 x −3 x(x − 3) = 2(x + 7) x = 7, −2 is an extraneous solution. Solution: x = 7 . x = −2 9. 2 log2 (x) − log2 (x − 1/3) = log2 (3) − 1 1 2 log (x) − log (x − 2 2 ) − log (3) = −1 2 3 ⎛ log2 ⎜ ⎜ ⎞ 2 x ⎟ = −1 ⎟ 1 ⎝ 3(x − ) ⎠ 3 2 x 1 = 3x − 1 2 x = 1, 1/2 10. log3 √2 (x) + log3 √2 (x − 3) = 2 log 3 √2 (x(x − 3)) = 2 x(x − 3) = 18 x = −3, 6 is an extraneous solution. Solution: x = 6 . x = −3 11. Pamini Thangarajah 10/21/2020 0E.53 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 log3 √2 (x) + log3 √2 (x + 3) = 2 log3 √2 (x(x + 3)) = 2 x(x + 3) = 18 x = 3, −6 is an extraneous solution. Solution: x = 3 . x = −6 Pamini Thangarajah 10/21/2020 0E.54 CC-BY-NC-SA https://math.libretexts.org/@go/page/10827 Common Mistakes The purpose of this page is to alert the reader to the common mistakes which might be made in simple calculations. The reader should care fully analyze each statement, making sure that they can justify the correct statement. Answer true or false for each of the statements given below. If the statement is false, write the correct version. Exercise 1 − − − − − − − − = √a + √b 1. √(a + b) Answer − −− − False. Leave √a + b as is. It cannot be simplified any further. Exercise 2 2. a −1/2 1 = a2 Answer False. a −1/2 1 = − − √a Exercise 3 3. (a + b) 2 2 =a 2 +b Answer False. (a + b) 2 2 =a 2 + 2ab + b Exercise 4 4. −(−2) 2 =4 Answer False. −(−2) 2 = −4 Exercise 5 5. −3 4 = 81 Answer False. −3 4 4 = −(3 ) = −81 Exercise 6 6. 2 3 =6 Answer False. 2 3 =8 Pamini Thangarajah 10/21/2020 1 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 Exercise 7 7. a+b a b = c + c c Answer True. Exercise 8 8. c c c = a+b + a b Answer c False. cannot be simplified any further. a+b Exercise 9 9. 5h 4h = h(h + 2) h +2 Answer 5h False. 5 = h(h + 2) h +2 Exercise 10 10. 0 is not defined. 3 Answer False. 0 =0 3 Exercise 11 11. 5 =0 0 Answer False. 5 0 is undefined. Exercise 12 12. xy x y = 2 2 2 Answer False. xy 2 cannot be simplified any further. Exercise 13 Pamini Thangarajah 10/21/2020 2 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 13. −2x = 0 ⇒ x = 2 Answer False. x = 0 Exercise 14 14. −3x = 2 ⇒ x = 2 + 3 = 5 Answer False. x = −2 3 Exercise 15 15. x 2 + 1 = 0 ⇒ x = ±1 Answer False. x 2 +1 = 0 does not have any real solutions. It does have complex solutions x = ±i . Exercise 16 16. (x − 2) 2 = 0 ⇒ x = ±2 Answer False. x = 2 Exercise 17 17. The solutions of the quadratic equation c 2 + 5c + 6 = 0 − − − − − − − are given by c = ±√(5c − 6) . Answer False. Factor the quadratic (c − 5)(c + 1) = 0 . Then c = 5, −1 . Exercise 18 18. (c − 1)(c − 3) = 1 ⇒ c − 1 = 1, or c − 3 = 1 Answer False. This is true if (c − 1)(c − 3) = 0 (not 1). Exercise 19 19. 2 ( 4 8 ) = 3 6 Answer False. 2 ⋅ 4 8 = 3 3 Exercise 20 Pamini Thangarajah 10/21/2020 3 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 1 20. 2 3 = 3 2 Answer 1 False. 2 1 = 3 6 Exercise 21 21. The only solution of the quadratic equation x 2 − 2x = 0 is x = 2 . Answer False. x = 0 is also a solution. Exercise 22 22. x −1 = 0 ⇒ x − 1 = 0, x +5 or x + 5 = 0 Answer False. Only x − 1 = 0 (if x + 5 = 0 , this would cause division by zero). Exercise 23 23. 3x 3 = 2 x 2 Answer True. Exercise 24 24. By factoring an expression such as x 2 we mean: Solve x 2 − 4, −4 = 0 which yields x 2 = 4 ⇒ x = ±2 Answer False. By factoring an expression such as x 2 −4 , we mean: x 2 − 4 = (x − 2)(x + 2) . Exercise 25 − − − − − − − − − − − − 25. The conjugate of √(x + 1) + 2 is √(x − 1) − 2 . Answer − − − − − False. The conjugate is √x + 1 −2 . Exercise 26 − − − − − − 26. (√(x + 1) ) 2 2 = (x + 1 ) Answer − − − − − False. (√x + 1 ) Pamini Thangarajah 2 = x +1 10/21/2020 4 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 Exercise 27 27. x − 1 is the same as 1 − x Answer False. x − 1 is the same as −(1 − x) . Exercise 28 28. (x − 1)(x − 3) = x −1 (3 − x) Answer (x − 1)(x − 3) False. = −(x − 1), x ≠ 3 3 −x Exercise 29 29. −x − 1 = −(x − 1) Answer False. −x − 1 = −(x + 1) Exercise 30 30. 4x + 4 4 +4 = 8 = x +8 9 9 Answer 4(x + 1) 4x + 4 False. = x +8 x +8 , which cannot be simplified any further. Exercise 31 31. 3h = 2h h Answer 3h False. = 3, h ≠ 0 h Exercise 32 32. −(x 2 2 − 5x + 1) = −x − 5x + 1 Answer False. −(x 2 2 − 5x + 1) = −x + 5x − 1 Exercise 33 33. 1 3 + 3x 4 = 7x Pamini Thangarajah 7x 10/21/2020 5 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 Answer False. 1 3 + 3x 16 = 7x 21x Exercise 34 34. If f (x) = x 2 , then f (2 + h) = 4 + h Answer False. f (2 + h) = (2 + h) 2 2 = 4 + 4h + h Exercise 35 35. The slope of the straight line which is perpendicular to the straight line 2x − 3y + 11 = 0 is 3/2. Answer False. The slope would be −3 2 . Exercise 36 36. Given that A = (1, 5) and B = (7, 3), the midpoint of the line segment AB is the point M is given by 1 −7 M =( 5 −3 , ) 2 (1) 2 Answer False. The midpoint would be M 1 +7 =( 5 +3 , 2 ) . 2 Exercise 37 − − − − − − − − 2 + 1) = 2x + 1 37. √(4x Answer − −−−− − 2 +1 False. √4x cannot be simplified any further. Exercise 38 38. sin(cos(x)) = sin(x) cos(x) Answer False. sin(cos x) cannot be simplified any further. Exercise 39 39. 2 2 x(cos(x) + sin(x)) = cos(x ) + sin(x ) 2 = x 2 cos +x sin Answer Pamini Thangarajah 10/21/2020 6 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 False. x(cos x + sin x) = x cos x + x sin x$ Exercise 40 40. 1 − cos(8x) = 2 sin 2 (8x) Answer False. 1 − cos(8x) cannot be simplified any further. Exercise 41 41. tan(x 2 2 ) = tan x Answer False. tan(x ) cannot be simplified any further. Furthermore, the same as multiplication. 2 2 tan(x ) is the composition of tan(x) with x . It is not 2 Exercise 42 −− − − 42. √√− a =a Answer −− − − False. √√− a =a 1/4 Exercise 43 − − 43. √8 3 2 2 3 – = (√8) Answer True. Exercise 44 44. − − √27 – = √9 = 3 9 Answer False. − − √27 – 3 √3 = 9 – √3 = 9 3 Exercise 45 45. a 2 − 1 = −a Answer False. This is an equation that can be solved for a but it is not true in general. Exercise 46 46. a −3/4 1 = 4/3 a Pamini Thangarajah 10/21/2020 7 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 Answer False. a −3/4 1 = 3/4 a Exercise 47 47. log 2 x = 2 log x Answer False. log 2 2 2 (x) = (log2 x ) Exercise 48 48. cos 3 3 x = cos x Answer False. cos 3 3 (x) = (cos x ) Exercise 49 49. Recall that log 1 = 0. Then, log(−1) = − log 1 = 0. Answer False. log a (b) where b is a negative number is always undefined. Therefore, log(−1) is undefined. Exercise 50 50. (x − 1) 2 2 − (y + 1 ) =4 represents a circle centered at (1, −1) and having radius 2. Answer False. The equation represents a hyperbola centered at (1, −1). Exercise 51 51. y 2 − 3x + 1 = 0 is an equation of a straight line with slope 3. Answer False. y − 3x + 1 = 0 is an equation of a straight line with slope equal to 3. Exercise 52 52. In the triangle given sin α = a c Pamini Thangarajah 10/21/2020 8 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 Answer False. sin α is defined for right angled triangles. Exercise 53 53. The area of the triangle is 1/2ab. Answer False. The area of a triangle is 1 , where b is the base and h is the height of the triangle. bh 2 Exercise 54 54. Two triangles are said to be similar if and only if they have the same area. Answer False. Two triangles are said to be similar if and only if their angles are equal. Exercise 55 55. Two triangles are said to be similar if and only if they have sides of the same length. Answer False. See question above. Exercise 56 56. Two triangles are said to be similar if and only if their angles are equal. Pamini Thangarajah 10/21/2020 9 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 Answer True. Exercise 57 57. An isoceles triangle is a triangle in which all the sides have the same length. Answer False. An isosceles triangle is a triangle in which two sides have the same length. Exercise 58 58. An equilateral triangle must also be right angled. Answer False. The angles of equilateral triangle are all 60 . ∘ Exercise 59 59. The centre of the circle (x − 1) 2 2 + (y + 3 ) =5 is c = (−1, 3). Answer False. The centre of the circle is (1, −3). Exercise 60 60. To solve the inequality, x −1 > 0, (2) x we proceed as follows: Multiply both sides by x to get x − 1 > 0 ⇒ x > 1 . Answer False. To solve use a sign chart. The process described does not work because we don't know if negative. If x is negative, then multiplication by x causes a change in the inequality. x is positive or Exercise 61 61. sin π = 180 ∘ Answer False. sin(π) = 0 Exercise 62 62. If x = 1 then sin x = sin 1 = 0.17452406. Answer False. sin(1) = 0.841470985 Pamini Thangarajah 10/21/2020 10 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 Exercise 63 63. If sin x = 1 2 then sin x = 30 . ∘ Answer True. Exercise 64 64. cos(−x) = cos x Answer False. cos(−x) = cos(x) Exercise 65 65. On the number line shown below, the collection of real numbers represented is [−∞, 2). Answer False. The collection is (−∞, 2) . Exercise 66 66. On the number line shown above, the collection of real numbers represented is (2, −∞). Answer False. See question above. Exercise 67 67. The interval (−5, 4] can be represented on the real number line as shown below. Answer False. The interval is [−5, 4). Exercise 68 − − 68. √x 4 3 4/3 =x Answer − − False. √x 4 3 3/4 =x Exercise 69 Pamini Thangarajah 10/21/2020 11 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 − − 69. √64 = ±8 Answer − − False. √64 = 8 Exercise 70 − − − 70. √−1 = −1 Answer − − − False. √−1 = i Exercise 71 − − 71. √x 3 2 2/3 =x Answer True. Exercise 72 72. The pair of straight lines 2x + y − 1 = 0 and 2y + x − 1 = 0 are parallel. Answer False. The lines are neither parallel nor perpendicular. Exercise 73 73. The point of intersection of the parallel lines 3x + 2y − 1 = 0, and 6x + 4y + 17 = 0 is (−1, 4). Answer False. There is no point of intersection of parallel lines. Exercise 74 74. The graphs of y = −2 and x = 5 are as shown below. Pamini Thangarajah 10/21/2020 12 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 Answer False. The graphs of x = −2 and y = 5 are shown. Exercise 75 75. The straight line 2x + 3y = 12 has an x−intercept of 6 and y−intercept of 4. Therefore, the line must contain the point (6, 4). Answer False. The intercepts are correct but the line would contain the points (6, 0) and (0, 4). Exercise 76 3 76. If sin x = , 5 then sec x = 5 . 3 Answer False. If sin(x) = 3/5, then csc(x) = 5/3 . Exercise 77 77. |x 2 2 | = −x if x < 0. Answer False. |x 2 2 | =x for all x . Exercise 78 78. | − x| = x if x < 0. Answer False. | − x| = x if x > 0 . Exercise 79 79. | − 3x| = −3x if x < 0. Answer Pamini Thangarajah 10/21/2020 13 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 True. Exercise 80 80. x 2 2 y = xy . Answer False. There is no way of fixing this. Exercise 81 −− − 81. √−x is not defined for any real number values of x. 3 Answer −− − False. √−x is defined for all real numbers. 3 Exercise 82 82. If x 3 = 2, then x = 2 3 = 8. Answer False. If x 3 =2 – , then x = √2 . 3 Exercise 83 − – 83. If √x = 8, then x = √8 = 2. 3 3 Answer False. If √− x = 8 , then x = 8 3 3 = 512 . Exercise 84 84. The equation √− x = −1 has no real number solutions. 3 Answer False. The solution is x = −1 . Exercise 85 85. The graph of y = cos x is the one shown below: Pamini Thangarajah 10/21/2020 14 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 Answer False. The graph of y = sin(x) is shown. Exercise 86 86. The inequality −1 ≤ x ≥ 2 is valid. Answer True. It is redundant as it is equivalent to x ≥ −1 . Exercise 87 87. The inequality −1 ≥ x ≥ 2 is valid. Answer True. Exercise 88 88. The radius of the circle 4x 2 + 4y 2 = 81 is equal to 9 units. Answer False. The radius is 9/2. Exercise 89 89. Every pair of straight lines is either parallel or perpendicular. Answer False. For example, y = x and y = 2x are neither parallel nor perpendicular. Exercise 90 90. The equation of the x−axis is x = 0. Answer False. The equation of the x− axis is y = 0 . Pamini Thangarajah 10/21/2020 15 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 Exercise 91 91. The equation of the y−axis is x = 0. Answer True. Exercise 92 92. x 2 + 1 = (x + 1)(x + 1). Answer False. x 2 +1 is not factorable. Exercise 93 93. x 3 2 + 1 = (x + 1)(x + 1). Answer False. x 3 2 + 1 = (x + 1)(x − x + 1) Exercise 94 94. x 4 − 1 = (x − 1)(x − 1)(x − 1)(x − 1). Answer False. x 4 2 − 1 = (x − 1)(x + 1)(x + 1) Exercise 95 95. The slope of the straight line 3y = 5x − 2 is 5. Answer False. The slope is 5/3. Exercise 96 96. The distance between the points (0, 0) and (5, 5) is 5 units. Answer – False. The distance is 5√2 units. Exercise 97 97. The distance between the points (−5, 0) and (5, 5) is 0 units. Answer False. The distance is 10 units. Exercise 98 Pamini Thangarajah 10/21/2020 16 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 98. The point-slope formula for the equation of a straight line may be expressed as slope of the straight line and the line contains the point (x , y ). 1 where x − x1 = m(y − y1 ) m is the 1 Answer False. The equation is y − y 1 − m(x − x1 ) . Exercise 99 99. x 2 − 10x − 24 = (x − 4)(x − 6) Answer False. x 2 − 10x + 24 = (x − 4)(x − 6) Exercise 100 100. x 2 − 10x + 24 = (x − 12)(x + 2) Answer False. x 2 − 10x − 24 = (x − 12)(x + 2) Exercise 101 101. If ax 2 then + bx + c = 0 a. x = b. x = − −−−−−− − 2 ±√(b − 4ac) 2a − −−−−−− − −b ± √(4ac − b2 ) 2a Answer Both are false. x = − − − − − − − 2 −b ± √b − 4ac 2a Exercise 102 102. Question: Simplify x +1 x +2 = x −1 x −4 Answer: by cross multiplying we get, (x + 1)(x − 4) = (x + 2)(x − 1) 2 ⇒ x 2 − 3x − 4 = x +x −2 ⇒ −4x = 2 −1 ⇒ x = 2 Answer False. We are asked to simplify an expression not an equation. Therefore are solution should be a fraction and not x = something. The answer is obtained by making the expression into a single expression: (x + 1)(x + 2) . (x − 1)(x − 4) Exercise 103 Pamini Thangarajah 10/21/2020 17 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 103. If x +1 =0 x −1 then (x − 1) = 0(x 2 2 + 3) ⇒ x − 1 = x +3 Answer False. If we have x − 1 = 0 ⋅ (x 2 + 3) , then x − 1 = 0 . Exercise 104 104. If log x =5 2 then one of the following is true: a. x = 5 2 b. x =5 c. 5 =2 2 x d. x = 2 5 e. 2 x =5 f. None of the above. Answer d. is true. Exercise 105 2 105. If f (x) = x −4 x −2 then f (2) = 4 −4 0 = 2 −2 = 1. 0 Answer False. f (2) is undefined because of division by zero. Exercise 106 106. 2 5 + x 2y + 5x = y x +y Answer True. Exercise 107 107. To sketch a straight line, you need at least three points which lie on the straight line. Answer False. To sketch a straight line you need two points. Exercise 108 108. A circle is completely determined if its centre and radius are known. Answer True. Pamini Thangarajah 10/21/2020 18 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 Exercise 109 − 109. √−− −x is undefined because of the negative sign. Answer −− − False. √−x is defined if x < 0 . Exercise 110 110. log(xy) = (log x)(log y) Answer False. log(xy) = log(x) + log(y) Exercise 111 111. 3 0 =0 Answer False. 3 0 =1 Pamini Thangarajah 10/21/2020 19 CC-BY-NC-SA https://math.libretexts.org/@go/page/13808 CHAPTER OVERVIEW 1: LIMIT AND CONTINUITY OF FUNCTIONS 1.0: LIBRARY OF FUNCTIONS 1.0E: EXERCISES 1.1: INTRODUCTION TO CONCEPT OF A LIMIT 1.1E: EXERCISES 1.2: ONE SIDED LIMITS AND VERTICAL ASYMPTOTES 1.2E: EXERCISES 1.3: LIMIT CALCULATIONS FOR ALGEBRAIC EXPRESSIONS 1.3E: EXERCISES 1.4: LIMITS AT INFINITY AND HORIZONTAL ASYMPTOTES : END BEHAVIOR 1.4E: EXERCISES 1.5: FORMAL DEFINITION OF A LIMIT (OPTIONAL) 1.5E: EXERCISES 1.6: CONTINUITY AND THE INTERMEDIATE VALUE THEOREM 1.6E: EXCERCISES 1.7: LIMIT OF TRIGONOMETRIC FUNCTIONS 1.7E: EXERCISES 1.8: LIMITS AND CONTINUITY OF INVERSE TRIGONOMETRIC FUNCTIONS 1.8E: EXERCISES 1.9: LIMIT OF EXPONENTIAL FUNCTIONS AND LOGARITHMIC FUNCTIONS 1.9E: EXERCISES 1E: REVIEW EXERCISES 1 10/21/2020 1.0: Library of functions Thinking out Loud What is the meaning of a real-valued function? In this section, we provide a formal definition of a function and examine several ways in which functions are represented— namely, through tables, formulas, and graphs. We study formal notation and terms related to functions. We also define composition of functions and symmetry properties. Most of this material will be a review for you, but it serves as a handy reference to remind you of some of the algebraic techniques useful for working with functions. Functions Given two sets A and B a set with elements that are ordered pairs (x, y) where x is an element of A and y is an element of B, is a relation from A to B . A relation from A to B defines a relationship between those two sets. A function is a special type of relation in which each element of the first set is related to exactly one element of the second set. The element of the first set is called the input; the element of the second set is called the output. Functions are used all the time in mathematics to describe relationships between two sets. For any function, when we know the input, the output is determined, so we say that the output is a function of the input. For example, the area of a square is determined by its side length, so we say that the area (the output) is a function of its side length (the input). The velocity of a ball thrown in the air can be described as a function of the amount of time the ball is in the air. The cost of mailing a package is a function of the weight of the package. Since functions have so many uses, it is important to have precise definitions and terminology to study them. Figure 1.0.1 : A function can be visualized as an input/output device Definition: Functions A function f consists of a set of inputs, a set of outputs, and a rule for assigning each input to exactly one output. The set of inputs is called the domain of the function. The set of outputs is called the range of the function. Figure 1.0.2 : A function maps every element in the domain to exactly one element in the range. Although each input can be sent to only one output, two different inputs can be sent to the same output. For example, consider the function f , where the domain is the set of all real numbers and the rule is to square the input. Then, the input x = 3 is assigned to the output 3 = 9 . 2 Since every nonnegative real number has a real-value square root, every nonnegative number is an element of the range of this function. Since there is no real number with a square that is negative, the negative real numbers are not elements of the range. We conclude that the range is the set of nonnegative real numbers. For a general function f with domain D, we often use x to denote the input and y to denote the output associated with x. When doing so, we refer to x as the independent variable and \(y as the dependent variable, because it depends on x. Using function notation, we write y = f (x), and we read this equation as “y equals f of x. ” For the squaring function described earlier, we write f (x) = x . 2 10/21/2020 1.0.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 The concept of a function can be visualized using Figures 1.0.1 - 1.0.3. Figure 1.0.3 : In this case, a graph of a function f has a domain of 1, 2, 3 and a range of 1, 2. The independent variable is x and the dependent variable is y . We can also visualize a function by plotting points (x, y) in the coordinate plane where y = f (x). The graph of a function is the set of all these points. For example, consider the function f , where the domain is the set D = 1, 2, 3 and the rule is f (x) = 3 − x . In Figure 1.0.4, we plot a graph of this function. Figure 1.0.4 : Here we see a graph of the function points (x, f (x)) for all x in the domain. f with domain 1, 2, 3 and rule f (x) = 3 − x . The graph consists of the Every function has a domain. However, sometimes a function is described by an equation, as in f (x) = x , with no specific domain given. In this case, the domain is taken to be the set of all real numbers x for which f (x) is a real number. For example, since any real number can be squared, if no other domain is specified, we consider the domain of f (x) = x to be the set of all real numbers. On the other hand, the square root function f (x) = √− x only gives a real output if x is nonnegative. Therefore, the domain of the function f (x) = √− x is the set of nonnegative real numbers, sometimes called the natural domain. 2 2 For the functions f (x) = x and f (x) = √x , the domains are sets with an infinite number of elements. Clearly we cannot list all these elements. When describing a set with an infinite number of elements, it is often helpful to use set-builder or interval notation. When using set-builder notation to describe a subset of all real numbers, denoted R , we write 2 {x|x has some property}. (1.0.1) We read this as the set of real numbers x such that x has some property. For example, if we were interested in the set of real numbers that are greater than one but less than five, we could denote this set using set-builder notation by writing {x|1 < x < 5}. (1.0.2) A set such as this, which contains all numbers greater than a and less than b, can also be denoted using the interval notation (a, b) . Therefore, (1, 5) = {x|1 < x < 5}. (1.0.3) The numbers 1 and 5 are called the endpoints of this set. If we want to consider the set that includes the endpoints, we would denote this set by writing [1, 5] = {x|1 < x < 5}. 10/21/2020 1.0.2 CC-BY-NC-SA (1.0.4) https://math.libretexts.org/@go/page/10145 We can use similar notation if we want to include one of the endpoints, but not the other. To denote the set of nonnegative real numbers, we would use the set-builder notation {x|0 ≤ x}. (1.0.5) The smallest number in this set is zero, but this set does not have a largest number. Using interval notation, we would use the symbol ∞, which refers to positive infinity, and we would write the set as [0, ∞) = {x|0 ≤ x}. (1.0.6) It is important to note that ∞ is not a real number. It is used symbolically here to indicate that this set includes all real numbers greater than or equal to zero. Similarly, if we wanted to describe the set of all nonpositive numbers, we could write (−∞, 0] = {x|x ≤ 0}. (1.0.7) Here, the notation −∞ refers to negative infinity, and it indicates that we are including all numbers less than or equal to zero, no matter how small. The set (−∞, ∞) = {x|x is any real number} (1.0.8) refers to the set of all real numbers.Some functions are defined using different equations for different parts of their domain. These types of functions are known as piecewise-defined functions. For example, suppose we want to define a function f with a domain that is the set of all real numbers such that f (x) = 3x + 1 for x ≥ and f (x) = x forx < 2 .We denote this function by writing 2 f (x) = { 3x + 1 x ≥2 2 x x <2 (1.0.9) When evaluating this function for an input x,the equation to use depends on whether x ≥ 2 or x < 2 . For example, since 5 > 2 , we use the fact that f (x) = 3x + 1 for x ≥ 2 and see that f (5) = 3(5) + 1 = 16 . On the other hand, for x = −1 , we use the fact that f (x) = x for x < 2 and see that f (−1) = 1 . 2 Example 1.0.1 : Evaluating Functions For the function f (x) = 3x 2 + 2x − 1 , evaluate: a. f (−2) – b. f (√2) c. f (a + h) Solution Substitute the given value for x in the formula for f (x). a. f (−2) = 3(−2 2 + 2(−2) − 1 = 12 − 4 − 1 = 7 – – – – b. f (√2) = 3(√2) + 2√− x − 1 = 6 + 2 √2 − 1 = 5 + 2 √2 c. f (a + h) = 3(a + h) + 2(a + h) − 1 = 3(a + 2ah + h ) + 2a + 2h − 1 ) 2 2 2 = 3a 2 + 6ah + 3 h 2 2 + 2a + 2h − 1 Exercise 1.0.1 For f (x) = x 2 − 3x + 5 , evaluate f (1) and f (a + h) . Hint Substitute 1 and a + h for x in the formula for f (x). Answer f (1) = 3 and f (a + h) = a 2 2 + 2ah + h − 3a − 3h + 5 Example 1.0.2 : Finding Domain and Range 10/21/2020 1.0.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 For each of the following functions, determine the i. domain and ii. range. a. f (x) = (x − 4) + 5 −−−− − b. f (x) = √3x + 2 − 1 c. f (x) = 3x − 2 2 Solution 1. Consider f (x) = (x − 4) 2 + 5. 1.Since f (x) = (x − 4) 2 +5 is a real number for any real number x, the domain of f is the interval (−∞, ∞). 2. Since (x − 4) ≥ 0 , we know f (x) = (x − 4) + 5 ≥ 5 . Therefore, the range must be a subset of {y|y ≥ 5 }. To show that every element in this set is in the range, we need to show that for a given y in that set, there is a real number x such that f (x) = (x − 4) + 5 = y . Solving this equation for x, we see that we need x such that 2 2 2 2 (x − 4 ) = y − 5. This equation is satisfied as long as there exists a real number x such that − −− − x − 4 = ±√y − 5 Since y ≥ 5 , the square root is well-defined. We conclude that for range is {y|y ≥ 5 }. − −− − x = 4 ± √y − 5 f (x) = y , , and therefore the −−−− − 2. Consider f (x) = √3x + 2 − 1 . 1.To find the domain of f,we need the expression 3x + 2 ≥ 0 . Solving this inequality, we conclude that the domain is {x|x ≥ −2/3 }. −−−− − −−−− − 2.To find the range of f,we note that since √3x + 2 ≥ 0 ,f (x) = √3x + 2 − 1 ≥ −1 . Therefore, the range of f must be a subset of the set {y|y ≥ −1}. To show that every element in this set is in the range of f , we need to show that for all y in this set, there exists a real number x in the domain such that f (x) = y . Let y ≥ −1 . Then, f (x) = y if and only if −−−− − √3x + 2 − 1 = y. Solving this equation for x, we see that x must solve the equation −−−− − √3x + 2 = y + 1. Since y ≥ −1 , such an x could exist. Squaring both sides of this equation, we have 3x + 2 = (y + 1) 2 . Therefore, we need 2 3x = (y + 1 ) − 2, which implies x = 1 3 2 (y + 1 ) − 2 3 . We just need to verify that x is in the domain of f . Since the domain of f consists of all real numbers greater than or equal to \frac{−2}{3}\), and 1 3 2 (y + 1 ) − 2 3 ≥− 2 3 , there does exist an x in the domain of f . We conclude that the range of f is {y|y ≥ −1}. 3.Consider f (x) = 3/(x − 2). 1.Since 3/(x − 2) is defined when the denominator is nonzero, the domain is {x|x ≠ 2}. 2.To find the range of f , we need to find the values of y such that there exists a real number x in the domain with the property that 3 x − 2 = y. Solving this equation for x, we find that 10/21/2020 1.0.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 x = Therefore, as long as y|y ≠ 0 }. y ≠0 , there exists a real number 3 y x + 2. in the domain such that f (x) = y . Thus, the range is { Exercise 1.0.2 −−−− − Find the domain and range for f (x) = √4 − 2x + 5. Hint Use 4 − 2x ≥ 0 . Answer Domain = {x ∣ x ≤ 2 } and range = {y ∣ y ≥ 5 } Representing Functions Typically, a function is represented using one or more of the following tools: A table A graph A formula We can identify a function in each form, but we can also use them together. For instance, we can plot on a graph the values from a table or create a table from a formula. Tables Functions described using a table of values arise frequently in real-world applications. Consider the following simple example. We can describe temperature on a given day as a function of time of day. Suppose we record the temperature every hour for a 24-hour period starting at midnight. We let our input variable x be the time after midnight, measured in hours, and the output variable y be the temperature x hours after midnight, measured in degrees Fahrenheit. We record our data in Table 1.0.1. Table 1.0.1 : Temperature as a Function of Time of Day Hour After Midnight Temperature(°F) Hour After Midnight Temperature(°F) 0 58 12 84 1 54 13 85 2 53 14 85 3 52 15 83 4 52 16 82 5 55 17 80 6 60 18 77 7 64 19 74 8 72 20 69 9 75 21 65 10 78 22 60 11 80 23 58 We can see from the table that temperature is a function of time, and the temperature decreases, then increases, and then decreases again. However, we cannot get a clear picture of the behavior of the function without graphing it. Graphs 10/21/2020 1.0.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Given a function f described by a table, we can provide a visual picture of the function in the form of a graph. Graphing the temperatures listed in Table 1.0.1 can give us a better idea of their fluctuation throughout the day. Figure shows the plot of the temperature function. Figure 1.0.5 : The graph of the data from Table shows temperature as a function of time. From the points plotted on the graph in Figure 1.0.5, we can visualize the general shape of the graph. It is often useful to connect the dots in the graph, which represent the data from the table. In this example, although we cannot make any definitive conclusion regarding what the temperature was at any time for which the temperature was not recorded, given the number of data points collected and the pattern in these points, it is reasonable to suspect that the temperatures at other times followed a similar pattern, as we can see in Figure 1.0.6 . Figure 1.0.6 : Connecting the dots in Figure shows the general pattern of the data. Algebraic Formulas Sometimes we are not given the values of a function in table form, rather we are given the values in an explicit formula. Formulas arise in many applications. For example, the area of a circle of radius r is given by the formula A(r) = πr . When an object is thrown upward from the ground with an initial velocity v ft/s, its height above the ground from the time it is thrown until it hits the ground is given by the formula s(t) = −16t + v t . When P dollars are invested in an account at an annual interest rate r compounded continuously, the amount of money after t years is given by the formula A(t) = P e . Algebraic formulas are important tools to calculate function values. Often we also represent these functions visually in graph form. 2 0 2 0 rt Given an algebraic formula for a function f , the graph of f is the set of points (x, f (x)), where x is in the domain of f and f (x) is in the range. To graph a function given by a formula, it is helpful to begin by using the formula to create a table of 10/21/2020 1.0.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 inputs and outputs. If the domain of f consists of an infinite number of values, we cannot list all of them, but because listing some of the inputs and outputs can be very useful, it is often a good way to begin. When creating a table of inputs and outputs, we typically check to determine whether zero is an output. Those values of x where f (x) = 0 are called the zeros of a function. For example, the zeros of f (x) = x − 4 are x = ±2 . The zeros determine where the graph of f intersects the x-axis, which gives us more information about the shape of the graph of the function. The graph of a function may never intersect the x-axis, or it may intersect multiple (or even infinitely many) times. 2 Another point of interest is the y -intercept, if it exists. The y -intercept is given by (0, f (0)). Since a function has exactly one output for each input, the graph of a function can have, at most, one y -intercept. If x=0 is in the domain of a function f , then f has exactly one y -intercept. If x = 0 is not in the domain of f , then f has no y -intercept. Similarly, for any real number c , if c is in the domain of f , there is exactly one output f (c), and the line x = c intersects the graph of f exactly once. On the other hand, if c is not in the domain of f ,f (c) is not defined and the line x = c does not intersect the graph of f . This property is summarized in the vertical line test. Vertical Line Test Given a function f , every vertical line that may be drawn intersects the graph of f no more than once. If any vertical line intersects a set of points more than once, the set of points does not represent a function. We can use this test to determine whether a set of plotted points represents the graph of a function (Figure 1.0.7). Figure 1.0.7 : (a) The set of plotted points represents the graph of a function because every vertical line intersects the set of points, at most, once. (b) The set of plotted points does not represent the graph of a function because some vertical lines intersect the set of points more than once. Example 1.0.3 : Finding Zeros and y-Intercepts of a Function Consider the function f (x) = −4x + 2. a. Find all zeros of f . b. Find the y -intercept (if any). c. Sketch a graph of f . Solution 1.To find the zeros, solve f (x) = −4x + 2 = 0 . We discover that f has one zero at x = 1/2. 2. The y-intercept is given by (0, f (0)) = (0, 2). 3. Given that f is a linear function of the form sketch the graph of f (Figure 1.0.8). 10/21/2020 f (x) = mx + b that passes through the points 1.0.7 CC-BY-NC-SA (1/2, 0) and , we can (0, 2) https://math.libretexts.org/@go/page/10145 Figure 1.0.8 : The function f (x) = −4x + 2 is a line with x -intercept (1/2, 0) and y -intercept (0, 2). Example 1.0.4 : Using Zeros and y-Intercepts to Sketch a Graph − − − − − Consider the function f (x) = √x + 3 +1 . a. Find all zeros of f . b. Find the y -intercept (if any). c. Sketch a graph of f . Solution − − − − − 1.To find the zeros, solve √x + 3 + 1 = 0 . This equation implies equation has no solutions, and therefore f has no zeros. − − − − − √x + 3 = −1 . Since − − − − − √x + 3 ≥ 0 for all x , this – 2.The y -intercept is given by (0, f (0)) = (0, √3 + 1) . 3.To graph this function, we make a table of values. Since we need choose values that make the square-root function easy to evaluate. x +3 ≥ 0 , we need to choose values of x -3 -2 1 f (x) 1 2 3 x ≥ −3 . We Making use of the table and knowing that, since the function is a square root, the graph of f should be similar to the graph of y = √− x , we sketch the graph (Figure 1.0.9). − − − − − Figure 1.0.9 : The graph of f (x) = √x + 3 +1 has a y -intercept but no x -intercepts. Exercise 1.0.4 Find the zeros of f (x) = x 3 2 − 5x + 6x. Hint Factor the polynomial. Answer 10/21/2020 1.0.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 x = 0, 2, 3 Example 1.0.5 : Finding the Height of a Free-Falling Object If a ball is dropped from a height of 100 ft, its height s at time t is given by the function s(t) = −16t + 100 , where s is measured in feet and t is measured in seconds. The domain is restricted to the interval [0, c], where t = 0 is the time when the ball is dropped and t = c is the time when the ball hits the ground. 2 a. Create a table showing the height s(t) when t = 0, 0.5, 1, 1.5, 2,and 2.5. Using the data from the table, determine the domain for this function. That is, find the time c when the ball hits the ground. b. Sketch a graph of s . Solution t 0 0.5 1 1.5 2 2.5 s(t) 100 96 84 64 36 0 Since the ball hits the ground when t = 2.5 , the domain of this function is the interval [0, 2.5]. 2. Figure 1.0.8 , the values of f (x) are getting smaller as x is getting larger. A function with this property is said to be − − − − − decreasing. On the other hand, for the function f (x) = √x + 3 + 1 graphed in Figure 1.0.9 , the values of f (x) are getting larger as the values of x are getting larger. A function with this property is said to be increasing. It is important to note, however, that a function can be increasing on some interval or intervals and decreasing over a different interval or intervals. For example, using our temperature function plotted above, we can see that the function is decreasing on the interval (0, 4), increasing on the interval (4, 14), and then decreasing on the interval (14, 23). We make the idea of a function increasing or decreasing over a particular interval more precise in the next definition. Definition: Increasing and Decreasing on an Interval We say that a function f is increasing on the interval I if for all x 1, f (x1 ) ≤ f (x2 ) We say f is strictly increasing on the interval I if for all x 1, x2 ∈ I , when x 1 < x2 . x2 ∈ I , f (x1 ) < f (x2 ) when x 1 We say that a function f is decreasing on the interval I if for all x 1, f (x1 ) ≥ f (x2 ) if x < x2 . x2 ∈ I , 1 < x2 . We say that a function f is strictly decreasing on the interval I if for all x 1, 10/21/2020 x2 ∈ I 1.0.9 CC-BY-NC-SA , https://math.libretexts.org/@go/page/10145 f (x1 ) > f (x2 ) if x 1 < x2 . For example, the function f (x) = 3x is increasing on the interval (−∞, ∞) because 3x other hand, the function f (x) = −x is decreasing on the interval (−∞, ∞) because −x 1.0.10). 1 3 3 1 Figure 1.0.10 : (a) The function f (x) = 3x is increasing on the interval decreasing on the interval (−∞, ∞) . (−∞, ∞) < 3 x2 3 > −x 2 whenever x whenever x . (b) The function 1 < x2 1 < x2 3 f (x) = −x . On the (Figure is Combining Functions Now that we have reviewed the basic characteristics of functions, we can see what happens to these properties when we combine functions in different ways, using basic mathematical operations to create new functions. For example, if the cost for a company to manufacture x items is described by the function C (x) and the revenue created by the sale of x items is described by the function R(x), then the profit on the manufacture and sale of x items is defined as P (x) = R(x) − C (x) . Using the difference between two functions, we created a new function. Alternatively, we can create a new function by composing two functions. For example, given the functions g(x) = 3x + 1 , the composite function f ∘ g is defined such that 2 (f ∘ g)(x) = f (g(x)) = (g(x)) 2 = (3x + 1 ) . 2 f (x) = x and (1.0.10) The composite function g∘f is defined such that 2 (g ∘ f )(x) = g(f (x)) = 3f (x) + 1 = 3 x + 1. (1.0.11) Note that these two new functions are different from each other. Combining Functions with Mathematical Operators To combine functions using mathematical operators, we simply write the functions with the operator and simplify. Given two functions f and g , we can define four new functions: 10/21/2020 (f + g)(x) = f (x) + g(x) Sum (f − g)(x) = f (x) − g(x) Difference (f ⋅ g)(x) = f (x)g(x) Product 1.0.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 f ( g f(x) )(x) = Quotient forg(x) ≠ 0 g(x) Example 1.0.6 : Combining Functions Using Mathematical Operations Given the functions f (x) = 2x − 3 and g(x) = x 2 −1 , find each of the following functions and state its domain. a. (f + g)(x) b. (f − g)(x) c. (f ⋅ g)(x) d. ( f g )(x) Solution 1. (f + g)(x) = (2x − 3) + (x 2 2 − 1) = x + 2x − 4. The domain of this function is the interval (−∞, ∞). 2.(f − g)(x) = (2x − 3) − (x 2 2 − 1) = −x + 2x − 2. The domain of this function is the interval (−∞, ∞). 3. (f ⋅ g)(x) = (2x − 3)(x 2 3 − 1) = 2 x 2 − 3x − 2x + 3. The domain of this function is the interval (−∞, ∞). 4. ( f g )(x) = 2x−3 2 x −1 . The domain of this function is {x ∣∣ x ≠ ±1 }. Exercise 1.0.6 For f (x) = x 2 +3 and g(x) = 2x − 5 , find (f /g)(x) and state its domain. Hint The new function (f /g)(x)(f /g)(x) is a quotient of two functions. For what values of x is the denominator zero? Answer 2 f ( g )(x) = x +3 2x−5 . The domain is {x|x ≠ 5 2 }. Function Composition When we compose functions, we take a function of a function. For example, suppose the temperature T on a given day is described as a function of time t (measured in hours after midnight) as in Table. Suppose the cost C , to heat or cool a building for 1 hour, can be described as a function of the temperature T . Combining these two functions, we can describe the cost of heating or cooling a building as a function of time by evaluating C (T (t)). We have defined a new function, denoted C ∘ T , which is defined such that (C ∘ T )(t) = C (T (t)) for all t in the domain of T . This new function is called a composite function. We note that since cost is a function of temperature and temperature is a function of time, it makes sense to define this new function (C ∘ T )(t) . It does not make sense to consider (T ∘ C )(t) , because temperature is not a function of cost. Definition: Composite Functions Consider the function f with domain A and range B , and the function g with domain D and range E . If B is a subset of D, then the composite function (g ∘ f )(x) is the function with domain A such that (g ∘ f )(x) = g(f (x)) (1.0.12) A composite function g ∘ f can be viewed in two steps. First, the function f maps each input x in the domain of f to its output f (x) in the range of f . Second, since the range of f is a subset of the domain of g , the output f (x) is an element in the domain 10/21/2020 1.0.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 of g , and therefore it is mapped to an output g(f (x)) in the range of g . In Figure 1.0.11, we see a visual image of a composite function. Figure 1.0.11 : For the composite function g ∘ f , we have (g ∘ f )(1) = 4,(g ∘ f )(2) = 5, and (g ∘ f )(3) = 4. Example 1.0.7 : Compositions of Functions Defined by Formulas Consider the functions f (x) = x 2 +1 and g(x) = 1/x . a. Find (g ∘ f )(x) and state its domain and range. b. Evaluate (g ∘ f )(4) ,(g ∘ f )(−1/2). c. Find (f ∘ g)(x) and state its domain and range. d. Evaluate (f ∘ g)(4) , (f ∘ g)(−1/2). Solution 1. We can find the formula for (g ∘ f )(x) in two different ways. We could write 2 (g ∘ f )(x) = g(f (x)) = g(x 1 + 1) = 2 x +1 . Alternatively, we could write (g ∘ f )(x) = g(f (x)) = Since 1 f (x) = 1 2 x +1 . for all real numbers x, the domain of (g ∘ f )(x) is the set of all real numbers. Since , the range is, at most, the interval (0, 1]. To show that the range is this entire interval, we let y=1/(x2+1) and solve this equation for x to show that for all y in the interval (0, 1], there exists a real number x such that y = 1/(x + 1) . Solving this equation for x, we see that x + 1 = 1/y , which implies that 2 x +1 ≠ 0 0 < 1/(x2 + 1) ≤ 1 2 2 − − − − − x = ±√ 1 y −1 If y is in the interval (0, 1], the expression under the radical is nonnegative, and therefore there exists a real number x such that 1/(x + 1) = y . We conclude that the range of g∘f is the interval (0, 1]. 2 2. (g ∘ f )(4) = g(f (4)) = g(4 2 (g ∘ f )(− 1 2 ) = g(f (− 1 2 + 1) = g(17) = )) = g((− 1 2 2 ) + 1) = g( 1 17 5 4 ) = 4 5 3. We can find a formula for (f ∘ g)(x) in two ways. First, we could write (f ∘ g)(x) = f (g(x)) = f ( 1 x ) =( 1 x 2 ) + 1. Alternatively, we could write 2 (f ∘ g)(x) = f (g(x)) = (g(x)) +1 = ( 1 x 2 ) + 1. The domain of f ∘ g is the set of all real numbers x such that x ≠ 0 . To find the range of f , we need to find all values y for which there exists a real number x ≠ 0 such that ( 1 x 2 ) + 1 = y. Solving this equation for x, we see that we need x to satisfy ( 1 x 2 ) = y − 1, which simplifies to 10/21/2020 1.0.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 1 x − −− − = ±√y − 1 Finally, we obtain x =± 1 √y−1 . − −− − Since 1/√y − 1 is a real number if and only if y > 1 , the range of f is the set {y ∣∣ y ≥ 1 }. 4.(f ∘ g)(4) = f (g(4)) = f ( (f ∘ g)(− 1 2 ) = f (g(− 1 2 1 4 ) =( 1 4 2 ) +1 = 2 )) = f (−2) = (−2 ) In Example, we can see that functions matters. 17 16 +1 = 5 (f ∘ g)(x) ≠ (g ∘ f )(x) . This tells us, in general terms, that the order in which we compose Exercise 1.0.7 Let f (x) = 2 − 5x . Let g(x) = √− x . Find (f ∘ g)(x) . Solution − (f ∘ g)(x) = 2 − 5 √x . Example 1.0.8 : Composition of Functions Defined by Tables Consider the functions f and g described by x -3 -2 -1 0 1 2 3 4 f(x) 0 4 2 4 -2 0 -2 4 x -4 -2 0 2 4 g(x) 1 0 3 0 5 a. Evaluate (g ∘ f )(3) ,(g ∘ f )(0) . b. State the domain and range of (g ∘ f )(x) . c. Evaluate (f ∘ f )(3),(f ∘ f )(1). d. State the domain and range of (f ∘ f )(x). Solution: 1. (g ∘ f )(3) = g(f (3)) = g(−2) = 0 (g ∘ f )(0) = g(4) = 5 2.The domain of g ∘ f is the set {−3, −2, −1, 0, 1, 2, 3, 4}. Since the range of f is the set {−2, 0, 2, 4}, the range of g ∘ f is the set {0, 3, 5}. 3. (f ∘ f )(3) = f (f (3)) = f (−2) = 4 (f ∘ f )(1) = f (f (1)) = f (−2) = 4 4.The domain of f ∘ f is the set {−3, −2, −1, 0, 1, 2, 3, 4}. Since the range of f is the set {−2, 0, 2, 4}, the range of f ∘ f is the set {0, 4}. Example 1.0.9 : Application Involving a Composite Function A store is advertising a sale of 20% off all merchandise. Caroline has a coupon that entitles her to an additional 15% off any item, including sale merchandise. If Caroline decides to purchase an item with an original price of x dollars, how much will she end up paying if she applies her coupon to the sale price? Solve this problem by using a composite function. Solution 10/21/2020 1.0.13 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Since the sale price is 20% off the original price, if an item is x dollars, its sale price is given by f (x) = 0.80x. Since the coupon entitles an individual to 15% off the price of any item, if an item is y dollars, the price, after applying the coupon, is given by g(y)=0.85y. Therefore, if the price is originally x dollars, its sale price will be f (x) = 0.80x and then its final price after the coupon will be g(f (x)) = 0.85(0.80x) = 0.68x. Exercise 1.0.9 If items are on sale for 10% off their original price, and a customer has a coupon for an additional 30% off, what will be the final price for an item that is originally x dollars, after applying the coupon to the sale price? Hint The sale price of an item with an original price of x dollars is f (x) = 0.90x. The coupon price for an item that is y dollars is g(y) = 0.70y. Solution (g ∘ f )(x) = 0.63x Symmetry of Functions The graphs of certain functions have symmetry properties that help us understand the function and the shape of its graph. For example, consider the function f (x) = x − 2x − 3 shown in Figure 1.0.12a. If we take the part of the curve that lies to the right of the y -axis and flip it over the y -axis, it lays exactly on top of the curve to the left of the y -axis. In this case, we say the function has symmetry about the y-axis. On the other hand, consider the function f (x) = x − 4x shown in Figure 1.0.12b. If we take the graph and rotate it 180° about the origin, the new graph will look exactly the same. In this case, we say the function has symmetry about the origin. 4 2 3 Figure 1.0.12 : (a) A graph that is symmetric about the y -axis. (b) A graph that is symmetric about the origin. If we are given the graph of a function, it is easy to see whether the graph has one of these symmetry properties. But without a graph, how can we determine algebraically whether a function f has symmetry? Looking at Figure again, we see that since f is symmetric about the y -axis, if the point (x, y) is on the graph, the point (−x, y) is on the graph. In other words, f (−x) = f (x). If a function f has this property, we say f is an even function, which has symmetry about the y -axis. For example, f (x) = x is even because 2 2 f (−x) = (−x ) 2 =x = f (x). In contrast, looking at Figure again, if a function f is symmetric about the origin, then whenever the point (x, y) is on the graph, the point (−x, −y) is also on the graph. In other words, f (−x) = −f (x). If f has this property, we say f is an odd function, which has symmetry about the origin. For example, f (x) = x is odd because 3 3 f (−x) = (−x ) 10/21/2020 3 = −x = −f (x). 1.0.14 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Definition: Even and Odd Functions If f (x) = f (−x) for all x in the domain of f , then f is an even function. An even function is symmetric about the y axis. If f (−x) = −f (x) for all x in the domain of f , then f is an odd function. An odd function is symmetric about the origin. Example 1.0.10 : Even and Odd Functions Determine whether each of the following functions is even, odd, or neither. a. f (x) = −5x + 7x − 2 b. f (x) = 2x − 4x + 5 c. f (x) = 4 2 5 3x 2 x +1 Solution To determine whether a function is even or odd, we evaluate f (−x) and compare it to f (x) and −f (x). 1. f (−x) = −5(−x ) 4 2.f (−x) = 2(−x ) 5 5 −f (x) = −2 x 2 + 7(−x ) 4 − 2 = −5 x 2 + 7x − 2 = f (x) . Therefore, f is even. + 4x + 5 . Now, f (−x) ≠ f (x). Furthermore, , we see that f (−x) ≠ −f (x). Therefore, f is neither even nor odd. 5 − 4(−x) + 5 = −2 x + 4x − 5 3.f (−x) = 3(−x)/((−x)2 + 1)= −3x/(x 2 2 + 1) = −[3x/(x + 1)] = −f (x) noting that . Therefore, f is odd. Exercise 1.0.10 Determine whether f(x)=4x^3−5x is even, odd, or neither. Hint Compare f (−x) with f (x) and −f (x) . Answer f(x)is odd. One symmetric function that arises frequently is the absolute value function, written as defined as f (x) = { −x x <0 x x ≥0 . The absolute value function is |x| (1.0.13) Some students describe this function by stating that it “makes everything positive.” By the definition of the absolute value function, we see that if x < 0 , then |x| = −x > 0 , and if x > 0 , then |x| = x > 0 . However, for x = 0 ,|x| = 0. Therefore, it is more accurate to say that for all nonzero inputs, the output is positive, but if x = 0 , the output |x| = 0. We conclude that the range of the absolute value function is {y|y ≥ 0 }. In Figure 1.0.13, we see that the absolute value function is symmetric about the y -axis and is therefore an even function. 10/21/2020 1.0.15 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Figure 1.0.13 : The graph of f (x) = |x| is symmetric about the y -axis. Example 1.0.11 : Working with the Absolute Value Function Find the domain and range of the function f (x) = 2|x − 3| + 4 . Solution Since the absolute value function is defined for all real numbers, the domain of this function is (−∞, ∞). Since |x − 3| ≥ 0 for all x, the function f (x) = 2|x − 3| + 4 ≥ 4 . Therefore, the range is, at most, the set {y|y ≥ 4 }. To see that the range is, in fact, this whole set, we need to show that for y ≥ 4 there exists a real number x such that 2|x − 3| + 4 = y A real number x satisfies this equation as long as |x − 3| = 1 2 (y − 4) Since y ≥ 4 , we know y − 4 ≥ 0 , and thus the right-hand side of the equation is nonnegative, so it is possible that there is a solution. Furthermore, |x − 3| = { −(x − 3) if x < 3 x −3 if x ≥ 3 . Therefore, we see there are two solutions: x =± 1 2 (y − 4) + 3 . The range of this function is y|y ≥ 4 . Exercise 1.0.11 : Domain and Range For the function f (x) = |x + 2| − 4 , find the domain and range. Hint |x + 2| ≥ 0 for all real numbers x . Answer Domain = (−∞, ∞) , range = {y|y ≥ −4 }. Real Valued functions Definition Let D ⊂ R . Then f : D → R is said to be a real valued function from D to R if for each element x ∈ D, there exists a unique (one and only one) element y ∈ R , such that y = f (x). In this case D is called domain of f , x is called an independent variable, and y is called an dependent variable, 10/21/2020 1.0.16 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Note that we will follow the following convention: for a given real valued function f , the domain of f is denoted by $ D(f) $ and defined by D(f ) = the set of all x such that there exists a y ∈ R such that y = f (x). Library of functions We have studied the general characteristics of functions, so now let’s examine some specific classes of functions. We begin by reviewing the basic properties of linear and quadratic functions, and then generalize to include higher-degree polynomials. By combining root functions with polynomials, we can define general algebraic functions and distinguish them from the transcendental functions we examine later in this chapter. We finish the section with examples of piecewise-defined functions and take a look at how to sketch the graph of a function that has been shifted, stretched, or reflected from its initial form. The following examples of functions will be used in this course: Linear Functions and Slope The easiest type of function to consider is a linear function. Linear functions have the form f (x) = ax + b , where a and b are constants. In Figure 1.0.1, we see examples of linear functions when a is positive, negative, and zero. Note that if a > 0 , the graph of the line rises as x increases. In other words, f (x) = ax + b is increasing on (−∞, ∞). If a < 0 , the graph of the line falls as x increases. In this case, f (x) = ax + b is decreasing on (−∞, ∞). If a = 0 , the line is horizontal. Figure 1.0.1 : These linear functions are increasing or decreasing on (∞, ∞) and one function is a horizontal line. As suggested by Figure, the graph of any linear function is a line. One of the distinguishing features of a line is its slope. The slope is the change in y for each unit change in x. The slope measures both the steepness and the direction of a line. If the slope is positive, the line points upward when moving from left to right. If the slope is negative, the line points downward when moving from left to right. If the slope is zero, the line is horizontal. To calculate the slope of a line, we need to determine the ratio of the change in y versus the change in x. To do so, we choose any two points (x , y ) and (x , y ) on the line and 1 calculate y2 − y1 1 2 2 . In Figure 1.0.2, we see this ratio is independent of the points chosen. x2 − x1 10/21/2020 1.0.17 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Figure : For any linear function, the slope on the line. 1.0.2 ( x2 , y2 ) is independent of the choice of points ( y2 − y1 )/( x2 − x1 ) ( x1 , y1 ) and Definition: Linear Functions Consider line L passing through points (x , y y and x,respectively. The slope of the line is 1 1) and (x 2, m = y2 ) . Let Δy = y 2 y2 − y1 − y1 and Δx = x 2 − x1 denote the changes in Δy = x2 − x1 (1.0.14) Δx We now examine the relationship between slope and the formula for a linear function. Consider the linear function given by the formula f (x) = ax + b . As discussed earlier, we know the graph of a linear function is given by a line. We can use our definition of slope to calculate the slope of this line. As shown, we can determine the slope by calculating (y − y )/(x − x ) for any points (x , y ) and (x , y ) on the line. Evaluating the function f at x = 0 , we see that (0, b) is a point on this line. Evaluating this function at x = 1 , we see that (1, a + b) is also a point on this line. Therefore, the slope of this line is 2 1 2 1 1 1 2 2 (a + b) − b = a. (1.0.15) 1 −0 We have shown that the coefficient a is the slope of the line. We can conclude that the formula f (x) = ax + b describes a line with slope a . Furthermore, because this line intersects the y -axis at the point (0, b), we see that the y-intercept for this linear function is (0, b). We conclude that the formula f (x) = ax + b tells us the slope, a, and the y -intercept, (0, b), for this line. Since we often use the symbol m to denote the slope of a line, we can write f (x) = mx + b (1.0.16)  slope-intercept form to denote the slope-intercept form of a linear function. Sometimes it is convenient to express a linear function in different ways. For example, suppose the graph of a linear function passes through the point (x , y ) and the slope of the line is m. Since any other point (x, f (x)) on the graph of f must satisfy the equation 1 1 f (x) − y1 m = , (1.0.17) x − x1 this linear function can be expressed by writing 10/21/2020 1.0.18 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 f (x) − y1 = m(x − x1 ) . (1.0.18)  point-slope equation We call this equation the point-slope equation for that linear function. Since every nonvertical line is the graph of a linear function, the points on a nonvertical line can be described using the slopeintercept or point-slope equations. However, a vertical line does not represent the graph of a function and cannot be expressed in either of these forms. Instead, a vertical line is described by the equation x = k for some constant k . Since neither the slope-intercept form nor the point-slope form allows for vertical lines, we use the notation ax + by = c ,  (1.0.19) standard form where a, b are both not zero, to denote the standard form of a line. Definition: point-slope equation, point-slope equation and the standard form of a line Consider a line passing through the point (x 1, y1 ) with slope m. The equation y − y1 = m(x − x1 ) (1.0.20) is the point-slope equation for that line. Consider a line with slope m and y -intercept (0, b). The equation y = mx + b (1.0.21) ax + by = c, (1.0.22) is an equation for that line in point-slope equation. The standard form of a line is given by the equation where a and b are both not zero. This form is more general because it allows for a vertical line, x = k . Example 1.0.1 : Finding the Slope and Equations of Lines Consider the line passing through the points (11, −4) and (−4, 5), as shown in Figure. Figure 1.0.3 : Finding the equation of a linear function with a graph that is a line between two given points. 1. Find the slope of the line. 2. Find an equation for this linear function in point-slope form. 3. Find an equation for this linear function in slope-intercept form. Solution 1. The slope of the line is 10/21/2020 1.0.19 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 m = y2 − y1 5 − (−4) 9 = x2 − x1 =− −4 − 11 3 =− 15 2. To find an equation for the linear function in point-slope form, use the slope line. If we choose the point (11, −4), we get the equation . 5 m = −3/5 and choose any point on the 3 f (x) + 4 = − (x − 11). 5 3. To find an equation for the linear function in slope-intercept form, solve the equation in part b. for this, we get the equation 3 f (x) = − . When we do f (x) 13 x+ . 5 5 Exercise 1.0.1 Consider the line passing through points (−3, 2) and (1, 4). a. Find the slope of the line. b. Find an equation of that line in point-slope form. c. Find an equation of that line in slope-intercept form. Hint The slope m = Δy/Δx. Answer a . m = 1/2 Answer b The point-slope form is y − 4 = 1 (x − 1) 2 . Answer c The slope-intercept form is y = 1 7 x+ 2 2 . Example 1.0.2 : Jessica leaves her house at 5:50 a.m. and goes for a 9-mile run. She returns to her house at 7:08 a.m. Answer the following questions, assuming Jessica runs at a constant pace. a. Describe the distance D (in miles) Jessica runs as a linear function of her run time t (in minutes). b. Sketch a graph of D. c. Interpret the meaning of the slope. Solution: a. At time t = 0 , Jessica is at her house, so D(0) = 0 . At time t = 78 minutes, Jessica has finished running 9 mi, so D(78) = 9 . The slope of the linear function is 9 −0 3 m = = 78 − 0 . 26 The y -intercept is (0, 0), so the equation for this linear function is 3 D(t) = t. 26 b. To graph D, use the fact that the graph passes through the origin and has slope m = 3/26. 10/21/2020 1.0.20 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 c. The slope m = 3/26 ≈ 0.115 describes the distance (in miles) Jessica runs per minute, or her average velocity. Example 1.0.1 Define f : R → R by f (x) = x + 1 . Then D(f ) = R . Exercise 1.0.3 Define f : R → R by f (x) = x − 1 . Find the domain. Answer D(f ) = R IR Polynomials A linear function is a special type of a more general class of functions: polynomials. A polynomial function is any function that can be written in the form n f (x) = an x n−1 + an−1 x + … + a1 x + a0 (1.0.23) for some integer n ≥ 0 and constants a , a + n − 1, … , a , where a ≠ 0 . In the case when n = 0 , we allow for a = 0 ; if a = 0 , the function f (x) = 0 is called the zero function. The value n is called the degree of the polynomial; the constant an n 0 n 0 0 10/21/2020 1.0.21 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 is called the leading coefficient. A linear function of the form f (x) = mx + b is a polynomial of degree 1 if m ≠ 0 and degree 0 if m = 0 . A polynomial of degree 0 is also called a constant function. A polynomial function of degree 2 is called a quadratic function. In particular, a quadratic function has the form 2 f (x) = ax + bx + c, (1.0.24) where a ≠ 0 . A polynomial function of degree 3 is called a cubic function. Power Functions Some polynomial functions are power functions. A power function is any function of the form f (x) = ax , where a and b are any real numbers. The exponent in a power function can be any real number, but here we consider the case when the exponent is a positive integer. (We consider other cases later.) If the exponent is a positive integer, then f (x) = ax is a polynomial. If n is even, then f (x) = ax is an even function because f (−x) = a(−x ) = ax if n is even. If n is odd, then f (x) = ax is an odd function because f (−x) = a(−x ) = −ax if n is odd (Figure 1.0.3). b n n n n n n n Figure 1.0.4 : (a) For any even integer n ,f (x) = ax is an even function. (b) For any odd integer function. n , n n f (x) = ax is an odd Quadratic functions Example 1.0.2 : Define f 10/21/2020 : R → R by f (x) = x . Then D(f ) = R . 2 1.0.22 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Exercise 1.0.2 Define f : R → R by f (x) = −x . Find the domain. 2 Answer D(f ) = R Cubic functions Example 1.0.3 : Define f : R → R by f (x) = x . Then D(f ) = R . 3 . Exercise 1.0.3 Define f : R → R by f (x) = x . Find the Domain 3 Answer D(f ) = R 10/21/2020 1.0.23 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Algebraic Functions By allowing for quotients and fractional powers in polynomial functions, we create a larger class of functions. An algebraic function is one that involves addition, subtraction, multiplication, division, rational powers, and roots. Two types of algebraic functions are rational functions and root functions. Just as rational numbers are quotients of integers, rational functions are quotients of polynomials. In particular, a rational function is any function of the form f (x) = p(x)/q(x),where p(x) and q(x) are polynomials. For example, 3x − 1 f (x) = 5x + 2 4 and g(x) = 2 x +1 are rational functions. A root function is a power function of the form f (x) = x , where n is a positive integer greater than one. For example, f(x)=x1/2=x√ is the square-root function and g(x) = x = √− x ) is the cube-root function. By allowing for − −−− − compositions of root functions and rational functions, we can create other algebraic functions. For example, f (x) = √4 − x is an algebraic function. 1/n 1/3 3 2 Example 1.0.5 : Finding Domain and Range for Algebraic Functions For each of the following functions, find the domain and range. a. 3x − 1 f (x) = 5x + 2 − −−− − b. f (x) = √4 − x 2 Solution 1.It is not possible to divide by zero, so the domain is the set of real numbers x such that x ≠ −2/5. To find the range, we need to find the values y for which there exists a real number x such that 3x − 1 y = 5x + 2 When we multiply both sides of this equation by 5x + 2 , we see that x must satisfy the equation 5xy + 2y = 3x − 1. From this equation, we can see that x must satisfy 2y + 1 = x(3 − 5y). If y=3/5, this equation has no solution. On the other hand, as long as y ≠ 3/5, 2y + 1 x = 3 − 5y satisfies this equation. We can conclude that the range of f is y|y ≠ 3/5. 2. To find the domain of f , we need 4 − x ≥ 0 . When we factor, we write 4 − x = (2 − x)(2 + x) ≥ 0 . This inequality holds if and only if both terms are positive or both terms are negative. For both terms to be positive, we need to find x such that 2 2 2 −x ≥ 0 These two inequalities reduce to 2 ≥ x and both terms to be negative, we need x ≥ −2 and 2 + x ≥ 0. . Therefore, the set 2 −x ≤ 0 x| − 2 ≤ x ≤ 2 and 2 + x ≥ 0. These two inequalities also reduce to 2 ≤ x and x ≥ −2 . There are no values of Thus, we can conclude the domain of this function is x| − 2 ≤ x ≤ 2. If −2 ≤ x ≤ 2 , then 0 ≤ 4 − x 2 ≤4 must be part of the domain. For −−−− − . Therefore, 0 ≤ √4 − x2 ≤2 x that satisfy both of these inequalities. , and the range of f is y|0 ≤ y ≤ 2. Exercise 1.0.3 10/21/2020 1.0.24 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Find the domain and range for the function f (x) = (5x + 2)/(2x − 1). Hint The denominator cannot be zero. Solve the equation y = (5x + 2)/(2x − 1) for x to find the range. Answer The domain is the set of real numbers x such that x ≠ 1/2. The range is the set {y|y ≠ 5/2}. The root functions f (x) = x have defining characteristics depending on whether n is odd or even. For all even integers n ≥ 2 , the domain of f (x) = x is the interval [0, ∞). For all odd integers n ≥ 1 , the domain of f (x) = x is the set of all real numbers. Since x = (−x ) for odd integers n ,f (x) = x is an odd function ifn is odd. See the graphs of root functions for different values of n in Figure. 1/n 1/n 1/n 1/n 1/n Figure 1.0.7 : (a) If n is even, the domain of and the function f (x) = [ ]x n 1/n n − f (x) = √x is [0, ∞) . (b) If n is odd, the domain of [ f (x) = ]x n is (−∞, ∞) is an odd function. Example 1.0.6 : Finding Domains for Algebraic Functions For each of the following functions, determine the domain of the function. a. 3 f (x) = 2 x b. f (x) = −1 2x + 5 2 3x + 4 −−−− − c. f (x) = √4 − 3x −−−− − d. f (x) = √2x − 1 3 Solution a. You cannot divide by zero, so the domain is the set of values x such that x − 1 ≠ 0 . Therefore, the domain is x|x ≠ ±1. b. You need to determine the values of x for which the denominator is zero. Since 3x + 4 ≥ 4 for all real numbers x, the denominator is never zero. Therefore, the domain is (−∞, ∞). c. Since the square root of a negative number is not a real number, the domain is the set of values x for which 4 − 3x ≥ 0 . Therefore, the domain is x|x ≤ 4/3. d. The cube root is defined for all real numbers, so the domain is the interval (−∞, ∞). 2 2 Exercise 1.0.4 Find the domain for each of the following functions: f (x) = (5 − 2x)/(x 2 + 2) −−−− − and g(x) = √5x − 1 . Hint 10/21/2020 1.0.25 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Determine the values of x when the expression in the denominator of expression inside the radical of g is nonnegative. f is nonzero, and find the values of x when the Answer The domain of f is (−∞, ∞) . The domain of g is x|x ≥ 1/5. Transcendental Functions Thus far, we have discussed algebraic functions. Some functions, however, cannot be described by basic algebraic operations. These functions are known as transcendental functions because they are said to “transcend,” or go beyond, algebra. The most common transcendental functions are trigonometric, exponential, and logarithmic functions. A trigonometric function relates the ratios of two sides of a right triangle. They are sinx, cosx, tanx, cotx, secx, andcscx. (We discuss trigonometric functions later in the chapter.) An exponential function is a function of the form f (x) = b , where the base b > 0, b ≠ 1 . A logarithmic function is a function of the form f (x) = log (x) for some constant b > 0, b ≠ 1, where log (x) = y if and only if b = x . (We also discuss exponential and logarithmic functions later in the chapter.) x b b y Example 1.0.7 : Classifying Algebraic and Transcendental Functions Classify each of the following functions, a. through c., as algebraic or transcendental. a. − −−− − √x3 + 1 f (x) = 4x + 2 b. f (x) = 2 c. f (x) = sin(2x) 2 x Solution a. Since this function involves basic algebraic operations only, it is an algebraic function. b. This function cannot be written as a formula that involves only basic algebraic operations, so it is transcendental. (Note that algebraic functions can only have powers that are rational numbers.) c. As in part b, this function cannot be written using a formula involving basic algebraic operations only; therefore, this function is transcendental. Exercise 1.0.5 : Is f (x) = x/2 an algebraic or a transcendental function? Answer Algebraic Piecewise-Defined Functions Sometimes a function is defined by different formulas on different parts of its domain. A function with this property is known as a piecewise-defined function. The absolute value function is an example of a piecewise-defined function because the formula changes with the sign of x: f (x) = { −x x <0 x x ≥0 . (1.0.25) Other piecewise-defined functions may be represented by completely different formulas, depending on the part of the domain in which a point falls. To graph a piecewise-defined function, we graph each part of the function in its respective domain, on the same coordinate system. If the formula for a function is different for x < a and x > a , we need to pay special attention to what happens at x = a when we graph the function. Sometimes the graph needs to include an open or closed circle to indicate the value of the function at x = a . We examine this in the next example. Example 1.0.8 : Graphing a Piecewise-Defined Function 10/21/2020 1.0.26 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Sketch a graph of the following piecewise-defined function: f (x) = { x + 3, x <1 2 (x − 2 ) , . (1.0.26) x ≥1 Solution Graph the linear function y = x + 3 on the interval (−∞, 1) and graph the quadratic function y = (x − 2) on the interval [1, ∞). Since the value of the function at x = 1 is given by the formula f (x) = (x − 2) , we see that f (1) = 1 . To indicate this on the graph, we draw a closed circle at the point (1, 1). The value of the function is given by f (x) = x + 2 for all x < 1 , but not at x = 1 . To indicate this on the graph, we draw an open circle at (1, 4). 2 2 Figure 1.0.8 : This piecewise-defined function is linear for x < 1 and quadratic for x ≥ 1. 2) Sketch a graph of the function f (x) = { 2 − x, x ≤2 x + 2, x >2 . Solution: Example 1.0.9 : Parking Fees Described by a Piecewise-Defined Function In a big city, drivers are charged variable rates for parking in a parking garage. They are charged $10 for the first hour or any part of the first hour and an additional $2 for each hour or part thereof up to a maximum of $30 for the day. The parking garage is open from 6 a.m. to 12 midnight. a. Write a piecewise-defined function that describes the cost C to park in the parking garage as a function of hours parked x. b. Sketch a graph of this function C (x). Solution 10/21/2020 1.0.27 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 1.Since the parking garage is open 18 hours each day, the domain for this function is x|0 < x ≤ 18. The cost to park a car at this parking garage can be described piecewise by the function 10, ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ 12, ⎪ ⎪ ⎪ 14, C (x) = ⎨ 0 <x ≤1 1 <x ≤2 2 <x ≤3 . 16, 3 <x ≤4 30, 10 < x ≤ 18 ⎪ ⎪ ⎪ ⎪ ⎪⋮ ⎪ ⎪ ⎩ ⎪ (1.0.27) 2.The graph of the function consists of several horizontal line segments. Exercise 1.0.6 The cost of mailing a letter is a function of the weight of the letter. Suppose the cost of mailing a letter is 49¢ for the first ounce and 21¢ for each additional ounce. Write a piecewise-defined function describing the cost C as a function of the weight x for 0 < x ≤ 3 , where C is measured in cents and x is measured in ounces. Hint The piecewise-defined function is constant on the intervals (0,1],(1,2],…. Answer ⎧ 49, 0 <x ≤1 C (x) = ⎨ 70, ⎩ 1 <x ≤2 91, 2 <x ≤3 (1.0.28) Transformations of Functions We have seen several cases in which we have added, subtracted, or multiplied constants to form variations of simple functions. In the previous example, for instance, we subtracted 2 from the argument of the function y = x to get the function f (x) = (x − 2) . This subtraction represents a shift of the function y = x two units to the right. A shift, horizontally or vertically, is a type of transformation of a function. Other transformations include horizontal and vertical scalings, and reflections about the axes. 2 2 2 A vertical shift of a function occurs if we add or subtract the same constant to each output y . For c > 0 , the graph of f (x) + c is a shift of the graph of f (x) up c units, whereas the graph of f (x) − c is a shift of the graph of f (x) down c units. For example, the graph of the function f (x) = x + 4 is the graph of y = x shifted up 4 units; the graph of the function f (x) = x − 4 is the graph of y = x shifted down 4 units (Figure 1.0.9). 3 3 10/21/2020 3 3 1.0.28 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Figure 1.0.9 : (a) For c > 0 , the graph of y = f (x) + c is a vertical shift up c units of the graph of the graph of y = f (x) − c is a vertical shift down c units of the graph of y = f (x). . (b) For y = f (x) c > 0 , A horizontal shift of a function occurs if we add or subtract the same constant to each input x. For c > 0 , the graph of f (x + c) is a shift of the graph of f (x) to the left c units; the graph of f (x − c) is a shift of the graph of f (x) to the right c units. Why does the graph shift left when adding a constant and shift right when subtracting a constant? To answer this question, let’s look at an example. Consider the function f (x) = |x + 3| and evaluate this function at x − 3 . Since f (x − 3) = |x| and x − 3 < x , the graph of f (x) = |x + 3| is the graph of y = |x| shifted left 3 units. Similarly, the graph of f (x) = |x − 3| is the graph of y = |x| shifted right 3 units (Figure 1.0.10). Figure 1.0.10 : (a) For c > 0 , the graph of y = f (x + c) is a horizontal shift left c units of the graph of c > 0 , the graph of y = f (x − c) is a horizontal shift right c units of the graph of y = f (x). . (b) For y = f (x) A vertical scaling of a graph occurs if we multiply all outputs y of a function by the same positive constant. For c > 0 , the graph of the function cf (x) is the graph of f (x) scaled vertically by a factor of c . If c > 1 , the values of the outputs for the function cf (x) are larger than the values of the outputs for the function f (x); therefore, the graph has been stretched vertically. If 0 < c < 1 , then the outputs of the function cf (x) are smaller, so the graph has been compressed. For example, the graph of the function f (x) = 3x is the graph of y = x stretched vertically by a factor of 3, whereas the graph of f (x) = x /3 is the graph of y = x compressed vertically by a factor of 3 (Figure 1.0.11b). 2 2 2 2 10/21/2020 1.0.29 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Figure 1.0.11 : (a) If c > 1 , the graph of y = cf (x) is a vertical stretch of the graph of y = f (x). (b) If of y = cf (x) is a vertical compression of the graph of y = f (x). 0 < c < 1 , the graph The horizontal scaling of a function occurs if we multiply the inputs x by the same positive constant. For c > 0 , the graph of the function f (cx) is the graph of f (x) scaled horizontally by a factor of c . If c > 1 , the graph of f (cx) is the graph of f (x) compressed horizontally. If 0 < c < 1 , the graph of f (cx) is the graph of f (x) stretched horizontally. For example, consider − − − − − the function f (x) = √2x and evaluate f at x/2. Since f (x/2) = √− x, the graph of f (x) = √2x is the graph of y = √x − − − − compressed horizontally. The graph of y = √x/2 is a horizontal stretch of the graph of y = √x (Figure 1.0.12). Figure 1.0.12 : (a) If c > 1 , the graph of y = f (cx) is a horizontal compression of the graph of y = f (x). (b) If 0 < c < 1 , the graph of y = f (cx) is a horizontal stretch of the graph of y = f (x). We have explored what happens to the graph of a function f when we multiply f by a constant c > 0 to get a new function cf (x). We have also discussed what happens to the graph of a function f when we multiply the independent variable x by c > 0 to get a new function f (cx). However, we have not addressed what happens to the graph of the function if the constant c is negative. If we have a constant c < 0 , we can write c as a positive number multiplied by −1; but, what kind of transformation do we get when we multiply the function or its argument by −1? When we multiply all the outputs by −1, we get a reflection about the x-axis. When we multiply all inputs by −1, we get a reflection about the y -axis. For example, the graph of f (x) = −(x + 1) is the graph of y = (x + 1) reflected about the x-axis. The graph of f (x) = (−x ) + 1 is the graph of y = x + 1 reflected about the y -axis (Figure 1.0.10). 3 3 3 3 10/21/2020 1.0.30 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Figure 1.0.13 : (a) The graph of y = −f (x) is the graph of y = f (x) reflected about the x -axis. (b) The graph of is the graph of y = f (x) reflected about the y -axis. y = f (−x) If the graph of a function consists of more than one transformation of another graph, it is important to transform the graph in the correct order. Given a function f (x), the graph of the related function y = cf (a(x + b)) + d can be obtained from the graph of y = f (x)by performing the transformations in the following order. Horizontal shift of the graph of y = f (x). If b > 0 , shift left. If b < 0 shift right. Horizontal scaling of the graph of y = f (x + b) by a factor of |a|. If a < 0 , reflect the graph about the y -axis. Vertical scaling of the graph of y = f (a(x + b)) by a factor of |c|. If c < 0 , reflect the graph about the x -axis. Vertical shift of the graph of y = cf (a(x + b)) . If d > 0 , shift up. If d < 0 , shift down. We can summarize the different transformations and their related effects on the graph of a function in the following table. Transformation of f (c > 0) Effect of the graph of f f (x) + c Vertical shift up c units f (x) − c Vertical shift down c units f (x + c) Shift left by c units f (x − c) Shift right by c units cf (x) Vertical stretch if c > 1 ; vertical compression if 0 < c < 1 f (cx) Horizontal stretch if 0 < c < 1 ; horizontal compression if c > 1 −f (x) Reflection about the x -axis −f (x) Reflection about the y -axis Example 1.0.10 : Transforming a Function For each of the following functions, a. and b., sketch a graph by using a sequence of transformations of a well-known function. a. f (x) = −|x + 2| − 3 b. f (x) = √− x +1 3 Solution: 1.Starting with the graph of y = |x|, shift 2 units to the left, reflect about the x-axis, and then shift down 3 units. 10/21/2020 1.0.31 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Figure 1.0.14 : The function y = |x| . f (x) = −|x + 2| − 3 can be viewed as a sequence of three transformations of the function 2. Starting with the graph of y=x√, reflect about the y-axis, stretch the graph vertically by a factor of 3, and move up 1 unit. − Figure 1.0.15 : The function f (x) = √− x + 1 can be viewed as a sequence of three transformations of the function y = √x . 3 Exercise 1.0.7 Describe how the function transformations 2 f (x) = −(x + 1 ) −4 can be graphed using the graph of 2 y =x and a sequence of Answer Shift the graph y = x to the left 1 unit, reflect about the x -axis, then shift down 4 units. 2 Below are some more examples: More Examples Rational functions Example 1.0.8 2 Define f 10/21/2020 : R → R by f (x) = x −1 x −1 . Then D(f ) = R ∖ {1} . 1.0.32 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Exercise 1.0.9 2 Define f : R → R by f (x) = x −1 x +1 . Find the domain. Answer D(f ) = R ∖ {−1} Example 1.0.10 : Define f : R → R by f (x) = 1 x −1 . Then D(f ) = R ∖ {1} . Exercise 1.0.11 Define f : R → R by f (x) = 1 x −1 . Find the domain. Answer 10/21/2020 1.0.33 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 D(f ) = R ∖ {−1} . Irrational functions upper half unit circle 1.0.12 Define f − −−− − : R → R by f (x) = √1 − x . Then D(f ) = [−1, 1]. 2 Exercise 1.0.13 Define f − −−− − : R → R by f (x) = √4 − x . Find the domain. 2 Answer D(f ) = [−2, 2]. Example 1.0.14 : Define f 10/21/2020 : R → R − − − − − by f (x) = √1 − x . Then D(f ) = [0, ∞) . 1.0.34 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Exercise 1.0.15 Define f : R → R − − − − − by f (x) = √1 − x . Find the domain. Answer D(f ) = [0, ∞) Piecewise-defined function Absolute valued function Example 1.0.16 Define f : R → R Note that − − √x2 = |x| by f (x) = |x|. . Exercise 1.0.17 Define f 10/21/2020 : R → R by f (x) = |x| + 1 1.0.35 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Answer Step functions Example 1.0.18 : Sign function Define f : R ∖ {0} → R by f (x) = |x| x . Example 1.0.19 : Unit Step function Define f : R → R by f (x) =. Key Concepts A function is a mapping from a set of inputs to a set of outputs with exactly one output for each input. If no domain is stated for a function y = f (x), the domain is considered to be the set of all real numbers x for which the function is defined. When sketching the graph of a function f , each vertical line may intersect the graph, at most, once. A function may have any number of zeros, but it has, at most, one y-intercept. To define the composition g ∘ f , the range of f must be contained in the domain of g . Even functions are symmetric about the y -axis whereas odd functions are symmetric about the origin. The power function f (x) = x is an even function if n is even and n ≠ 0 , and it is an odd function if n is odd. The root function f (x) = x has the domain [0, ∞) if n is even and the domain (−∞, ∞) if n is odd. If n is odd, then f (x) = x is an odd function. The domain of the rational function f (x) = p(x)/q(x), where p(x) and q(x) are polynomial functions, is the set of x such that q(x) ≠ 0 . Functions that involve the basic operations of addition, subtraction, multiplication, division, and powers are algebraic functions. All other functions are transcendental. Trigonometric, exponential, and logarithmic functions are examples of transcendental functions. A polynomial function f with degree n ≥ 1 satisfies f (x) → ±∞ as x → ±∞ . The sign of the output as x → ∞ depends on the sign of the leading coefficient only and on whether n is even or odd. Vertical and horizontal shifts, vertical and horizontal scalings, and reflections about the x- and y -axes are examples of transformations of functions. n 1/n 1/n Key Equations Point-slope equation of a line y − y1 = m(x − x1 ) Slope-intercept form of a line y = mx + b 10/21/2020 1.0.36 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Standard form of a line ax + by = c Polynomial function n x f (x) = an n−1 + an−1 x + ⋯ + a1 x + a0 Composition of two functions g ∘ f )(x) = g(f (x)) Absolute value function f (x) = { −x x <0 x x ≥0 Glossary absolute value function f (x) = { −x x <0 x x ≥0 composite function given two functions f and g , a new function, denoted g ∘ f , such that (g ∘ f )(x) = g(f (x)) decreasing on the interval II a function decreasing on the interval I if, for all x 1, x2 ∈ I , f (x1 ) ≥ f (x2 ) if x 1 < x2 dependent variable the output variable for a function domain the set of inputs for a function even function a function is even if f (−x) = f (x) for all x in the domain of f function a set of inputs, a set of outputs, and a rule for mapping each input to exactly one output graph of a function the set of points (x, y) such that x is in the domain of f and y = f (x) increasing on the interval II a function increasing on the interval I if for allx1, x2 ∈ I , f (x1) ≤ f (x2) if x1 < x2 independent variable the input variable for a function odd function a function is odd if f (−x) = −f (x) for all x in the domain of f range the set of outputs for a function symmetry about the origin the graph of a function f is symmetric about the origin if (−x, −y) is on the graph of f whenever (x, y) is on the graph 10/21/2020 1.0.37 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 symmetry about the y-axis the graph of a function f is symmetric about the y-axis if (−x, y) is on the graph of f whenever (x, y) is on the graph table of values a table containing a list of inputs and their corresponding outputs vertical line test given the graph of a function, every vertical line intersects the graph, at most, once zeros of a function when a real number x is a zero of a function f , f (x) = 0 algebraic function a function involving any combination of only the basic operations of addition, subtraction, multiplication, division, powers, and roots applied to an input variable x cubic function a polynomial of degree 3; that is, a function of the form f (x) = ax + bx + cx + d , where a ≠ 0 degree for a polynomial function, the value of the largest exponent of any term linear function 3 2 a function that can be written in the form f (x) = mx + b logarithmic function a function of the form f (x) = log (x) for some base b > 0, b ≠ 1 such that y = log mathematical model A method of simulating real-life situations with mathematical equations b (x) b if and only if b y =x piecewise-defined function a function that is defined differently on different parts of its domain point-slope equation equation of a linear function indicating its slope and a point on the graph of the function polynomial function a function of the form f (x) = a x + a x +… +a x +a power function a function of the form f (x) = x for any positive integer n ≥ 1 quadratic function a polynomial of degree 2; that is, a function of the form f (x) = ax rational function n n n−1 n−1 1 0 n 2 + bx + c where a ≠ 0 a function of the form f (x) = p(x)/q(x) , where p(x) and q(x) are polynomials root function a function of the form f (x) = x for any integer n ≥ 2 slope the change in y for each unit change in x slope-intercept form 1/n equation of a linear function indicating its slope and y-intercept transcendental function a function that cannot be expressed by a combination of basic arithmetic operations transformation of a function a shift, scaling, or reflection of a function Contributors 10/21/2020 1.0.38 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 1.0.39 CC-BY-NC-SA https://math.libretexts.org/@go/page/10145 1.0E: Exercises For the following exercises, (a) determine the domain and the range of each relation, and (b) state whether the relation is a function. Exercise 1.0.1 x y x y -3 9 1 1 -2 4 2 4 -1 1 3 9 0 0 Answer a. Domain = {−3, −2, −1, 0, 1, 2, 3}, range = {0, 1, 4, 9} b. Yes, a function Exercise 1.0.2 x y x y 1 -3 1 1 2 -2 2 2 3 -1 3 3 0 0 Answer a. Domain = {0, 1, 2, 3}, range = {−3, −2, −1, 0, 1, 2, 3} b. No, not a function Exercise 1.0.3 10/21/2020 x y x y 3 3 15 1 1.0E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10622 Answer x 5 y 2 21 x y 3 8 3 1 15 33 1 3 10 5 0 2 21 2 8 1 33 3 2 a. Domain = {3, 5, 8, 10, 15, 21, 33 }, range = {0, 1, 2, 3} 10 0 b. Yes, a function For the following exercises, find the values for each function, if they exist, then simplify. a. f (0) b. f (1) c. f (3) d. f (−x) e. f (a) f. f (a + h) Exercise 1.0.4 f (x) = 5x − 2 Answer a. −2 b. 3 c. 13 d. −5x − 2 e. 5a − 2 f. 5a + 5h − 2 Exercise 1.0.5 2 f (x) = x Answer a. Undefined b. 2 c. 23 d. − 2 x e 2 a f. 2 a+h Exercise 1.0.6 −−−− − f (x) = √6x + 5 Answer – − − − − − − − − − − − −−−−−−−− − − −−− − a. √5 b. √11 c. √23 d. √−6x + 5 e. √6a + 5 f. √6a + 6h + 5 Exercise 1.0.7 f (x) = |x − 7| + 8 Answer a. 15 b. 14 c. 12 d. |x + 7| + 8 e. |a − 7| + 8 f. |a + h − 7| + 8 Exercise 1.0.8 f (x) = x−2 3x+7 Answer a. \frac{-2}{7} b. −.1 c. 10/21/2020 1 17 d. − x+2 −3x+7 e a−2 3a+7 f. a+h−2 3a+3h+7 1.0E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10622 Exercise 1.0.9 f (x) = 9 Answer a. 9 b. 9 c. 9 d. 9 e. 9 f. 9 For the following exercises, find the domain, range, and all zeros/intercepts, if any, of the functions. Exercise 1.0.10 −−−− − g(x) = √8x − 1 Answer x ≥ 1 8 ; y ≥ 0; x = 1 8 ; no y-intercept Exercise 1.0.11 − − − − − f (x) = −1 + √x + 2 Answer – x ≥ −2; y ≥ −1; x = −1; y = −1 + √2 Exercise 1.0.12 g(x) = 3 x−4 Answer x ≠ 4; y ≠ 0 ; no x-intercept; y = − 3 4 Exercise 1.0.13 −− − g(x) = √ 7 x−5 Answer x > 5; y > 0 ; no intercepts Exercise 1.0.14 f (x) = x x2 −16 Answer x ≠ ±4 ; x = 0, y = 0 Contributors 10/21/2020 1.0E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10622 Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 1.0E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10622 1.1: Introduction to concept of a limit Learning Objectives Using correct notation, describe the limit of a function. Use a table of values to estimate the limit of a function or to identify when the limit does not exist. Use a graph to estimate the limit of a function or to identify when the limit does not exist. Define one-sided limits and provide examples. Explain the relationship between one-sided and two-sided limits. Using correct notation, describe an infinite limit. Define a vertical asymptote. Limits Two key problems led to the initial formulation of calculus: (1) the tangent problem, or how to determine the slope of a line tangent to a curve at a point; and (2) the area problem, or how to determine the area under a curve. The concept of a limit or limiting process, essential to the understanding of calculus, has been around for thousands of years. In fact, early mathematicians used a limiting process to obtain better and better approximations of areas of circles. Yet, the formal definition of a limit—as we know and understand it today—did not appear until the late 19th century. We, therefore, begin our quest to understand limits, as our mathematical ancestors did, by using an intuitive approach. We begin our exploration of limits by taking a look at the graphs of the functions 2 x −4 , f (x) = x −2 |x − 2| , and g(x) = x −2 1 h(x) = 2 , (x − 2) which are shown in Figure 1.1.1. In particular, let’s focus our attention on the behavior of each graph at and around x = 2 . Figure 1.1.1 : These graphs show the behavior of three different functions around x = 2 . Each of the three functions is undefined at x = 2 , but if we make this statement and no other, we give a very incomplete picture of how each function behaves in the vicinity of x = 2 . To express the behavior of each graph in the vicinity of 2 more completely, we need to introduce the concept of a limit. 10/21/2020 1.1.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/11015 Intuitive Definition of a Limit Let’s first take a closer look at how the function f (x) = (x − 4)/(x − 2) behaves around x = 2 in Figure 1.1.1. As the values of x approach 2 from either side of 2, the values of y = f (x) approach 4. Mathematically, we say that the limit of f (x) as x approaches 2 is 4. Symbolically, we express this limit as 2 lim f (x) = 4 . x→2 From this very brief informal look at one limit, let’s start to develop an intuitive definition of the limit. We can think of the limit of a function at a number a as being the one real number L that the functional values approach as the x-values approach a, provided such a real number L exists. Stated more carefully, we have the following definition: Definition(Intuitive): Limit Let f (x) be a function defined at all values in an open interval containing a , with the possible exception of a itself, and let L be a real number. If all values of the function f (x) approach the real number L as the values of x(≠ a) approach the number a, then we say that the limit of f (x) as x approaches a is L. (More succinct, as x gets closer to a , f (x) gets closer and stays close to L.) Symbolically, we express this idea as lim f (x) = L. (1.1.1) x→a We can estimate limits by constructing tables of functional values and by looking at their graphs. This process is described in the following Problem-Solving Strategy. Problem-Solving Strategy: Evaluating a Limit Using a Table of Functional Values 1. To evaluate , we begin by completing a table of functional values. We should choose two sets of x-values— lim f (x) x→a one set of values approaching a and less than a , and another set of values approaching a and greater than a . Table demonstrates what your tables might look like. 1.1.1 Table 1.1.1 x f(x) x f(x) a − 0.1 f (a − 0.1) a + 0.1 f (a + 0.1) a − 0.01 f (a − 0.01) a + 0.001 f (a + 0.001) a − 0.001 f (a − 0.001) a + 0.0001 f (a + 0.001) a − 0.0001 f (a − 0.0001) a + 0.00001 f (a + 0.0001) Use additional values as necessary. Use additional values as necessary. 2. Next, let’s look at the values in each of the f (x) columns and determine whether the values seem to be approaching a single value as we move down each column. In our columns, we look at the sequence f (a − 0.1) , f (a − 0.01) , f (a − 0.001), f (a − 0.0001), and so on, and f (a + 0.1), f (a + 0.01), f (a + 0.001), f (a + 0.0001), and so on. (Note: Although we have chosen the x-values a ± 0.1, a ± 0.01, a ± 0.001, a ± 0.0001 , and so forth, and these values will probably work nearly every time, on very rare occasions we may need to modify our choices.) 3. If both columns approach a common y -value L, we state lim f (x) = L . We can use the following strategy to confirm x→a the result obtained from the table or as an alternative method for estimating a limit. 4. Using a graphing calculator or computer software that allows us graph functions, we can plot the function f (x), making sure the functional values of f (x) for x-values near a are in our window. We can use the trace feature to move along the graph of the function and watch the y -value readout as the x-values approach a. If the y -values approach L as our xvalues approach a from both directions, then lim f (x) = L. We may need to zoom in on our graph and repeat this x→a process several times. We apply this Problem-Solving Strategy to compute a limit in Examples 1.1.1A and 1.1.1B. 10/21/2020 1.1.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/11015 Example 1.1.1A : Evaluating a Limit Using a Table of Functional Values Evaluate lim x→0 sin x x using a table of functional values. Solution We have calculated the values of f (x) = sin x x for the values of x listed in Table 1.1.2. Table 1.1.2 si n x x si n x x x x -0.1 0.998334166468 0.1 0.998334166468 -0.01 0.999983333417 0.01 0.999983333417 -0.001 0.999999833333 0.001 0.999999833333 -0.0001 0.999999998333 0.0001 0.999999998333 Note: The values in this table were obtained using a calculator and using all the places given in the calculator output. As we read down each sin x x column, we see that the values in each column appear to be approaching one. Thus, it is fairly reasonable to conclude that sin x lim x→0 =1 . A calculator-or computer-generated graph of x sin x f (x) = would be x similar to that shown in Figure 1.1.2, and it confirms our estimate. Figure 1.1.2 : The graph of f (x) = (sin x)/x confirms the estimate from Table. Example 1.1.1B : Evaluating a Limit Using a Table of Functional Values Evaluate lim x→4 − √x − 2 x −4 using a table of functional values. Solution As before, we use a table—in this case, Table 1.1.3—to list the values of the function for the given values of x. Table 1.1.3 x √x−2 x x−4 10/21/2020 √x−2 x−4 3.9 0.251582341869 4.1 0.248456731317 3.99 0.25015644562 4.01 0.24984394501 3.999 0.250015627 4.001 0.249984377 3.9999 0.250001563 4.0001 0.249998438 3.99999 0.25000016 4.00001 0.24999984 1.1.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/11015 After inspecting this table, we see that the functional values less than 4 appear to be decreasing toward 0.25 whereas the functional values greater than 4 appear to be increasing toward 0.25. We conclude that this estimate using the graph of f (x) = − √x − 2 x −4 − √x − 2 lim x→4 = 0.25 . We confirm x −4 shown in Figure 1.1.3. Figure 1.1.3 : The graph of √x−2 x−4 confirms the estimate from table Exercise 1.1.1 1 Estimate lim x→1 x −1 x −1 using a table of functional values. Use a graph to confirm your estimate. Hint Use 0.9, 0.99, 0.999, 0.9999, 0.99999 and 1.1, 1.01, 1.001, 1.0001, 1.00001 as your table values. Answer 1 lim x→1 x −1 = −1 x −1 At this point, we see from Examples 1.1.1A and 1.1.1b that it may be just as easy, if not easier, to estimate a limit of a function by inspecting its graph as it is to estimate the limit by using a table of functional values. In Example 1.1.2, we evaluate a limit exclusively by looking at a graph rather than by using a table of functional values. Example 1.1.2 : Evaluating a Limit Using a Graph For g(x) shown in Figure 1.1.4, evaluate lim g(x) . x→−1 Figure 1.1.4 : The graph of g(x) includes one value not on a smooth curve. 10/21/2020 1.1.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/11015 Solution: Despite the fact that g(−1) = 4 , as the x-values approach −1 from either side, the g(x) values approach 3. Therefore, lim g(x) = 3 . Note that we can determine this limit without even knowing the algebraic expression of the function. x→−1 Based on Example 1.1.2, we make the following observation: It is possible for the limit of a function to exist at a point, and for the function to be defined at this point, but the limit of the function and the value of the function at the point may be different. Exercise 1.1.2 Use the graph of h(x) in Figure 1.1.5 to evaluate lim h(x), if possible. x→2 Figure 1.1.5 : Hint What y-value does the function approach as the x -values approach 2? Solution lim h(x) = −1. x→2 Looking at a table of functional values or looking at the graph of a function provides us with useful insight into the value of the limit of a function at a given point. However, these techniques rely too much on guesswork. We eventually need to develop alternative methods of evaluating limits. These new methods are more algebraic in nature and we explore them in the next section; however, at this point we introduce two special limits that are foundational to the techniques to come. Two Important Limits Let a be a real number and c be a constant. i. lim x = a x→a ii. lim c = c x→a We can make the following observations about these two limits. i. For the first limit, observe that as x approaches a , so does f (x), because f (x) = x. Consequently, lim x = a . x→a ii. For the second limit, consider Table 1.1.4. Table 1.1.4 10/21/2020 1.1.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/11015 x f(x) = c x f(x) = c a − 0.1 c a + 0.1 c a − 0.01 c a + 0.01 c a − 0.001 c a + 0.001 c a − 0.0001 c a + 0.0001 c Observe that for all values of x (regardless of whether they are approaching a ), the values f (x) remain constant at c . We have no choice but to conclude lim c = c . x→a The Existence of a Limit As we consider the limit in the next example, keep in mind that for the limit of a function to exist at a point, the functional values must approach a single real-number value at that point. If the functional values do not approach a single value, then the limit does not exist. Example 1.1.3 : Evaluating a Limit That Fails to Exist Evaluate lim sin(1/x) using a table of values. x→0 Solution Table 1.1.5 lists values for the function sin(1/x) for the given values of x. Table 1.1.5 x sin(1/x) x sin(1/x) -0.1 0.544021110889 0.1 −0.544021110889 -0.01 0.50636564111 0.01 −0.50636564111 -0.001 −0.8268795405312 0.001 0.8268795405312 -0.0001 0.305614388888 0.0001 −0.305614388888 -0.00001 −0.035748797987 0.00001 0.035748797987 -0.000001 0.349993504187 0.000001 −0.349993504187 After examining the table of functional values, we can see that the y -values do not seem to approach any one single value. It appears the limit does not exist. Before drawing this conclusion, let’s take a more systematic approach. Take the following sequence of x-values approaching 0: 2 2 , π 2 , 3π 2 , 2 , 5π 7π 2 , 9π , …. 11π The corresponding y -values are 1, −1, 1, −1, 1, −1, . . . . At this point we can indeed conclude that lim sin(1/x) does not exist. (Mathematicians frequently abbreviate “does not x→0 exist” as DNE. Thus, we would write lim sin(1/x) DNE.) The graph of f (x) = sin(1/x) is shown in Figure 1.1.6 and it x→0 gives a clearer picture of the behavior of sin(1/x) as wildly between −1 and 1 as x approaches 0. 10/21/2020 x approaches 0. You can see that 1.1.6 CC-BY-NC-SA sin(1/x) oscillates ever more https://math.libretexts.org/@go/page/11015 Figure 1.1.6 : The graph of f (x) = sin(1/x) oscillates rapidly between −1 and 1 as x approaches 0. Exercise 1.1.3 2 Use a table of functional values to evaluate lim x→2 ∣ ∣x − 4∣ ∣ x −2 , if possible. Hint Use x -values 1.9, 1.99, 1.999, 1.9999, 1.9999 and 2.1, 2.01, 2.001, 2.0001, 2.00001 in your table. Answer 2 ∣ ∣x − 4 ∣ ∣ lim x→2 x −2 does not exist. Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 1.1.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/11015 1.1E: Exercises Exercise 1.1E. 1 Terms and Concepts 1. In your own words, what does it mean to "find the limit of f (x) as x approaches 3"? 2. An expression of the form is called _____. 3. T/F: The limit of f (x) as x approaches 5 is f (5). 4. Describe three situations where lim f (x) does not exist. 0 0 x→c Exercise 1.1E. 2 2 For exercises 1 - 2, consider the function f (x) = x −1 . |x − 1| 1) [T] Complete the following table for the function. Round your solutions to four decimal places. x f(x) x f(x) 0.9 a. 1.1 e. 0.99 b. 1.01 f. 0.999 c. 1.001 g. 0.9999 d. 1.0001 h. 2) What do your results in the preceding exercise indicate about the two-sided limit lim f (x)? Explain your response. x→1 Answer lim f (x) does not exist because x→1 . lim f (x) = −2 ≠ lim f (x) = 2 − x→1 + x→1 Exercise 1.1E. 3 Consider the function x 1 f (x) = (1 + ) x . Make a table showing f (x) for x = 1, 2, 3, . . . . . . Round your solutions to five decimal places. What can you say about the value of the function f (x) as x increases indefinitely? Answer x 1 limx→∞ (1 + 10/21/2020 ) = 2.7183 = e. x 1.1E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/11018 1.2: One sided Limits and Vertical Asymptotes One Sided Limits We begin our exploration of limits by taking a look at the graphs of the following functions f (x) = x + 1 2 x −1 g(x) = ,x ≠1 x −1 2 ⎧ x −1 ⎪ ⎪ ⎪ x −1 h(x) = ⎨ if x ≠ 1 ⎪ ⎪ ⎩ ⎪ 0 if x = 1 which are shown in Figure 1.2.1. In particular, let’s focus our attention on the behavior of each graph at and around x = 1 . Example 1.2.1 : Consider, f (x) = x + 1 Evaluate lim f (x). (1.2.1) x→1 lim f (x) = 2. (1.2.2) x→1 Example 1.2.2 : 2 Consider, g(x) = 10/21/2020 x −1 ,x ≠1 x −1 , 1.2.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10144 Evaluate lim g(x). (1.2.3) x→1 lim g(x) = 2. (1.2.4) x→1 See the figure above. Note that g(2) is undefined. Example 1.2.3 : 2 ⎧ x ⎪ ⎪ ⎪ Consider, h(x) = ⎨ −1 if x ≠ 1 x −1 ⎪ ⎪ ⎩ ⎪ 0 if x = 1 Evaluate lim h(x). (1.2.5) x→1 lim h(x) = 2. (1.2.6) x→1 See the figure above. Note that h(2) = 0 . We can now present an informal definition of one-sided limits. Definition: One-sided Limits We define two types of one-sided limits. Limit from the left: Let f (x) be a function defined at all values in an open interval of the form z , and let L be a real number. If the values of the function f (x) approach the real number L as the values of x (where x < a ) approach the number a, then we say that L is the limit of f (x) as x approaches a from the left. Symbolically, we express this idea as lim f (x) = L. (1.2.7) x→a− Limit from the right: Let f (x) be a function defined at all values in an open interval of the form (a, c), and let L be a real number. If the values of the function f (x) approach the real number L as the values of x (where x > a ) approach the number a , then we say that L is the limit of f (x) as x approaches a from the right. Symbolically, we express this idea as lim f (x) = L. (1.2.8) x→a+ 10/21/2020 1.2.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10144 Example 1.2.1 Example 1.2.4 : Evaluating One-Sided Limits For the function f (x) = { a. x + 1, 2 x − 4, if x < 2 if x ≥ 2 , evaluate each of the following limits. lim f (x) − x→2 b. lim f (x) + x→2 Solution From Figure , we can conclude that a. 1.2.7 lim f (x) = 3 and b. − lim f (x) = 0 . Therefore, the (two-sided) limit of f (x) does not exist at + x→2 x =2 x→2 . Figure 1.2.7 : The graph of f (x) = { x + 1, 2 x if x < 2 − 4, if x ≥ 2 has a break at x = 2 . Example 1.2.4 : |x − 2| g(x) = x −2 Let us now consider the relationship between the limit of a function at a point and the limits from the right and left at that point. It seems clear that if the limit from the right and the limit from the left have a common value, then that common value is the limit of the function at that point. Similarly, if the limit from the left and the limit from the right take on different values, the limit of the function does not exist. These conclusions are summarized in Note. Relating One-Sided and Two-Sided Limits Let f (x) be a function defined at all values in an open interval containing a , with the possible exception of Then, lim f (x) = L a itself, and let L be a real number. (1.2.9) x→a if and only if lim f (x) = L − x→a 10/21/2020 and lim f (x) = L . + x→a 1.2.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10144 App: Intuitive Notion of the Limit (You can change the values by dragging) Limits: Intuitive Notion of the Limit - One-Sided Limits Infinite Limits `f(x)` = `f(x)` = Evaluating the limit of a function at a point or evaluating the limit of a function from the right and left at a point helps us to characterize the behavior of floor(x) around + 0.1 sin(15 x) - value. 0.2 x + As 0.5 we shall see, we can also describe the behavior of functions that doExamples a function a given not have finite limits. Graph We now turn our attention to h(x) = 1/(x − 2) , the third and final function introduced at the beginning of this section (see Figure 1.2.1(c)). From its graph we see that as the values of x approach 2, the values of h(x) = 1/(x − 2) become larger and larger and, in fact, become infinite. Mathematically, we say that the limit of h(x) as x approaches 2 is positive infinity. Symbolically, we express this idea as 2 2 lim h(x) = +∞. x→2 More generally, we define infinite limits as follows: Definitions: Infinite Limits We define three types of infinite limits. Infinite limits from the left: Let f (x) be a function defined at all values in an open interval of the form (b, a). i. If the values of f (x) increase without bound as the values of approaches a from the left is positive infinity and we write x (where x <a ) approach the number a , then we say that the limit as lim f (x) = +∞. x (1.2.10) − x→a ii. If the values of f (x) decrease without bound as the values of approaches a from the left is negative infinity and we write x (where x <a ) approach the number a , then we say that the limit as lim f (x) = −∞. x (1.2.11) − x→a Infinite limits from the right: Let f (x) be a function defined at all values in an open interval of the form (a, c). i. If the values of f (x) increase without bound as the values of approaches a from the left is positive infinity and we write 10/21/2020 x (where x >a ) approach the number a , then we say that the limit as 1.2.4 CC-BY-NC-SA x https://math.libretexts.org/@go/page/10144 lim f (x) = +∞. (1.2.12) + x→a ii. If the values of f (x) decrease without bound as the values of approaches a from the left is negative infinity and we write (where x x >a ) approach the number a , then we say that the limit as lim f (x) = −∞. x (1.2.13) + x→a Two-sided infinite limit: Let f (x) be defined for all x ≠ a in an open interval containing a i. If the values of f (x) increase without bound as the values of approaches a is positive infinity and we write (where x x ≠a ) approach the number a , then we say that the limit as lim f (x) = +∞. x (1.2.14) x→a ii. If the values of f (x) decrease without bound as the values of approaches a is negative infinity and we write x (where x ≠a ) approach the number a , then we say that the limit as lim f (x) = −∞. x (1.2.15) x→a It is important to understand that when we write statements such as lim f (x) = +∞ x→a or lim f (x) = −∞ we are describing the behavior of the x→a function, as we have just defined it. We are not asserting that a limit exists. For the limit of a function f (x) to exist at a , it must approach a real number L as x approaches a . That said, if, for example, lim f (x) = +∞ , we always write lim f (x) = +∞ rather than lim f (x) DNE. x→a x→a x→a Example 1.2.5 : Recognizing an Infinite Limit Evaluate each of the following limits, if possible. Use a table of functional values and graph f (x) = 1/x to confirm your conclusion. a. 1 lim x − x→0 b. 1 lim x + x→0 c. 1 lim x→0 x Solution Begin by constructing a table of functional values. Table 1.2.7 1 x x x 1 x -0.1 -10 0.1 10 -0.01 -100 0.01 100 -0.001 -1000 0.001 1000 -0.0001 -10,000 0.0001 10,000 -0.00001 -100,000 0.00001 100,000 -0.000001 -1,000,000 0.000001 1,000,000 a. The values of 1/x decrease without bound as x approaches 0 from the left. We conclude that 1 lim − x→0 = −∞. x b. The values of 1/x increase without bound as x approaches 0 from the right. We conclude that 1 lim + x→0 c. Since 1 lim − x→0 = −∞ x and 1 lim + x→0 = +∞ x = +∞. x have different values, we conclude that 1 lim x→0 DNE. x The graph of f (x) = 1/x in Figure 1.2.8 confirms these conclusions. 10/21/2020 1.2.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10144 Figure 1.2.8 : The graph of f (x) = 1/x confirms that the limit as x approaches 0 does not exist. Exercise 1.2.5 Evaluate each of the following limits, if possible. Use a table of functional values and graph f (x) = 1/x to confirm your conclusion. 2 a. 1 lim 2 − x x→0 b. 1 lim 2 x + x→0 c. 1 lim x→0 2 x Hint Follow the procedures from Example 1.2.4. Answer a. 1 lim − x→0 b. 2 = +∞ ; = +∞ ; x 1 lim + x→0 c. lim x→0 2 x 1 = +∞ 2 x It is useful to point out that functions of the form f (x) = 1/(x − a) , where n is a positive integer, have infinite limits as x approaches a from either the left or right (Figure 1.2.9). These limits are summarized in the above definitions. n Figure 1.2.9 : The function f (x) = 1/(x − a) has infinite limits at a . n Infinite Limits from Positive Integers If n is a positive even integer, then 1 lim x→a 10/21/2020 n = +∞. (1.2.16) (x − a) 1.2.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10144 If n is a positive odd integer, then 1 lim + x→a n = +∞ (1.2.17) = −∞. (1.2.18) (x − a) and 1 lim − x→a n (x − a) We should also point out that in the graphs of f (x) = 1/(x − a) , points on the graph having x-coordinates very near to a are very close to the vertical line x = a . That is, as x approaches a , the points on the graph of f (x) are closer to the line x = a . The line x = a is called a vertical asymptote of the graph. We formally define a vertical asymptote as follows: n Definition: Vertical Asymptotes Let f (x) be a function. If any of the following conditions hold, then the line x = a is a vertical asymptote of f (x). lim f (x) = +∞ (1.2.19) − x→a lim f (x) = −∞ (1.2.20) − x→a lim f (x) = +∞ (1.2.21) + x→a lim f (x) = −∞ (1.2.22) + x→a lim f (x) = +∞ (1.2.23) x→a lim f (x) = −∞ (1.2.24) x→a Example 1.2.6 : Finding a Vertical Asymptote Evaluate each of the following limits using Equations , , and 1.2.16 1.2.17 1.2.18 above. Identify any vertical asymptotes of the function 4 f (x) = 1/(x + 3 ) . a. 1 lim 4 − (x + 3) x→−3 b. 1 lim 4 + (x + 3) x→−3 c. 1 lim 4 (x + 3) x→−3 Solution We can use the above equations directly. a. 1 lim 4 − b. 1 lim 4 + = +∞ (x + 3) x→−3 c. = +∞ (x + 3) x→−3 1 lim x→−3 4 = +∞ (x + 3) The function f (x) = 1/(x + 3) has a vertical asymptote of x = −3 . 4 Exercise 1.2.6 Evaluate each of the following limits. Identify any vertical asymptotes of the function f (x) = 1 3 . (x − 2) a. 1 lim (x − 2)3 − x→2 b. 1 lim 3 + (x − 2) x→2 c. 1 lim x→2 (x − 2)3 Answer a 1 lim − x→2 = −∞ 3 (x − 2) 10/21/2020 1.2.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10144 Answer b 1 lim = +∞ 3 + (x − 2) x→2 Answer c 1 DNE. The line x = 2 is the vertical asymptote of f (x) = 1/(x − 2) 3 lim x→2 3 . (x − 2) In the next example we put our knowledge of various types of limits to use to analyze the behavior of a function at several different points. Example 1.2.7 : Behavior of a Function at Different Points Use the graph of f (x) in Figure 1.2.10 to determine each of the following values: a. ; lim f (x) − b. f (x ; lim f (x) − d. f (x) + ; lim f (x) lim f (x); f (1) + x→1 x→1 ; lim f (x) − ; lim f (x) lim f (x); f (3) + x→3 lim f (x); f (−2) x→−2 x→−2 x→1 lim f (x); f (−4) x→−4 ; lim − x→−2 c. f (x) x→−4 ); lim ; lim + x→−4 x→3 x→3 Figure 1.2.10 : The graph shows f (x). Solution Using the definitions above and the graph for reference, we arrive at the following values: a. lim f (x) = 0 ; − b. lim f (x) = 3 − lim f (x) = 6 − ; ; lim f (x) = 3 + ; lim f (x) = 3 − ; ; lim f (x) = 3; f (−2) ; lim f (x) DNE; f (1) = 6 lim f (x) = −∞ + x→3 is undefined x→−2 x→1 x→1 lim f (x) = −∞ lim f (x) = 0; f (−4) = 0 x→−4 x→−2 + x→1 d. f (x) = 0 x→−4 x→−2 c. lim + x→−4 x→3 ; lim f (x) = −∞ ; f (3) is undefined x→3 Exercise 1.2.7 Evaluate lim f (x) for f (x) shown here: x→1 Figure 1.2.11 . The graph of a piecewise function f . Hint 10/21/2020 1.2.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10144 Compare the limit from the right with the limit from the left. Answer lim f (x) does not exist x→1 Example 1.2.8 : Einstein’s Equation In the Chapter opener we mentioned briefly how Albert Einstein showed that a limit exists to how fast any object can travel. Given Einstein’s equation for the mass of a moving object m0 m = − −−− − 2 √1 − , v 2 c what is the value of this bound? Figure 1.2.12 . (Crefit:NASA) Solution Our starting point is Einstein’s equation for the mass of a moving object, m = m0 − −−− − 2 √1 − , v 2 c where m is the object’s mass at rest, v is its speed, and c is the speed of light. To see how the mass changes at high speeds, we can graph the ratio of masses m/m as a function of the ratio of speeds, v/c (Figure 1.2.13). 0 0 Figure 1.2.13 : This graph shows the ratio of masses as a function of the ratio of speeds in Einstein’s equation for the mass of a moving object. We can see that as the ratio of speeds approaches 1—that is, as the speed of the object approaches the speed of light—the ratio of masses increases without bound. In other words, the function has a vertical asymptote at v/c = 1 . We can try a few values of this ratio to test this idea. Table 1.2.8 − − − − − − v/c √1 − v2 c2 m/mo 0.99 0.1411 7.089 0.999 0.0447 22.37 0.9999 0.0141 70.7 Thus, according to Table 1.2.8:, if an object with mass 100 kg is traveling at 0.9999c, its mass becomes 7071 kg. Since no object can have an infinite mass, we conclude that no object can travel at or more than the speed of light. Key Concepts A table of values or graph may be used to estimate a limit. If the limit of a function at a point does not exist, it is still possible that the limits from the left and right at that point may exist. If the limits of a function from the left and right exist and are equal, then the limit of the function is that common value. We may use limits to describe infinite behavior of a function at a point. 10/21/2020 1.2.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10144 Key Equations Intuitive Definition of the Limit lim f (x) = L x→a Two Important Limits lim x = a lim c = c x→a x→a One-Sided Limits lim f (x) = L lim f (x) = L − + x→a x→a Infinite Limits from the Left lim f (x) = +∞ lim f (x) = −∞ − − x→a x→a Infinite Limits from the Right lim f (x) = +∞ lim f (x) = −∞ + + x→a x→a Two-Sided Infinite Limits lim f (x) = +∞ : x→a lim f (x) = −∞ x→a lim f (x) = +∞ and − : lim f (x) = −∞ − x→a lim f (x) = +∞ + x→a x→a and lim f (x) = −∞ + x→a Glossary infinite limit A function has an infinite limit at a point a if it either increases or decreases without bound as it approaches a intuitive definition of the limit If all values of the function f (x) approach the real number L as the values of x(≠ a) approach a, f (x) approaches L one-sided limit A one-sided limit of a function is a limit taken from either the left or the right vertical asymptote A function has a vertical asymptote at x = a if the limit as x approaches a from the right or left is infinite Source Calculus Applets using GeoGebra by Marc Renault is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CCBY-SA-NC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 1.2.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10144 1.2E: Exercises Exercise 1.2E. 1 In exercises 1 - 4, consider the graph of the functiony = f (x) shown here. Which of the statements about y = f (x) are true and which are false? Explain why a statement is false. 1) lim f (x) = 0 x→10 2) lim f (x) = 3 + x→−2 Answer False; lim f (x) = +∞ + x→−2 3) lim f (x) = f (−8) x→−8 4) lim f (x) = 5 x→6 Answer False; lim f (x) DNE since x→6 lim f (x) = 2 − x→6 and lim f (x) = 5 . + x→6 Exercise 1.2E. 2 In exercises 1 - 5, use the following graph of the function necessary. 10/21/2020 y = f (x) to find the values, if possible. Estimate when 1.2E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10658 1) lim f (x) − x→1 2) lim f (x) + x→1 Answer 2 3) lim f (x) x→1 4) lim f (x) x→2 Answer 1 5) f (1) Exercise 1.2E. 3 In exercises 1 - 4, use the graph of the function y = f (x) shown here to find the values, if possible. Estimate when necessary. 10/21/2020 1.2E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10658 1) lim f (x) − x→0 Answer 1 2) lim f (x) + x→0 3) lim f (x) x→0 Answer DNE 4) lim f (x) x→2 Exercise 1.2E. 4 In exercises 1 - 6, use the graph of the function y = f (x) shown here to find the values, if possible. Estimate when necessary. 1) lim f (x) − x→−2 Answer 0 2) lim f (x) + x→−2 3) lim f (x) x→−2 Answer DNE 4) lim f (x) − x→2 5) lim f (x) + x→2 Answer 10/21/2020 1.2E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10658 2 6) lim f (x) x→2 Exercise 1.2E. 5 1. In exercises a-c, use the graph of the function y = g(x) shown here to find the values, if possible. Estimate when necessary. a) lim g(x) − x→0 Answer 3 b) lim g(x) + x→0 c) lim g(x) x→0 Answer DNE 2. In exercises a-c, use the graph of the function when necessary. 10/21/2020 y = h(x) shown here to find the values, if possible. Estimate 1.2E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10658 a) lim h(x) − x→0 b) lim h(x) + x→0 Answer 0 c) lim h(x) x→0 Exercise 1.2E. 6 In exercises a-e, use the graph of the function necessary. a) y = f (x) shown here to find the values, if possible. Estimate when lim f (x) − x→0 Answer −2 b) lim f (x) + x→0 10/21/2020 1.2E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10658 c) lim f (x) x→0 Answer DNE d) lim f (x) x→1 e) lim f (x) x→2 Answer 0 Exercise 1.2E. 7 In exercises 1 - 5, sketch the graph of a function with the given properties. 1) lim f (x) = 1, 2) lim lim f (x) = 3, − x→2 f (x) = 0, lim f (x) = 6, is not defined. x→4 lim f (x) = −∞, − x→−∞ x =4 + x→4 x→−1 lim f (x) = ∞, + lim f (x) = f (0), f (0) = 1, lim f (x) = −∞ x→∞ x→0 x→−1 Answer Answers may vary 3) 1 lim f (x) = 2, − x→−∞ 4) lim lim f (x) = −∞, x→3 f (x) = 2, x→−∞ lim f (x) = −∞, x→−2 lim f (x) = ∞, + lim f (x) = 2, x→∞ x→3 lim f (x) = 2, f (0) = − 3 f (0) = 0 x→∞ Answer Answer may vary 10/21/2020 1.2E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10658 5) lim f (x) = 0, lim f (x) = ∞, − x→−∞ lim f (x) = −∞, f (0) = −1, + x→−1 lim f (x) = −∞, − x→−1 x→1 lim f (x) = ∞, + x→1 lim f (x) = 0 x→∞ Exercise 1.2E. 8 1) Shock waves arise in many physical applications, ranging from supernovas to detonation waves. A graph of the density of a shock wave with respect to distance, x, is shown here. We are mainly interested in the location of the front of the shock, labeled X in the diagram. SF a. Evaluate . lim ρ(x) + x→X SF b. Evaluate . lim x→X ρ(x) − SF c. Evaluate . Explain the physical meanings behind your answers. lim ρ(x) x→XSF Answer a. ρ b. ρ c. DNE unless ρ = ρ . As you approach X from the right, you are in the high-density area of the shock. When you approach from the left, you have not experienced the “shock” yet and are at a lower density. 2 1 1 2 SF 2) A track coach uses a camera with a fast shutter to estimate the position of a runner with respect to time. A table of the values of position of the athlete versus time is given here, where x is the position in meters of the runner and t is time in seconds. What is lim x(t)? What does it mean physically? t→2 10/21/2020 1.2E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10658 t(sec) x(m) 1.75 4.5 1.95 6.1 1.99 6.42 2.01 6.58 2.05 6.9 2.25 8.5 Exercise 1.2E. 9 Terms and Concepts 1. What are the ways in which a limit may fail to exist? 2. T/F: If lim f (x) = 5 , then lim f (x) = 5 x→1− 3. T/F: If lim f (x) = 5 x→1 , then x→1− 4. T/F: If lim f (x) = 5 , then x→1 5. T/F: If lim f (x) = −∞ − x→1 lim f (x) = 5 x→1+ lim f (x) = 5 x→1− , then lim f (x) = ∞ . + x→1 6. T/F: If lim f (x) = ∞ , then f has a vertical asymptote at x = 5 . x→5 Answer Under construction. Exercise 1.2E. 10 : Finding limits using Graphs Evaluate each expression using the given graph of f (x). 1. Answer (a)2,(b)2,(c) 2,(d) 1, (e)DNE, (f)1 2. 10/21/2020 1.2E.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10658 Answer Under construction. 3. Answer Under construction. 4. Answer Under construction. 10/21/2020 1.2E.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10658 Answer Under construction. 5.. Answer Under construction. 6. Answer Under construction. 7. 10/21/2020 1.2E.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10658 Answer Under construction. Exercise 1.2E. 11 : Infinite limits In Exercises 1-6, evaluate the given limits using the graph of the function. 1. f (x) = 1 2 (x+1) (a) lim f (x) − x→−1 (b) lim f (x) + x→−1 2. f (x) = (a) 1 2 (x−3)(x−5) lim f (x) − x→3 (b) lim f (x) + x→3 (c) lim f (x) x→3 (d) lim f (x) − x→5 (e) lim f (x) + x→5 (f) lim f (x) x→5 3. f (x) = 10/21/2020 1 ex +1 1.2E.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10658 (a) lim f (x) x→−∞ (b) lim f (x) x→∞ (c) lim f (x) − x→0 (d) lim f (x) + x→0 4. f (x) = x 2 (a) lim sin(πx) f (x) x→−∞ (b) lim f (x) x→∞ 5. f (x) = cos(x) (a) lim f (x) x→−∞ (b) lim f (x) x→∞ 6. f (x) = 2 x 10/21/2020 + 10 1.2E.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/10658 (a) lim f (x) x→−∞ (b) lim f (x) x→∞ Answer Under construction. Exercise 1.2E. 11 : One sided limits 3 ⎧ ⎪ x +1 if x < 0 0 if x = 0 − − − − − √x + 1 − 2 if x > 0 Given f (x) = ⎨ ⎩ ⎪ Identify each limit: 1. lim f (x) − x→0 2. lim f (x) + x→0 3. lim f (x) x→0 4. lim f (x) x→−1 5. lim f (x) x→3 Hint: Draw a graph. Answer: 11, -1, DNE, 0, 0 Solution: 1. 3 lim f (x) = 0 + 11 − x→0 −−− − lim f (x) = √0 + 1 − 2 = −1 2. + x→0 3. Since lim f (x) ≠ lim f (x) − + x→0 4. x→0 3 lim f (x) = (−1 ) , lim f (x) = DNE. \ x→0 +1 = 0 x→−1 5. −−− − lim f (x) = √3 + 1 − 2 = 0 x→3 Exercise 1.2E. 12 : Infinite limits y +6 lim − y→6 y 2 − 36 Answer Since y +6 lim − y→6 y 2 12 = y +6 lim − y→6 = −∞ y 2 − 36 0 − 36 , and 5.9+6 2 < 0, (5.9 ) −36 . Contributors 10/21/2020 1.2E.13 CC-BY-NC-SA https://math.libretexts.org/@go/page/10658 Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution Noncommercial (BY-NC) License. http://www.apexcalculus.com/ Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 1.2E.14 CC-BY-NC-SA https://math.libretexts.org/@go/page/10658 1.3: Limit calculations for algebraic expressions Learning Objectives Recognize the basic limit laws. Use the limit laws to evaluate the limit of a function. Evaluate the limit of a function by factoring. Use the limit laws to evaluate the limit of a polynomial or rational function. Evaluate the limit of a function by factoring or by using conjugates. In the previous section, we evaluated limits by looking at graphs or by constructing a table of values. In this section, we establish laws for calculating limits and learn how to apply these laws. In the Student Project at the end of this section, you have the opportunity to apply these limit laws to derive the formula for the area of a circle by adapting a method devised by the Greek mathematician Archimedes. We begin by restating two useful limit results from the previous section. These two results, together with the limit laws, serve as a foundation for calculating many limits. Evaluating Limits with the Limit Laws The first two limit laws were stated previously and we repeat them here. These basic results, together with the other limit laws, allow us to evaluate the limits of many algebraic functions. Basic Limit Results For any real number a and any constant c , I. lim x = a x→a II. lim c = c x→a Example 1.3.1 : Evaluating a Basic Limit Evaluate each of the following limits using Note. a. lim x x→2 b. lim 5 x→2 Solution a. The limit of x as x approaches a is a: lim x = 2 . x→2 b. The limit of a constant is that constant: lim 5 = 5 . x→2 We now take a look at the limit laws, the individual properties of limits. The proofs that these laws hold are omitted here. Limit Laws Let f (x) and g(x) be defined for all x ≠ a over some open interval containing a . Assume that L and M are real numbers such that lim f (x) = L and lim g(x) = M . Let c be a constant. Then, each of the following statements holds: x→a x→a Sum law for limits: lim(f (x) + g(x)) = lim f (x) + lim g(x) = L + M x→a x→a (1.3.1) x→a Difference law for limits: lim(f (x) − g(x)) = lim f (x) − lim g(x) = L − M x→a 10/21/2020 x→a (1.3.2) x→a 1.3.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10246 Constant multiple law for limits: lim cf (x) = c ⋅ lim f (x) = cL x→a (1.3.3) x→a Product law for limits: lim(f (x) ⋅ g(x)) = lim f (x) ⋅ lim g(x) = L ⋅ M x→a x→a (1.3.4) x→a Quotient law for limits: lim f (x) f (x) lim x→a L x→a = = g(x) (1.3.5) M lim g(x) x→a for M ≠0 . Power law for limits: n lim (f (x)) n = ( lim f (x)) x→a n =L (1.3.6) x→a for every positive integer n . Root law for limits: − − − − − − − − − − − − n − n lim √f (x) = n lim f (x) = √L √ x→a (1.3.7) x→a for all L if n is odd and for L ≥ 0 if n is even. We now practice applying these limit laws to evaluate a limit. Example 1.3.2A : Evaluating a Limit Using Limit Laws Use the limit laws to evaluate lim (4x + 2). x→−3 Solution Let’s apply the limit laws one step at a time to be sure we understand how they work. We need to keep in mind the requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied. lim (4x + 2) = x→−3 lim 4x + lim 2 x→−3 Apply the sum law. x→−3 = 4 ⋅ lim x + lim 2 x→−3 Apply the constant multiple law. x→−3 = 4 ⋅ (−3) + 2 = −10. Apply the basic limit results and simplify. Example 1.3.2B : Using Limit Laws Repeatedly Use the limit laws to evaluate 2 2x lim x→2 − 3x + 1 x3 + 4 . Solution To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied. 10/21/2020 1.3.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10246 2 2 2x lim x→2 lim(2 x − 3x + 1 3 x − 3x + 1) x→2 3 = 3 +4 lim(x Apply the quotient law, make sure that (2 ) + 4 ≠ 0. + 4) x→2 2 2 ⋅ lim x − 3 ⋅ lim x + lim 1 x→2 = x→2 3 lim x x→2 Apply the sum law and constant multiple law. + lim 4 x→2 x→2 2 2 ⋅ ( lim x) − 3 ⋅ lim x + lim 1 x→2 = x→2 x→2 Apply the power law. 3 ( lim x) + lim 4 x→2 x→2 2(4) − 3(2) + 1 = 3 (2 ) 1 = . Apply the basic limit laws and simplify. 4 +4 Exercise 1.3.2 − − − − − Use the limit laws to evaluate lim(2x − 1)√x + 4 . In each step, indicate the limit law applied. x→6 Hint Begin by applying the product law. Answer − − 11 √10 Limits of Polynomial and Rational Functions By now you have probably noticed that, in each of the previous examples, it has been the case that lim f (x) = f (a) . This is x→a not always true, but it does hold for all polynomials for any choice of which the rational function is defined. a and for all rational functions at all values of a for Limits of Polynomial and Rational Functions Let p(x) and q(x) be polynomial functions. Let a be a real number. Then, lim p(x) = p(a) (1.3.8) x→a p(x) lim x→a p(a) = (1.3.9) q(x) q(a) when q(a) ≠ 0 . To see that this theorem holds, consider the polynomial n p(x) = cn x n−1 + cn−1 x + ⋯ + c1 x + c0 . (1.3.10) By applying the sum, constant multiple, and power laws, we end up with lim p(x) x→a n = lim(cn x x→a n−1 + cn−1 x + ⋯ + c1 x + c0 ) n = cn ( lim x) x→a n = cn a n−1 + cn−1 ( lim x) x→a n−1 + cn−1 a + ⋯ + c1 ( lim x) + lim c0 x→a x→a + ⋯ + c1 a + c0 = p(a) It now follows from the quotient law that if p(x) and q(x) are polynomials for which q(a) ≠ 0 , 10/21/2020 1.3.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10246 then p(x) lim p(a) = x→a q(x) . (1.3.11) q(a) Example 1.3.3 : Evaluating a Limit of a Rational Function 2 Evaluate the lim 2x − 3x + 1 5x + 4 x→3 . Solution 2 Since 3 is in the domain of the rational function f (x) = 2x − 3x + 1 5x + 4 , we can calculate the limit by substituting 3 for x into the function. Thus, 2 2x − 3x + 1 lim 10 = 5x + 4 x→3 . 19 Exercise 1.3.3 Evaluate 3 lim (3 x − 2x + 7) . x→−2 Hint Use LIMITS OF POLYNOMIAL AND RATIONAL FUNCTIONS as reference Answer −13 Additional Limit Evaluation Techniques As we have seen, we may evaluate easily the limits of polynomials and limits of some (but not all) rational functions by direct substitution. However, as we saw in the introductory section on limits, it is certainly possible for lim f (x) to exist when f (a) x→a is undefined. The following observation allows us to evaluate many limits of this type: If for all x ≠ a, f (x) = g(x) over some open interval containing a , then lim f (x) = lim g(x). x→a 2 To understand this idea better, consider the limit lim x→1 x (1.3.12) x→a −1 . x −1 The function 2 x (x − 1)(x + 1) −1 f (x) = = x −1 and the function 1.3.1. 10/21/2020 g(x) = x + 1 x −1 are identical for all values of x ≠1 . The graphs of these two functions are shown in Figure 1.3.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10246 Figure 1.3.1 : The graphs of f (x) and g(x) are identical for all x ≠ 1 . Their limits at 1 are equal. We see that 2 x x→1 The limit has the form (x − 1)(x + 1) −1 lim = lim x −1 , where x→1 lim f (x)g(x) lim f (x) = 0 x→a x→a = lim(x + 1) = 2. x −1 and x→1 lim g(x) = 0 . (In this case, we say that f (x)/g(x) has the x→a indeterminate form 0/0.) The following Problem-Solving Strategy provides a general outline for evaluating limits of this type. Problem-Solving Strategy: Calculating a Limit When f(x)/g(x) has the Indeterminate Form 0/0 1. First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately using the limit laws. 2. We then need to find a function that is equal to h(x) = f (x)/g(x) for all x ≠ a over some interval containing a. To do this, we may need to try one or more of the following steps: a. If f (x) and g(x) are polynomials, we should factor each function and cancel out any common factors. b. If the numerator or denominator contains a difference involving a square root, we should try multiplying the numerator and denominator by the conjugate of the expression involving the square root. c. If f (x)/g(x) is a complex fraction, we begin by simplifying it. 3. Last, we apply the limit laws. The next examples demonstrate the use of this Problem-Solving Strategy. Example 1.3.4 illustrates the factor-and-cancel technique; Example 1.3.5 shows multiplying by a conjugate. In Example 1.3.6, we look at simplifying a complex fraction. Example 1.3.4 : Evaluating a Limit by Factoring and Canceling 2 Evaluate lim x→3 x 2 2x − 3x . − 5x − 3 Solution 2 Step 1. The function x f (x) = 2 2x − 3x is undefined for x =3 . In fact, if we substitute 3 into the function we get − 5x − 3 , which is undefined. Factoring and canceling is a good strategy: 0/0 2 x lim x→3 2 Step 2. For all x ≠ 3, x 2 2x 10/21/2020 − 3x 2 2x x = − 5x − 3 2x + 1 − 3x x(x − 3) = lim − 5x − 3 x→3 (x − 3)(2x + 1) . Therefore, 1.3.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10246 x(x − 3) x lim x→3 = lim (x − 3)(2x + 1) x→3 . 2x + 1 Step 3. Evaluate using the limit laws: x 3 lim x→3 = 2x + 1 . 7 Exercise 1.3.4 2 Evaluate x lim + 4x + 3 x2 − 9 x→−3 . Hint Follow the steps in the Problem-Solving Strategy Answer 1 3 Example 1.3.5 : Evaluating a Limit by Multiplying by a Conjugate Evaluate − − − − − √x + 2 − 1 lim x→−1 . x +1 Solution Step 1. − − − − − √x + 2 − 1 x +1 has the form 0/0 at −1. Let’s begin by multiplying by − − − − − √x + 2 + 1 , the conjugate of − − − − − √x + 2 − 1 , on the numerator and denominator: − − − − − √x + 2 − 1 lim x→−1 − − − − − √x + 2 − 1 = x +1 lim − − − − − √x + 2 + 1 ⋅ x +1 x→−1 − − − − − √x + 2 + 1 . Step 2. We then multiply out the numerator. We don’t multiply out the denominator because we are hoping that the (x + 1) in the denominator cancels out in the end: = x +1 . − − − − − (x + 1)(√ x + 2 + 1) lim x→−1 Step 3. Then we cancel: = lim x→−1 1 . − − − − − √x + 2 + 1 Step 4. Last, we apply the limit laws: 1 lim x→−1 − − − − − √x + 2 + 1 1 = . 2 Exercise 1.3.5 Evaluate lim x→5 − − − − − √x − 1 − 2 x −5 . Hint Follow the steps in the Problem-Solving Strategy 10/21/2020 1.3.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10246 Answer 1 4 Example 1.3.6 : Evaluating a Limit by Simplifying a Complex Fraction 1 1 − Evaluate lim x +1 2 . x −1 x→1 Solution 1 1 − Step 1. x +1 2 has the form 0/0 at 1. We simplify the algebraic fraction by multiplying by 2(x + 1)/2(x + 1) : x −1 1 1 1 2 x +1 − lim x→1 x +1 1 − = lim x −1 2(x + 1) 2 ⋅ x −1 x→1 . 2(x + 1) Step 2. Next, we multiply through the numerators. Do not multiply the denominators because we want to be able to cancel the factor (x − 1) : 2 − (x + 1) = lim x→1 . 2(x − 1)(x + 1) Step 3. Then, we simplify the numerator: −x + 1 = lim x→1 . 2(x − 1)(x + 1) Step 4. Now we factor out −1 from the numerator: −(x − 1) = lim x→1 . 2(x − 1)(x + 1) Step 5. Then, we cancel the common factors of (x − 1) : −1 = lim x→1 . 2(x + 1) Step 6. Last, we evaluate using the limit laws: −1 lim x→1 1 =− 2(x + 1) . 4 Exercise 1.3.6 1 +1 Evaluate lim x→−3 x +2 x +3 . Hint Follow the steps in the Problem-Solving Strategy and Answer −1 10/21/2020 1.3.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10246 Example does not fall neatly into any of the patterns established in the previous examples. However, with a little creativity, we can still use these same techniques. Example 1.3.7 : Evaluating a Limit When the Limit Laws Do Not Apply Evaluate lim ( x→0 1 5 + x ) . x(x − 5) Solution: Both 1/x and 5/x(x − 5) fail to have a limit at zero. Since neither of the two functions has a limit at zero, we cannot apply the sum law for limits; we must use a different strategy. In this case, we find the limit by performing addition and then applying one of our previous strategies. Observe that 1 5 x −5 +5 + x = x = x(x − 5) x(x − 5) . x(x − 5) Thus, 1 lim ( x→0 5 x + x 1 ) = lim x(x − 5) x→0 = lim x(x − 5) x→0 1 =− x −5 . 5 Exercise 1.3.7 Evaluate lim ( x→3 1 4 − x −3 2 x ) . − 2x − 3 Hint Use the same technique as Example 1.3.7. Don’t forget to factor x 2 − 2x − 3 before getting a common denominator. Answer 1 4 Let’s now revisit one-sided limits. Simple modifications in the limit laws allow us to apply them to one-sided limits. For example, to apply the limit laws to a limit of the form lim h(x), we require the function h(x) to be defined over an open − x→a interval of the form (b, a); for a limit of the form , we require the function h(x) to be defined over an open interval lim h(x) + x→a of the form (a, c). Example 1.3.8A illustrates this point. Example 1.3.8A : Evaluating a One-Sided Limit Using the Limit Laws Evaluate each of the following limits, if possible. a. − − − − − lim √x − 3 − x→3 b. − − − − − lim √x − 3 + x→3 Solution − − − − − Figure illustrates the function f (x) = √x − 3 and aids in our understanding of these limits. 10/21/2020 1.3.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10246 − − − − − Figure 1.3.2 : The graph shows the function f (x) = √x − 3 . − − − − − a. The function f (x) = √x − 3 is defined over the interval [3, +∞). Since this function is not defined to the left of 3, we − − − − − − − − − − cannot apply the limit laws to compute lim √x − 3 . In fact, since f (x) = √x − 3 is undefined to the left of 3, − x→3 − − − − − lim √x − 3 does not exist. − x→3 − − − − − b. Since f (x) = √x − 3 is defined to the right of 3, the limit laws do apply to − − − − − lim √x − 3 . By applying these limit laws + x→3 we obtain − − − − − lim √x − 3 = 0 . + x→3 In Example 1.3.8B we look at one-sided limits of a piecewise-defined function and use these limits to draw a conclusion about a two-sided limit of the same function. Example 1.3.8B : Evaluating a Two-Sided Limit Using the Limit Laws For f (x) = { a. 4x − 3, if x < 2 2 (x − 3 ) , if x ≥ 2 , evaluate each of the following limits: lim f (x) − x→2 b. lim f (x) + x→2 c. lim f (x) x→2 Solution Figure illustrates the function f (x) and aids in our understanding of these limits. Figure 1.3.3 : This graph shows a function f (x). a. Since f (x) = 4x − 3 for all x in (−∞, 2), replace f (x) in the limit with 4x − 3 and apply the limit laws: lim f (x) = lim (4x − 3) = 5 − x→2 − x→2 b. Since f (x) = (x − 3) for all x in (2, +∞), replace f (x) in the limit with (x − 3) and apply the limit laws: 2 2 2 lim f (x) = lim (x − 3 ) + x→2 c. Since lim f (x) = 5 − x→2 10/21/2020 and lim f (x) = 1 + x→2 = 1. − x→2 , we conclude that lim f (x) does not exist. x→2 1.3.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10246 f (x) We now turn our attention to evaluating a limit of the form lim x→a , where lim f (x) = K , where K ≠ 0 and x→a g(x) lim g(x) = 0 . x→a That is, f (x)/g(x) has the form K/0, K ≠ 0 at a. Example 1.3.9 : Evaluating a Limit of the Form K/0, K Evaluate x −3 lim 2 x − x→2 Using the Limit Laws ≠ 0 . − 2x Solution Step 1. After substituting in x = 2 , we see that this limit has the form −1/0. That is, as x approaches 2 from the left, the numerator approaches −1 ; and the denominator approaches 0 . Consequently, the magnitude of x −3 x(x − 2) becomes infinite. To get a better idea of what the limit is, we need to factor the denominator: x −3 lim − x→2 x −3 = lim 2 x − − 2x x→2 x(x − 2) Step 2. Since x − 2 is the only part of the denominator that is zero when 2 is substituted, we then separate 1/(x − 2) from the rest of the function: x −3 = lim 1 ⋅ x − x→2 x −2 Step 3. Using the Limit Laws, we can write: x −3 = ( lim x→2 x −3 Step 4. lim +∞ 1 =− x − x→2 2 1 and lim − x→2 = −∞ x −2 1 ) ⋅ ( lim x − − x→2 ). x −2 . Therefore, the product of (x − 3)/x and 1/(x − 2) has a limit of : x −3 lim − x→2 2 x = +∞. − 2x Exercise 1.3.9 x +2 Evaluate lim x→1 2 . (x − 1) Solution Use the methods from Example 1.3.9. Answer +∞ Example 1.3.10 Evaluate lim x→4 (x − 3) . (4 − x)(x + 3) Solution 10/21/2020 1.3.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10246 If we substitute (x − 3) into x =4 , we get (4 − x)(x + 3) (x − 3) lim x→4 (4 − x)(x + 3) 1 . In this case we use the following sign chart to decide 0 . Notice that the point(s) where the denominator zero is x =4 and x = −3 , and the point(s) where the numerator zero is x = 3 . (−∞, −3) (−3, 3) (x − 3) − − (4 − x) + (x + 3) − (3, 4) (4, ∞) + + + + + + − + + Thus (x − 3) lim − x→4 =∞ , and (x − 3) lim + (4 − x)(x + 3) x→4 = −∞ . Therefore (4 − x)(x + 3) - (x − 3) lim x→4 = DN E. (4 − x)(x + 3) Exercise 1.3.10 Evaluate lim x→3 (x + 3) . (4 − x)(x − 3) Answer Add texts here. Do not delete this text first. Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 1.3.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10246 1.3E: Exercises Exercise 1.3E. 1 In exercises 1 - 4, use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s). 1) lim (4x 2 − 2x + 3) x→0 Answer Use constant multiple law and difference law: 2 lim (4 x 2 − 2x + 3) = 4 lim x x→0 2) lim 3 x 2 + 3x − 2 lim x + lim 3 = 0 + 0 + 3 = 3 x→0 x→0 +5 4 − 7x x→1 3) x→0 −−−−−−−− − 2 lim √x − 6x + 3 x→−2 Answer −−−−−−−−−−−−− − −−−−−−−− − − − 2 2 lim √x − 6x + 3 = √ lim (x − 6x + 3) = √19 Use root law: x→−2 4) x→−2 2 lim (9x + 1 ) x→−1 Exercise 1.3E. 2 In exercises 1 - 4, use direct substitution to evaluate the limit of each continuous function. 1) lim x 2 x→7 Answer 2 lim x = 49 x→7 2) 2 lim (4 x − 1) x→−2 3) lim x→0 1 1 + sin x Answer 1 lim x→0 4) lim = 1 1 + sin x 2 − 7x x→1 x +6 Answer 2 − 7x lim x→1 5 = − x +6 7 Exercise 1.3E. 3 10/21/2020 1.3E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10659 In exercises 1 - 9, use direct substitution to show that each limit leads to the indeterminate form evaluate the limit analytically. 2 x 1) lim 0/0 . Then, − 16 x −4 x→4 Answer 2 x − 16 lim 16 − 16 0 = = x −4 x→4 4 −4 2 then, lim x ; 0 (x + 4)(x − 4) − 16 = lim x −4 x→4 = lim(x + 4) = 4 + 4 = 8 x −4 x→4 x→4 x −2 2) lim 2 x→2 x − 2x 3x − 18 3) lim 2x − 12 x→6 Answer 3x − 18 lim then, lim x→6 2 (1 + h ) 12 − 12 ; 0 3(x − 6) 3x − 18 = lim 2x − 12 3 = lim 2(x − 6) x→6 x→6 3 = 2 2 −1 h h→0 t −9 5) lim t→9 0 = 2x − 12 x→6 4) lim 18 − 18 = √t − 3 Answer t −9 9 −9 lim x→9 0 = = √t − 3 0 t −9 then, lim t −9 (t − 9)(√t + 3) √t + 3 = lim t→9 1 ; 3 −3 √t − 3 t→9 = lim √t − 3 t→9 √t + 3 – = lim(√t + 3) = √9 + 3 = 6 t −9 t→9 1 − a+h 6) lim 3 7) lim x→1 x 2 x , where a is a real-valued constant −1 −1 2 8) a h h→0 2x + 3x − 2 lim x→1/2 2x − 1 Answer 2 2x 1 + 3x − 2 lim = 2x − 1 x→1/2 2 then, 2x + 3 2 −2 1 −1 ; 0 (2x − 1)(x + 2) = 2x − 1 0 = + 3x − 2 lim x→1/2 9) 2 lim x→1/2 1 = 2x − 1 lim (x + 2) = x→1/2 5 +2 = 2 2 − − − − − √x + 4 − 1 lim x→−3 x +3 Exercise 1.3E. 4 10/21/2020 1.3E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10659 In exercises 1 - 4, use direct substitution to obtain an undefined expression. Then, use the method used in Example 9 of this section to simplify the function and determine the limit. 2 1) 2x lim − x→−2 2 x + 7x − 4 +x −2 Answer −∞ 2) + x→−2 + 7x − 4 2 +x −2 x 2 3) 2 2x lim 2x lim 2 − x x→1 + 7x − 4 +x −2 Answer −∞ 2 4) 2x lim 2 + x x→1 + 7x − 4 +x −2 Exercise 1.3E. 5 In exercises 1 - 8, assume that lim f (x) = 4, lim g(x) = 9 x→6 x→6 , and lim h(x) = 6 . Use these three facts and the limit x→6 laws to evaluate each limit. 1) lim 2f (x)g(x) x→6 Answer lim 2f (x)g(x) = 2 ( lim f (x)) ( lim g(x)) = 2(4)(9) = 72 x→6 2) lim x→6 x→6 g(x) − 1 f (x) x→6 3) lim (f (x) + x→6 1 g(x)) 3 Answer 1 lim (f (x) + 1 g(x)) = lim f (x) + 3 x→6 x→6 1 lim g(x) = 4 + 3 x→6 (9) = 7 3 3 4) lim (h(x)) x→6 2 − −−−−−−− − 5) lim √g(x) − f (x) x→6 Answer − −−−−−−− − lim √g(x) − f (x) = x→6 − −−−−−−−−−−−−− − −−− − – lim g(x) − lim f (x) = √9 − 4 = √5 √ x→6 x→6 6) lim x ⋅ h(x) x→6 7) lim[(x + 1) ⋅ f (x)] x→6 Answer 10/21/2020 1.3E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10659 lim[(x + 1)f (x)] = ( lim(x + 1)) ( lim f (x)) = 7(4) = 28 x→6 x→6 x→6 8) lim(f (x) ⋅ g(x) − h(x)) x→6 Exercise 1.3E. 6 [T] In exercises 1 - 3, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits. 1) f (x) = { a. 2 x , x ≤3 x + 4, x >3 lim f (x) − x→3 b. lim f (x) + x→3 Answer a. 9; b.7 1) g(x) = { a. 3 x − 1, 1, x ≤0 x >0 lim g(x) − x→0 b. lim g(x) + x→0 3) h(x) = { a. 2 x − 2x + 1, 3 − x, x <2 x ≥2 lim h(x) − x→2 b. lim h(x) + x→2 Exercise 1.3E. 7 In exercises 1 - 8, use the following graphs and the limit laws to evaluate each limit. 10/21/2020 1.3E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10659 1) lim (f (x) + g(x)) + x→−3 2) lim (f (x) − 3g(x)) − x→−3 Answer lim (f (x) − 3g(x)) = − 3) lim f (x) − 3 lim g(x) = 0 + 6 = 6 − x→−3 x→−3 f (x)g(x) x→0 4) lim − x→−3 3 2 + g(x) lim x→−5 f (x) Answer 2 + ( lim g(x)) 2 + g(x) lim x→−5 x→−5 = f (x) 2 +0 = lim f (x) =1 2 x→−5 10/21/2020 1.3E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10659 5) lim(f (x)) 2 x→1 − −−−−−−− − 6) lim √f (x) − g(x) x→1 Answer − −−−−−−− − 3 lim √f (x) − g(x) = x→1 7) 3 − −−−−−−−−−−−−− − −−− − 3 – 3 lim f (x) − lim g(x) = √2 + 5 = √7 √ x→1 x→1 lim (x ⋅ g(x)) x→−7 8) lim [x ⋅ f (x) + 2 ⋅ g(x)] x→−9 Answer lim (xf (x) + 2g(x)) = ( lim x) ( lim f (x)) + 2 lim g(x) = (−9)(6) + 2(4) = −46 x→−9 x→−9 x→−9 x→−9 Exercise 1.3E. 8 1) [T] In physics, the magnitude of an electric field generated by a point charge at a distance r in a vacuum is governed by Coulomb’s law: E(r) = q 4πε0 r2 , where E represents the magnitude of the electric field, q is the charge of the particle, r is the distance between the particle and where the strength of the field is measured, and 2 8.988 × 109N ⋅ m / C 2 . 1 4πε0 a. Use a graphing calculator to graph E(r) given that the charge of the particle is q = 10 −10 b. Evaluate lim E(r) is Coulomb’s constant: . . What is the physical meaning of this quantity? Is it physically relevant? Why are you + r→0 evaluating from the right? Answer a. b. ∞. The magnitude of the electric field as you approach the particle q becomes infinite. It does not make physical sense to evaluate the negative distance. 2) [T] The density of an object is given by its mass divided by its volume: ρ = m/V . 10/21/2020 1.3E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10659 a. Use a calculator to plot the volume as a function of density (V of mass 8 kg (m = 8 ). b. Evaluate lim V (ρ) = m/ρ) , assuming you are examining something and explain the physical meaning. + x→0 Exercise 1.3E. 9 Evaluate the following: 1. lim x 2 − 3x + 7 x→3 7 x −3 2. lim ( ) x +5 x→π 3 3. lim 4 x −8x x→3 4. lim ln(1 + x) x→0 2 x 5. lim + 3x + 5 2 x→π 5x 6. lim 1 −x x→π 2 x 7. lim x→6 2 x 8. lim x→0 x + 2x 2 x − 2x 2 9. lim − 4x − 12 − 13x + 42 2 x + 6x − 16 2 x x→2 − 3x + 2 2 10. − 2x − 3 3x + 1 x lim x→2 − 5x − 14 x2 + 10x + 16 2 x 11. lim x→−2 2 x 2 x 12. lim x→−1 − 5x − 14 + 10x + 16 + 9x + 8 x2 − 6x − 7 Answer Under Construction Exercise 1.3E. 10 − − − − − √x + 4 − 2 lim x x→0 Answer − − − − − √x + 4 − 2 lim x→0 = lim x→0 = lim x→0 = lim x→0 = lim x→0 10/21/2020 −−− − √0 + 4 − 2 = 0 =[ x 0 ] 0 − − − − − − − − − − (√x + 4 − 2)(√x + 4 + 2) − − − − − x(√x + 4 + 2) ((x + 4) − 4) − − − − − x(√x + 4 + 2) x − − − − − x(√x + 4 + 2) 1 − − − − − (√x + 4 + 2) 1 = −−− − (√0 + 4 + 2) 1 = 4 . 1.3E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10659 Exercise 1.3E. 11 Evaluate the following limits: 1 1. lim x→4 |4 − x| Answer ∞ 2. − −−− − 2 √1 − x lim − x→−1 Answer 0 3. − −−− − √1 − x2 lim + x→−1 Answer 0 |x − 2| 4. lim x→2 2 x +x −6 Answer dne 1 5. lim x→2 x − 1 2 x −2 Answer 1 4 6. f (x) = { 2 x if x ≤ 1 2x if x > 1 Answer Under Construction Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution Noncommercial (BY-NC) License. http://www.apexcalculus.com/ Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 1.3E.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10659 1.4: Limits at Infinity and Horizontal Asymptotes : End Behavior Think Out Loud What do mean by the end behavior? Learning Objectives Calculate the limit of a function as x increases or decreases without bound. Recognize a horizontal asymptote on the graph of a function. Estimate the end behaviour of a function as x increases or decreases without bound. Recognize an oblique asymptote on the graph of a function. We have learned about lim f (x) = L, where a is a real number. In this section we would like to explore a to be ∞ or −∞ . x→a Limits at Infinity and Horizontal Asymptotes Recall that lim f (x) = L means f (x) becomes arbitrarily close to L as long as x is sufficiently close to a . We can extend this x→a idea to limits at infinity. For example, consider the function f (x) = 2 + . As can be seen graphically in Figure 1.4.1 and numerically in Table 1.4.1, as the values of x get larger, the values of f (x) approach 2. We say the limit as x approaches ∞ of f (x) is 2 and write lim f (x) = 2 . Similarly, for x < 0 , as the values |x| get larger, the values of f (x) approaches 2 . We say 1 x x→∞ the limit as x approaches −∞ of f (x) is 2 and write lim f (x) = 2 . x→a Figure 1.4.1 :The function approaches the asymptote y = 2 as x approaches ±∞ . Table 1.4.1 : Values of a function f as x → ±∞ x 2 + 1 x x 2 + 1 x 10 100 1,000 10,000 2.1 2.01 2.001 2.0001 −10 −100 −1000 −10,000 1.9 1.99 1.999 1.9999 More generally, for any function f , we say the limit as x → ∞ of f (x) is L if f (x) becomes arbitrarily close to L as long as x is sufficiently large. In that case, we write lim f (x) = L. Similarly, we say the limit as x → −∞ of f (x) is L if f (x) x→a becomes arbitrarily close to L as long as x <0 and |x| is sufficiently large. In that case, we write lim f (x) = L . We now x→−∞ look at the definition of a function having a limit at infinity. 10/20/2020 1.4.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10254 Definition: Limit at Infinity (Informal) If the values of f (x) become arbitrarily close to L as x becomes sufficiently large, we say the function infinity and write f has a limit at lim f (x) = L. (1.4.1) x→∞ If the values of f (x) becomes arbitrarily close to L for x < 0 as |x| becomes sufficiently large, we say that the function f has a limit at negative infinity and write lim f (x) = L. (1.4.2) x→−∞ If the values f (x) are getting arbitrarily close to some finite value L as x → ∞ or x → −∞ , the graph of f approaches the line y = L . In that case, the line y = L is a horizontal asymptote of f (Figure 1.4.2). For example, for the function 1 f (x) = , since x lim f (x) = 0 , the line y = 0 is a horizontal asymptote of f (x) = x→∞ 1 . x Figure 1.4.2 : (a) As x → ∞ , the values of f are getting arbitrarily close to L. The line y = L is a horizontal asymptote of f . (b) As x → −∞ , the values of f are getting arbitrarily close to M . The line y = M is a horizontal asymptote of f . Definition: Horizontal Asymptote If lim f (x) = L x→∞ or lim f (x) = L , we say the line y = L is a horizontal asymptote of f . x→−∞ A function cannot cross a vertical asymptote because the graph must approach infinity (or −∞ ) from at least one direction as x approaches the vertical asymptote. However, a function may cross a horizontal asymptote. In fact, a function may cross a horizontal asymptote an unlimited number of times. For example, the function intersects the horizontal asymptote decreasing amplitude. y =1 (cos x) f (x) = +1 x shown in Figure 1.4.3 an infinite number of times as it oscillates around the asymptote with ever- Figure 1.4.3 : The graph of f (x) = (cos x)/x + 1 crosses its horizontal asymptote y = 1 an infinite number of times. Example 1.4.1 Consider f (x) = 1/x , as shown in Figure appropriate, and descriptive, to state that 2 10/20/2020 . Note how, as 1.4.4 x approaches 0, 1.4.2 CC-BY-NC-SA f (x) grows very, very large. It seems https://math.libretexts.org/@go/page/10254 1 lim x→0 = ∞. x2 (1.4.3) Also note that as x gets very large, f (x) gets very, very small. We could represent this concept with notation such as 1 lim x→∞ 2 = 0. (1.4.4) x Figure 1.4.4:: Graphing f (x) = 1/x for values of x near 0 . 2 Limit Laws Let f (x) = k . Then lim f (x) = k , and lim x→∞ f (x) = k as shown in the following graph. x→−∞ Figure 1.4.1 :Graphing of f (x) = 1. (Copyright; Pamini Thangarajah via Onenote) The algebraic limit laws that we introduced in Introduction to Limits also apply to limits at infinity. We illustrate how to use these laws to compute several limits at infinity. Example 1.4.2 : Computing Limits at Infinity For each of the following functions f , valuate lim f (x) and x→∞ where f (x) = 5 − lim . Determine the horizontal asymptote(s) for f , f (x) x→−∞ 2 2 x Solution Using the algebraic limit laws, we have 2 lim (5 − x→∞ Similarly, lim f (x) = 5 x→∞ 2 1 ) = lim 5 − 2 ( lim x . Therefore, f (x) = x→∞ 5 −2 2 x→∞ 1 ) ⋅ ( lim x x→∞ ) = 5 − 2 ⋅ 0 = 5. x has a horizontal asymptote of y = 5 and f approaches this horizontal x asymptote as x → ±∞ as shown in the following graph. 10/20/2020 1.4.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10254 Figure 1.4.4 : This function approaches a horizontal asymptote as x → ±∞. Infinite Limits at Infinity Sometimes the values of a function f become arbitrarily large as x → ∞ (or as x → −∞ ). In this case, we write lim f (x) = ∞ (or lim f (x) = ∞ ). On the other hand, if the values of f are negative but become arbitrarily large in x→∞ x→−∞ magnitude as x → ∞ (or as x → −∞ ), we write lim f (x) = −∞ (or x→∞ For example, consider the function arbitrarily large. Therefore, lim x 3 lim f (x) = −∞ ). x→−∞ . As seen in Table 1.4.2 and Figure 1.4.8, as x → ∞ the values f (x) become = ∞ . On the other hand, as x → −∞ , the values of f (x) = x are negative but become 3 f (x) = x 3 x→∞ arbitrarily large in magnitude. Consequently, 3 lim x = −∞. x→−∞ Table 1.4.2 x x 3 x x 3 10 20 50 100 1000 1000 8000 125,000 1,000,000 1,000,000,000 −10 −20 −50 −100 −1000 −1000 −8000 −125,000 −1,000,000 −1,000,000,000 Values of a power function as x → ±∞ Figure 1.4.8 : For this function, the functional values approach infinity as x → ±∞. Definition: Infinite Limit at Infinity (Informal) We say a function f has an infinite limit at infinity and write lim f (x) = ∞. (1.4.5) x→∞ if f (x) becomes arbitrarily large for write x sufficiently large. We say a function has a negative infinite limit at infinity and lim f (x) = −∞. (1.4.6) x→∞ 10/20/2020 1.4.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10254 if f (x) < 0 and |f (x)| becomes arbitrarily large for x sufficiently large. Similarly, we can define infinite limits as x → −∞. End Behavior The behavior of a function as x → ±∞ is called the function’s end behavior. At each of the function’s ends, the function could exhibit one of the following types of behavior: 1. The function f (x) approaches a horizontal asymptote y = L . 2. The function f (x) → ∞ or f (x) → −∞. 3. The function does not approach a finite limit, nor does it approach oscillatory behavior. ∞ or −∞ . In this case, the function may have some Let’s consider several classes of functions here and look at the different types of end behaviors for these functions. Limit at infinity for Polynomial Functions Consider the power function f (x) = x where n is a positive integer. From Figure 1.4.5 and Figure 1.4.6, we see that n n lim x = ∞; n = 1, 2, 3, … (1.4.7) x→∞ and lim x→−∞ n x ={ ∞, n = 2, 4, 6, … −∞, n = 1, 3, 5, … . Figure 1.4.5 : For power functions with an even power of n , (1.4.8) n lim x = ∞ = x→∞ Figure 1.4.6 : For power functions with an odd power of n , n lim x x→∞ 10/20/2020 1.4.5 CC-BY-NC-SA = ∞ lim n x . x→−∞ and lim n x = −∞. x→−∞ https://math.libretexts.org/@go/page/10254 Using these facts, it is not difficult to evaluate n lim c x and x→∞ c >0 n lim cx , where c is any constant and n is a positive integer. If x→−∞ , the graph of y = cx is a vertical stretch or compression of y = x n n n lim c x and n = lim x x→∞ x→∞ n lim cx = x→−∞ and therefore , if c > 0 . n lim x x→−∞ If c < 0, the graph of y = cx is a vertical stretch or compression combined with a reflection about the x-axis, and therefore n n lim c x If c = 0, y = cx n = 0, and n = − lim x x→∞ x→∞ in which case n lim c x n lim cx =− x→−∞ =0 = lim x→∞ lim if c < 0. n x x→−∞ n cx . x→−∞ Example 1.4.3 : Limits at Infinity for Power Functions For each function f , evaluate lim f (x) and x→∞ . lim f (x) x→−∞ a. f (x) = −5x b. f (x) = 2x 3 4 Solution a. Since the coefficient of x is −5, the graph of f (x) = −5x involves a vertical stretch and reflection of the graph of y =x about the x-axis. Therefore, lim (−5x ) = −∞ and lim (−5x ) = ∞ . 3 3 3 3 3 x→∞ x→−∞ b. Since the coefficient of x is 2, the graph of lim 2 x = ∞ and lim 2 x = ∞ . 4 4 4 f (x) = 2x is a vertical stretch of the graph of 4 y =x . Therefore, 4 x→∞ x→−∞ Exercise 1.4.3 Let f (x) = −3x . Find 4 . lim f (x) x→∞ Hint The coefficient −3 is negative. Answer −∞ We now look at how the limits at infinity for power functions can be used to determine lim f (x) for any polynomial x→±∞ function f . Consider a polynomial function n f (x) = an x of degree n ≥ 1 so that a n ≠ 0. n + an−1 x 1 0 −1 +… +a x +a (1.4.9) Factoring, we see that n f (x) = an x (1 + an−1 1 an +… + x a1 1 an xn−1 + a0 n ). (1.4.10) a As x → ±∞, all the terms inside the parentheses approach zero except the first term. We conclude that lim x→±∞ For example, the function f (x) = 5x 1.4.3. 3 10/20/2020 2 − 3x +4 f (x) = lim n an x . (1.4.11) x→±∞ behaves like g(x) = 5x as x → ±∞ as shown in Figure 1.4.7 and Table 3 1.4.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10254 Figure 1.4.7 : The Limit at infinity of a polynomial is determined by the behavior of the term with the largest exponent. Table 1.4.3 : A polynomial’s Limit at infinity is determined by the term with the largest exponent x f(x) = 5 x 3 − 3x g(x) = 5x 2 +4 3 x f(x) = 5 x 3 − 3x g(x) = 5x 2 +4 3 10 100 1000 4704 4,970,004 4,997,000,004 5000 5,000,000 5,000,000,000 −10 −100 −000 −5296 −5,029,996 −5,002,999,996 −5000 −5,000,000 −5,000,000,000 Summary Basic Results Basic Results 1. lim k =k , where k is a constant. x→±∞ 2. n lim x , for all n ∈ N . =∞ x→∞ 3. lim n x =∞ , when n is even. x→−∞ 4. n lim x = −∞ , when n is odd. x→∞ 5. lim x→±∞ 6. lim n an x n + an−1 x 1 x→±∞ =0 n x 1 0 −1 +… +a x +a n = limx→±∞ an x . , for all n ∈ N . Limit at infinity for Rational Functions The Limit at infinity for rational functions and functions involving radicals is a little more complicated than for polynomials. In Example, we show that the limits at infinity of a rational function p(x) f (x) = q(x) depend on the relationship between the degree of the numerator and the degree of the denominator. To evaluate the limits at infinity for a rational function, we divide the numerator and denominator by the highest power of x appearing in the denominator. This determines which term in the overall expression dominates the behavior of the function at large values of x. Example 1.4.5 : Determining Limit at infinity for Rational Functions For each of the following functions, determine the limits as x → ∞ and x → −∞. Then, use this information to describe the Limit at infinity of the function. a. 3x − 1 f (x) = b. f (x) = 2x + 5 2 3x 3 4x 10/20/2020 (Note: The degree of the numerator and the denominator are the same.) + 2x (Note: The degree of numerator is less than the degree of the denominator.) − 5x + 7 1.4.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10254 2 c. 3x + 4x f (x) = x +2 in the denominator is x. Therefore, dividing the numerator and denominator by x and applying the algebraic limit laws, we see that Solution a. The highest power of x in the denominator is x. Therefore, dividing the numerator and denominator by x and applying the algebraic limit laws, we see that 3 − 1/x 3x − 1 lim = 2x + 5 x→±∞ lim x→±∞ 2 + 5/x limx→±∞ (3 − 1/x) = limx→±∞ (2 + 5/x) limx→±∞ 3 − limx→±∞ 1/x = limx→±∞ 2 + limx→±∞ 5/x 3 −0 = 3 = 2 +0 3 Since lim x→±∞ f (x) = 2 , we know that y = 3 2 . 2 is a horizontal asymptote for this function as shown in the following graph. Figure 1.4.8 : The graph of this rational function approaches a horizontal asymptote as x → ±∞. b. Since the largest power of x appearing in the denominator is doing so and applying algebraic limit laws, we obtain 2 3x lim x→±∞ 3 4x , divide the numerator and denominator by 2 3/x + 2/x + 2x = − 5x + 7 3 x lim x→±∞ 2 4 − 5/ x 3 x . After 3 ⋅ 0 +2 ⋅ 0 3 + 7/ x = = 0. 4 −5 ⋅ 0 +7 ⋅ 0 Therefore f has a horizontal asymptote of y = 0 as shown in the following graph. 10/20/2020 1.4.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10254 Figure 1.4.9 : The graph of this rational function approaches the horizontal asymptote y = 0 as x → ±∞. c. Dividing the numerator and denominator by x, we have 2 3x + 4x lim x→±∞ 3x + 4 = x +2 lim x→±∞ . 1 + 2/x As x → ±∞ , the denominator approaches 1. As x → ∞ , the numerator approaches +∞ . As x → −∞ , the numerator approaches −∞ . Therefore lim f (x) = ∞ , whereas lim f (x) = −∞ as shown in the following figure. x→∞ x→−∞ Figure 1.4.10 : As → ∞ , the values f (x) → ∞. As x → −∞ , the values f (x) → −∞. Exercise 1.4.5 2 Evaluate 3x lim x→±∞ 2 5x + 2x − 1 2 and use these limits to determine the Limit at infinity of f (x) = − 4x + 7 3x 2 5x + 2x − 2 . − 4x + 7 Hint Divide the numerator and denominator by x . 2 Answer 10/20/2020 1.4.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10254 3 5 We can summarize the results of Example to make the following conclusion regarding Limit at infinity for rational functions. Consider a rational function n p(x) f (x) = = q(x) where a ≠0 n and b m an x m bm x n−1 + an−1 x + … + a1 x + a0 m−1 + bm−1 x , + … + b1 x + b0 ≠ 0. 1. If the degree of the numerator is the same as the degree of the denominator (n = m), then f has a horizontal asymptote of y = a /b as x → ±∞. 2. If the degree of the numerator is less than the degree of the denominator (n < m), then f has a horizontal asymptote of y = 0 as x → ±∞. 3. If the degree of the numerator is greater than the degree of the denominator (n > m), then f does not have a horizontal asymptote. The limits at infinity are either positive or negative infinity, depending on the signs of the leading terms. In addition, using long division, the function can be rewritten as n m p(x) f (x) = r(x) = g(x) + q(x) where the degree of r(x) is less than the degree of , (1.4.12) q(x) . As a result, q(x) lim . Therefore, the values of r(x)/q(x) = 0 x→±∞ approach zero as x → ±∞ . If the degree of p(x) is exactly one more than the degree of ), the function g(x) is a linear function. In this case, we call g(x) an oblique asymptote. [f (x) − g(x)] n = m +1 q(x) (i.e., Now let’s consider the Limit at infinity for functions involving a radical. Example 1.4.6 : Determining Limit at infinity for a Function Involving a Radical Find the limits as x → ∞ and x → −∞ for f (x) = 3x − 2 − −−−− − √4 x2 + 5 . Solution Let’s use the same strategy as we did for rational functions: divide the numerator and denominator by a power of x. To − −−−− − determine the appropriate power of x, consider the expression √4x + 5 in the denominator. Since 2 − − − − −−−− − √ 4 x2 + 5 ≈ √4x2 = 2|x| for large values of x in effect x appears just to the first power in the denominator. Therefore, we divide the numerator and − − denominator by |x|. Then, using the fact that |x| = x for x > 0, |x| = −x for x < 0 , and |x| = √x for all x, we calculate the limits as follows: 2 (1/|x|)(3x − 2) 3x − 2 lim x→∞ − −−−− − = lim − −−−− − x→∞ 2 √ 4 x2 + 5 (1/|x|)√ 4 x + 5 (1/x)(3x − 2) = lim x→∞ − − − − − − − − − − − − − 2 2 √ (1/ x )(4 x + 5) 3 − 2/x = lim x→∞ 10/20/2020 − − − − − − − √ 4 + 5/x2 3 = – √4 1.4.10 CC-BY-NC-SA 3 = 2 https://math.libretexts.org/@go/page/10254 3x − 2 lim x→−∞ − −−−− − √ 4 x2 + 5 (1/|x|)(3x − 2) = lim x→−∞ − −−−− − (1/|x|)√ 4 x2 + 5 (−1/x)(3x − 2) = lim x→−∞ − − − − − − − − − − − − − 2 2 √ (1/ x )(4 x + 5) −3 + 2/x = lim x→−∞ Therefore, x → −∞ approaches the horizontal asymptote as shown in the following graph. f (x) y = −3 −3 . − − − − − − − = – = 2 2 √4 √ 4 + 5/x 3 2 as x → ∞ and the horizontal asymptote y =− 3 2 as Figure 1.4.11 : This function has two horizontal asymptotes and it crosses one of the asymptotes. Exercise 1.4.6 − −−−− − √3 x2 + 4 Evaluate lim x +6 x→∞ . Hint Divide the numerator and denominator by |x|. Answer – ±√3 Note Below we will show two ways of solving limits at the infinity of rational functions. Example 1.4.7 Find 2x + 3 . lim x→∞ 3x − 2 Solution: Method 1: Divide both numerator and denominator by highest power of x of the polynomial in the denominator. 2x + 3 lim x→∞ 3x − 2 2x + 3 ( ) x = lim 3x − 2 x→∞ ( ) x 2x ( = lim ( ) x 3x x→∞ 2 − x 10/20/2020 3 + x ) x 1.4.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10254 3 (2 + ) x = lim 2 x→∞ (3 − ) x 2 = . 3 Method 2: 2x + 3 lim 2x = lim 3x − 2 x→∞ 2 = x→∞ 3x . 3 Exercise 1.4.7 2x + 3 Find lim 3x − 2 x→−∞ . Answer 2x + 3 2x lim x→−∞ = 3x − 2 lim x→−∞ = df rac23. 3x Example 1.4.8 x +9 Find lim x→−∞ − −−−−−−−− − √4 x2 + 3x + 2 . Solution x +9 lim x→−∞ − −−−−−−−− − √4 x2 + 3x + 2 x = lim x→−∞ − − − √4x2 x |x| = lim x→−∞ − − − √4x2 |x| x = − − −x lim x→−∞ − − − √4x2 , Note that √x 2 = |x| . − − √x2 −1 = . 2 Exercise 1.4.8 x +9 Find lim x→∞ − −−−−−−−− − √4 x2 + 3x + 2 . Answer 1 = . 2 Example 1.4.9 : 10/20/2020 1.4.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/10254 Find − −−−− − − −−−− − 2 2 lim √x + 3x − √x + 4x . x→∞ Solution: Exercise 1.4.9 Find lim − −−−− − − −−−− − 2 2 √x + 3x − √x + 4x x→−∞ Answer −1 2 . Example 1.4.10 : Find the horizontal asymptote(s) if any of the function x +9 f (x) = − −−−−−−−− − √4 x2 + 3x + 2 . Solution: −1 y = 2 , and y == 1 2 are the horizontal asymptotes . See Example 1.4.3 and Exercise 1.4.3. Determining End Behavior for Transcendental Functions The six basic trigonometric functions are periodic and do not approach a finite limit as x → ±∞. For example, sin x oscillates between 1 and −1 (Figure 1.4.19). The tangent function x has an infinite number of vertical asymptotes as x → ±∞ ; therefore, it does not approach a finite limit nor does it approach ±∞ as x → ±∞ as shown in Figure 1.4.20. Figure 1.4.19 : The function f (x) = sin x oscillates between 1 and −1 as x → ±∞ Figure 1.4.20 : The function f (x) = tan x does not approach a limit and does not approach ±∞ as x → ±∞ Recall that for any base b > 0, b ≠ 1, the function y = b is an exponential function with domain (−∞, ∞) and range (0, ∞). If b > 1, y = b is increasing over (−∞, ∞). If 0 1 . Therefore, f (x) = e is increasing on `(−∞, ∞) and the range is `(0, ∞). x x x x 10/20/2020 x 1.4.13 CC-BY-NC-SA https://math.libretexts.org/@go/page/10254 The exponential function Figure 1.4.21. f (x) = e x approaches ∞ as and approaches x → ∞ 0 as x → −∞ as shown in Table 1.4.4 and Table 1.4.4 : End behavior of the natural exponential function x x e −5 −2 0 2 5 0.00674 0.135 1 7.389 148.413 Figure 1.4.21 : The exponential function approaches zero as x → −∞ and approaches ∞ as x → ∞. Recall that the natural logarithm function f (x) = ln(x) is the inverse of the natural exponential function y = e . Therefore, the domain of f (x) = ln(x) is (0, ∞) and the range is (−∞, ∞). The graph of f (x) = ln(x) is the reflection of the graph of y =e about the line y = x . Therefore, ln(x) → −∞ as x → 0 and ln(x) → ∞ as x → ∞ as shown in Figure 1.4.22 and Table 1.4.5. x x + Table 1.4.5 : End behavior of the natural logarithm function x 0.01 0.1 1 10 100 ln(x) −4.605 −2.303 0 2.303 4.605 Figure 1.4.22 : The natural logarithm function approaches ∞ as x → ∞. Example 1.4.7 : Determining End Behavior for a Transcendental Function Find the limits as x → ∞ and x → −∞ for f (x) = 2 + 3e 7 − 5e x x and describe the end behavior of f . Solution To find the limit as x → ∞, divide the numerator and denominator by e : x 2 + 3e lim f (x) = lim x→∞ x→∞ 7 − 5e x x x (2/ e ) + 3 = lim x→∞ As shown in Figure 1.4.21, e x → ∞ x (7/ e ) − 5. as x → ∞ . Therefore, 2 lim x→∞ 10/20/2020 e x 7 = 0 = lim x→∞ e x . 1.4.14 CC-BY-NC-SA https://math.libretexts.org/@go/page/10254 3 We conclude that lim (x) = − 5 x→∞f , and the graph of f approaches the horizontal asymptote y = − the limit as x → −∞ , use the fact that e x → 0 as x → −∞ to conclude that 2 lim f (x) = 7 x→∞ approaches the horizontal asymptote y = 3 5 as x → ∞. To find , and therefore the graph of as x → −∞ . 2 7 Exercise 1.4.7 Find the limits as x → ∞ and x → −∞ for f (x) = 3e x −4 5 ex + 2 . Hint lim e x and lim =∞ x→∞ x→∞ e x = 0. Answer 3 lim f (x) = x→∞ , 5 lim f (x) = −2 x→−∞ Key Concepts The limit of f (x) is L as x → ∞ (or as x → −∞) if the values f (x) become arbitrarily close to L as xbecomes sufficiently large. The limit of f (x) is ∞ as x → ∞ if f (x) becomes arbitrarily large as x becomes sufficiently large. The limit of f (x) is −∞ as x → ∞ if f (x) < 0 and |f (x)| becomes arbitrarily large as x becomes sufficiently large. We can define the limit of f (x) as x approaches −∞ similarly. For a polynomial function p(x) = a x + a x +… +a x +a , where a ≠ 0 , the Limit at infinity is determined by the leading term a x . If n ≠ 0, p(x) approaches ∞ or −∞ at each end. n n n−1 n−1 1 0 n n n For a rational function p(x) f (x) = q(x), the Limit at infinity is determined by the relationship between the degree of p and the degree of q. If the degree of p is less than the degree of q, the line y = 0 is a horizontal asymptote for f . If the degree of p is equal to the degree of q, then the line y = is a horizontal asymptote, where a and b are the an n bn n leading coefficients of p and q, respectively. If the degree of p is greater than the degree of q, then f approaches ∞ or −∞ at each end. Glossary End behaviour the behavior of a function as x → ∞ and x → −∞ horizontal asymptote if lim f (x) = L or lim x→∞ x→−∞ f (x) = L , then y = L is a horizontal asymptote of f infinite limit at infinity a function that becomes arbitrarily large as x becomes large limit at infinity a function that becomes arbitrarily large as x becomes large oblique asymptote the line y = mx + b if f (x) approaches it as x → ∞ orx → −∞ Contributors 10/20/2020 1.4.15 CC-BY-NC-SA https://math.libretexts.org/@go/page/10254 Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/ Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/20/2020 1.4.16 CC-BY-NC-SA https://math.libretexts.org/@go/page/10254 1.4E: Exercises Exercise 1.4E. 1 : Limit at infinity − −−− − 2 lim √x + 3 − x 1. Find x→∞ Answer − −−− − 2 lim √x + 3 − x To find , First multiply by the conjugate x→∞ − −−− − − −−− − (√x2 + 3 − x)(√x2 + 3 + x) − −−− − 2 √ lim x + 3 − x = lim − −−− − x→∞ x→∞ √x2 + 3 + x Then reduce the numerator − −−− − 2 lim √x + 3 − x = lim x→∞ 3 − −−− − √x2 + 3 + x x→∞ − − Now divide by √x 2 = |x| , − −−− − 2 lim √x + 3 − x = lim x→∞ x→∞ 3/|x| − −−− − − − (√x2 + 3 + x)/ √x2 Since x → ∞ , replace |x| by x . Thus, \( \displaystyle \lim_{x \to \infty} \sqrt{x^2+3}-x = \lim_{x \to \infty} \frac{3/x}{(\sqrt{1+3/\sqrt{x^2}}+x/x) This result will be: = 3(0)/2 =0 − −−−− − √x4 + 6x 2. Find lim 2 x→−∞ x −6 Answer − − Factor out the highest degree variable in the numerator and the denominator, √x 4 2 − −−−− − √x4 + 6x lim x→−∞ x √1 + = 2 x −6 2 2 = |x | = x . −−−− − 6x 3 x lim x→−∞ 2 (x )(1 − 6 2 x ) Evaluate: −−− − √1 + 0 = 1 −0 1 = 1 =1 3. Find 7 −t lim t→∞ − − − − − − √2 + 2t2 Answer Factor out the highest degree variable in the numerator and denominator 10/21/2020 1.4E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10660 = t( 7 t→∞ − 1) t lim − − − − − 2 |t| √ 2 +2 t Evaluate: 0 −1 = −−− − √0 + 2 −1 = – √2 7 −t 4.Find lim − − − − − − √2 + 2t2 t→−∞ Answer Factor out the highest degree variable in the numerator and denominator (t)( = lim 7 − 1) t − − − − − t→∞ 2 |t| √ 2 +2 t 0 −1 = −−− − (−1)√0 + 2 1 = – √2 − − −−− − 2 √4 y − 2 5.Find lim 2 −y y→−∞ Answer Factor out the highest degree variable in the numerator and denominator −−−− − y √4 − = lim y→−∞ |y|( 2 y 2 y 2 − 1) When y → −∞ approaches to −∞ . |y| Evaluate: −−− − √4 − 0 −(0 − 1) 2 1 2 − −−−−−− − 8 6.Find 5 lim √ 5s 3 − 4s 2 s8 − 1 s→∞ Answer Because the limit is approaching infinity, only the highest degree polynomials need to be considered. This then simplifies the limit to: − − − − = 8 5 5s lim √ s→∞ 10/21/2020 8 2s 1.4E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10660 5 − − 5 =√ 2 − −−−−−−−−− − 7.Find 3 + 2x + 5x2 3 lim √ 2 x→∞ 1 + 4x Answer Because the limit is approaching infinity, only the highest degree polynomials need to be considered. This then simplifies the limit to: − − − − 2 5x 3 = lim √ 2 x→∞ 3 − − 5 =√ 4x . 4 5 11 − t 8.Find lim 5 t→∞ 11 t +1 Answer Because the limit is approaching infinity, only the highest degree polynomials need to be considered. This then simplifies the limit to: 5 −t = lim 5 t→∞ 11t −1 = 11 3 9.Find lim x −3 x→−∞ Answer =0 2 3x 10.Find lim − 2x 2 5x x→∞ +1 Answer Because the limit is approaching infinity, only the highest degree polynomials need to be considered. This then simplifies the limit to: 2 3x = lim x→−∞ 3 2 5x . = 5 − −−−−−−−− − 2 √7 y + 6y − 8 11.Find lim y→−∞ 5 −y . Answer − −−−−−−−− − 2 √7 y + 6y − 8 lim y→−∞ 10/21/2020 −− − 2 √7y = 5 −y lim y→−∞ −y 1.4E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10660 √7y √7y 2 √y |y| lim y→−∞ −y = lim y→−∞ |y| 2 2 −y – = √7 −y Exercise 1.4E. 2 : Horizontal Asymptotes Find horizontal asymptote(s) of the function y = 5 + 3x −−−−−−− −. √ x2 + x − 4 (1.4E.1) Answer y =3 and y = −3 . Contributors Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution Noncommercial (BY-NC) License. http://www.apexcalculus.com/ Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 1.4E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10660 1.5: Formal Definition of a Limit (optional) Learning Objectives Describe the epsilon-delta definition of a limit. Apply the epsilon-delta definition to find the limit of a function. Describe the epsilon-delta definitions of one-sided limits and infinite limits. Use the epsilon-delta definition to prove the limit laws. By now you have progressed from the very informal definition of a limit in the introduction of this chapter to the intuitive understanding of a limit. At this point, you should have a very strong intuitive sense of what the limit of a function means and how you can find it. In this section, we convert this intuitive idea of a limit into a formal definition using precise mathematical language. The formal definition of a limit is quite possibly one of the most challenging definitions you will encounter early in your study of calculus; however, it is well worth any effort you make to reconcile it with your intuitive notion of a limit. Understanding this definition is the key that opens the door to a better understanding of calculus. Quantifying Closeness Before stating the formal definition of a limit, we must introduce a few preliminary ideas. Recall that the distance between two points a and b on a number line is given by |a − b |. The statement |f (x) − L |<ε may be interpreted as: The distance between f (x) and L is less than ε . The statement 0 < |x − a| < δ may be interpreted as: x ≠ a and the distance between x and a is less than δ . It is also important to look at the following equivalences for absolute value: The statement |f (x) − L| < ε is equivalent to the statement L − ε < f (x) < L + ε . The statement 0 < |x − a| < δ is equivalent to the statement a − δ < x < a + δ and x ≠ a . With these clarifications, we can state the formal epsilon-delta definition of the limit. Definition: Finite Limits (Formal) Let f (x) be defined for all x ≠ a over an open interval containing a . Let L be a real number. Then lim f (x) = L (1.5.1) x→a if, for every ε > 0 , there exists a δ > 0 , such that if 0 < |x − a| < δ , then |f (x) − L| < ε . This definition may seem rather complex from a mathematical point of view, but it becomes easier to understand if we break it down phrase by phrase. The statement itself involves something called a universal quantifier (for every ε > 0 ), an existential quantifier (there exists a δ > 0 ), and, last, a conditional statement (if 0 < |x − a| < δ , then |f (x) − L| < ε) . Let’s take a look at Table 1.5.1, which breaks down the definition and translates each part. Table 1.5.1 Definition Translation 1. For every ε > 0 , 1. For every positive distance ε from L, 2. there exists a δ > 0, 2. There is a positive distance δ from a, 3. such that 3. such that 4. if 0 < |x − a| < δ , then |f (x) − L| < ε . 4. if x is closer than δ to a and x ≠ a , then f (x) is closer than ε to L. We can get a better handle on this definition by looking at the definition geometrically. Figure shows possible values of δ for various choices of ε > 0 for a given function f (x), a number a, and a limit L at a. Notice that as we choose smaller values of ε (the distance between the function and the limit), we can always find a δ small enough so that if we have chosen an x value within δ of a, then the value of f (x) is within ε of the limit L. 10/21/2020 1.5.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10281 Figure 1.5.1 : These graphs show possible values of δ, given successively smaller choices of ε. Note Visit the following applet to experiment with finding values of δ for selected values of ε : The epsilon-delta definition of limit Example 1.5.1 shows how you can use this definition to prove a statement about the limit of a specific function at a specified value. Example 1.5.2 : Proving a Statement about the Limit of a Specific Function Prove that lim (2x + 1) = 3 . x→1 Solution Let ε > 0 . The first part of the definition begins “For every ε > 0 .”This means we must prove that whatever follows is true no matter what positive value of ε is chosen. By stating “Let ε >0 ,” we signal our intent to do so. Choose δ = . ε 2 The definition continues with “there exists a δ > 0 . ” The phrase “there exists” in a mathematical statement is always a signal for a scavenger hunt. In other words, we must go and find δ . So, where exactly did δ = ε/2 come from? There are two basic approaches to tracking down δ . One method is purely algebraic and the other is geometric. We begin by tackling the problem from an algebraic point of view. Since ultimately we want |(2x + 1) − 3| < ε , we begin by manipulating this expression: |(2x + 1) − 3| < ε is equivalent to |2x − 2| < ε , which in turn is equivalent to |2||x − 1| < ε . Last, this is equivalent to |x − 1| < ε/2 . Thus, it would seem that δ = ε/2 is appropriate. We may also find δ through geometric methods. Figure demonstrates how this is done. 10/21/2020 1.5.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10281 Figure 1.5.2 : This graph shows how we find δ geometrically. Assume 0 < |x − 1| < δ . When δ has been chosen, our goal is to show that if 0 < |x − 1| < δ , then |(2x + 1) − 3| < ε . To prove any statement of the form “If this, then that,” we begin by assuming “this” and trying to get “that.” Thus, |(2x + 1) − 3| = |2x − 2| property of absolute value = |2(x − 1)| = |2||x − 1| |2| = 2 = 2|x − 1| <2⋅δ =2⋅ ε 2 here’s where we use the assumption that 0 < |x − 1| < δ =ε here’s where we use our choice of δ = ε/2 Analysis In this part of the proof, we started with |(2x + 1) − 3| and used our assumption 0 < |x − 1| < δ in a key part of the chain of inequalities to get |(2x + 1) − 3| to be less than ε. We could just as easily have manipulated the assumed inequality 0 < |x − 1| < δ to arrive at |(2x + 1) − 3| < ε as follows: 0 < |x − 1| < δ ⇒ |x − 1| < δ ⇒ −δ < x − 1 < δ ⇒ − ε 2 < x −1 < ε 2 ⇒ −ε < 2x − 2 < ε ⇒ −ε < 2x − 2 < ε ⇒ |2x − 2| < ε ⇒ |(2x + 1) − 3| < ε. Therefore, lim (2x + 1) = 3. (Having completed the proof, we state what we have accomplished.) x→1 After removing all the remarks, here is a final version of the proof: Let ε > 0 . Choose δ = ε/2 . Assume 0 < |x − 1| < δ . Thus, 10/21/2020 1.5.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10281 |(2x + 1) − 3| = |2x − 2| = |2(x − 1)| = |2||x − 1| = 2|x − 1| <2⋅δ ε =2⋅ 2 = ε. Therefore, lim (2x + 1) = 3 . x→1 The following Problem-Solving Strategy summarizes the type of proof we worked out in Example 1.5.2. The following Problem-Solving Strategy summarizes the type of proof we worked out in Example. Problem-Solving Strategy: Proving That lim f(x) = L for a Specific Function f(x) x→a 1. Let’s begin the proof with the following statement: Let ε > 0 . 2. Next, we need to obtain a value for δ . After we have obtained this value, we make the following statement, filling in the blank with our choice of δ : Choose δ =_______. 3. The next statement in the proof should be (at this point, we fill in our given value for a ): Assume 0 < |x − a| < δ . 4. Next, based on this assumption, we need to show that |f (x) − L| < ε , where f (x) and L are our function f (x) and our limit L. At some point, we need to use 0 < |x − a| < δ . 5. We conclude our proof with the statement: Therefore, lim f (x) = L. x→a Example 1.5.3 : Proving a Statement about a Limit Complete the proof that lim (4x + 1) = −3 by filling in the blanks. x→−1 Let _____. Choose δ =_______. Assume 0 < |x−_______|< δ . Thus, |________−________|=_____________________________________ε . Solution We begin by filling in the blanks where the choices are specified by the definition. Thus, we have Let ε > 0 . Choose δ =_______. Assume 0 < |x − (−1)| < δ . (or equivalently, 0 < |x + 1| < δ .) Thus, |(4x + 1) − (−3)| = |4x + 4| = |4||x + 1| < 4δ _______ε . Focusing on the final line of the proof, we see that we should choose δ = . ε 4 We now complete the final write-up of the proof: Let ε > 0 . Choose δ = . ε 4 Assume 0 < |x − (−1)| < δ (or equivalently, 0 < |x + 1| < δ .) Thus, |(4x + 1) − (−3)| = |4x + 4]| = |4||x + 1| < 4δ = 4(ε/4) = ε . 10/21/2020 1.5.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10281 Exercise 1.5.1 Complete the proof that lim (3x − 2) = 4 by filling in the blanks. x→2 Let _______. Choose δ =_______. Assume 0 < |x−____| <____. Thus, |_______−____|=______________________________ε . Therefore, lim (3x − 2) = 4 . x→2 Hint Follow the outline in the Problem-Solving Strategy that we worked out in full in Example 1.5.3. Answer Let ε > 0 ; choose δ = ; assume 0 < |x − 2| < δ . ε 3 Thus, |(3x − 2) − 4| = |3x − 6| = |3| ⋅ |x − 2| < 3 ⋅ δ = 3 ⋅ (ε/3) = ε . Therefore, lim(3x − 2) = 4 . x→2 In Examples 1.5.1 and 1.5.2, the proofs were fairly straightforward, since the functions with which we were working were linear. In Example 1.5.4, we see how to modify the proof to accommodate a nonlinear function. Example 1.5.4 : Proving a Statement about the Limit of a Specific Function (Geometric Approach) Prove that lim x 2 =4 . x→2 Solution 1. Let ε > 0 . The first part of the definition begins “For every ε > 0 ,” so we must prove that whatever follows is true no matter what positive value of ε is chosen. By stating “Let ε > 0 ,” we signal our intent to do so. 2. Without loss of generality, assume ε ≤ 4 . Two questions present themselves: Why do we want ε ≤ 4 and why is it okay to make this assumption? In answer to the first question: Later on, in the process of solving for δ , we will discover − −− − that δ involves the quantity √4 − ε . Consequently, we need ε ≤ 4 . In answer to the second question: If we can find δ > 0 that “works” for ε ≤ 4 , then it will “work” for any ε > 4 as well. Keep in mind that, although it is always okay to put an upper bound on ε, it is never okay to put a lower bound (other than zero) on ε . − −− − − −− − 3. Choose δ = min2 − √4 − ε , √4 + ε −2 . Figure 1.5.3 shows how we made this choice of δ . Figure 1.5.3 : This graph shows how we find δ geometrically for a given ε for the proof in Example. 10/21/2020 1.5.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10281 4. We must show: If 0 < |x − 2| < δ , then |x 2 , so we must begin by assuming − 4| < ε 0 < |x − 2| < δ. We don’t really need 0 < |x − 2| (in other words, x ≠ 2 ) for this proof. Since 0 < |x − 2| < δ ⇒ |x − 2| < δ , it is okay to drop 0 < |x − 2| . |x − 2| < δ. Hence, −δ < x − 2 < δ. − −− − − −− − 2 − √4 − ε , √4 + ε − 2 − −− − − −− − Recall that δ = min { }. Thus, δ ≥ 2 − √4 − ε and consequently −(2 − √4 − ε ) ≤ −δ . We − −− − − −− − also use δ ≤ √4 + ε − 2 here. We might ask at this point: Why did we substitute 2 − √4 − ε for δ on the left-hand side − −− − of the inequality and √4 + ε − 2 on the right-hand side of the inequality? If we look at Figure 1.5.3, we see that − −− − − −− − 2 − √4 − ε corresponds to the distance on the left of 2 on the x-axis and √4 + ε − 2 corresponds to the distance on the right. Thus, − −− − − −− − −(2 − √4 − ε ) ≤ −δ < x − 2 < δ ≤ √4 + ε − 2. We simplify the expression on the left: − −− − − −− − −2 + √4 − ε < x − 2 < √4 + ε − 2 . Then, we add 2 to all parts of the inequality: − −− − − −− − √4 − ε < x < √4 + ε . We square all parts of the inequality. It is okay to do so, since all parts of the inequality are positive: 2 4 −ε < x < 4 + ε. We subtract 4 from all parts of the inequality: 2 −ε < x − 4 < ε. Last, 2 |x − 4| < ε. 5. Therefore, 2 lim x = 4. x→2 Exercise 1.5.2 Find δ corresponding to ε > 0 for a proof that lim √− x = 3. x→9 Hint Draw a graph similar to the one in Example 1.5.4. Answer Choose δ = min{9 − (3 − ε) 2 2 , (3 + ε) − 9} . The geometric approach to proving that the limit of a function takes on a specific value works quite well for some functions. Also, the insight into the formal definition of the limit that this method provides is invaluable. However, we may also approach limit proofs from a purely algebraic point of view. In many cases, an algebraic approach may not only provide us with additional insight into the definition, it may prove to be simpler as well. Furthermore, an algebraic approach is the primary tool used in proofs of statements about limits. For Example 1.5.5, we take on a purely algebraic approach. Example 1.5.5 :Proving a Statement about the Limit of a Specific Function (Algebraic Approach) 10/21/2020 1.5.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10281 Prove that lim 2 (x − 2x + 3) = 6. x→−1 Solution Let’s use our outline from the Problem-Solving Strategy: 1. Let ε > 0 . 2. Choose δ = min{1, ε/5}. This choice of δ may appear odd at first glance, but it was obtained by taking a look at our ultimate desired inequality: ∣∣(x − 2x + 3) − 6 ∣< ε . This inequality is equivalent to |x + 1| ⋅ |x − 3| < ε . At this point, the temptation simply to choose δ = is very strong. Unfortunately, our choice of δ must depend on ε only and no other variable. If we can replace |x − 3| by a numerical value, our problem can be resolved. This is the place where assuming δ ≤ 1 comes into play. The choice of δ ≤ 1 here is arbitrary. We could have just as easily used any other positive number. In some proofs, greater care in this choice may be necessary. Now, since δ ≤ 1 and |x + 1| < δ ≤ 1 , we are able to show that |x − 3| < 5 . Consequently, |x + 1| ⋅ |x − 3| < |x + 1| ⋅ 5 . At this point we realize that we also need δ ≤ ε/5 . Thus, we choose δ = min{1, ε/5}. 2 ε x−3 3. Assume 0 < |x + 1| < δ . Thus, ε |x + 1| < 1 and |x + 1| < . 5 Since |x + 1| < 1 , we may conclude that −1 < x + 1 < 1 . Thus, by subtracting obtain −5 < x − 3 < −1 . Consequently, |x − 3| < 5 . This gives us ∣(x2 − 2x + 3) − 6 ∣ = |x + 1| ⋅ |x − 3| < ∣ ∣ 4 from all parts of the inequality, we ε ⋅ 5 = ε. 5 Therefore, lim 2 (x − 2x + 3) = 6. x→−1 Exercise 1.5.3 Complete the proof that lim x 2 =1 . x→1 Let ε > 0 ; choose δ = min{1, ε/3}; assume 0 < |x − 1| < δ . Since |x − 1| < 1 , we may conclude that −1 < x − 1 < 1 . Thus, 1 < x + 1 < 3 . Hence, |x + 1| < 3 . Hint Use Example 1.5.5as a guide. Answer 2 ∣ ∣x − 1 ∣= |x − 1| ⋅ |x + 1| < ε/3 ⋅ 3 = ε You will find that, in general, the more complex a function, the more likely it is that the algebraic approach is the easiest to apply. The algebraic approach is also more useful in proving statements about limits. Proving Limit Laws We now demonstrate how to use the epsilon-delta definition of a limit to construct a rigorous proof of one of the limit laws. The triangle inequality is used at a key point of the proof, so we first review this key property of absolute value. Definition: The Triangle Inequality The triangle inequality states that if a and b are any real numbers, then |a + b| ≤ |a| + |b| . 10/21/2020 1.5.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10281 Proof We prove the following limit law: If lim f (x) = L and lim g(x) = M , then lim (f (x) + g(x)) = L + M . x→a x→a x→a Let ε > 0 . Choose δ >0 so that if 0 < |x − a| < δ , then |f (x) − L| < ε/2 . Choose δ >0 so that if 0 < |x − a| < δ , then |g(x) − M | < ε/2 . 1 2 1 2 Choose δ = min{δ 1, δ2 } . Assume 0 < |x − a| < δ . Thus, 0 < |x − a| < δ1 and 0 < |x − a| < δ . 2 Hence, |(f (x) + g(x)) − (L + M )| = |(f (x) − L) + (g(x) − M )| ≤ |f (x) − L| + |g(x) − M | ε < . ε + 2 =ε 2 □ We now explore what it means for a limit not to exist. The limit lim f (x) does not exist if there is no real number L for which x→a lim f (x) = L . Thus, for all real numbers x→a L , lim f (x) ≠ L . To understand what this means, we look at each part of the x→a definition of lim f (x) = L together with its opposite. A translation of the definition is given in Table 1.5.2. x→a Table 1.5.2 Definition Opposite 1. There exists ε > 0 so that 1. For every ε > 0 , 2. there exists a δ > 0, so that 2. for every δ > 0, 3. if 0 3. There is an x satisfying 0 < |x − a| < δ , then |f (x) − L| < ε . Translation of the Definition of lim x→a f (x) = L < |x − a| < δ so that |f (x) − L| ≥ ε . and its Opposite Finally, we may state what it means for a limit not to exist. The limit lim f (x) does not exist if for every real number L, there x→a exists a real number ε > 0 so that for all δ > 0 , there is an x satisfying 0 < |x − a| < δ , so that this in Example 1.5.6 to show that a limit does not exist. |f (x) − L| ≥ ε . Let’s apply Example 1.5.6 : Showing That a Limit Does Not Exist Show that lim x→0 10/21/2020 |x| x does not exist. The graph of f (x) = |x|/x is shown here: 1.5.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10281 Solution Suppose that L is a candidate for a limit. Choose ε = 1/2 . Let δ > 0 . Either L ≥ 0 or L < 0 . If L ≥ 0 , then let x = −δ/2. Thus, |x − 0| =∣ − δ 2 − 0 ∣= δ 2 <δ and ∣ ∣ ∣− ∣ ∣ − δ 2 ∣ ∣ δ ∣ − L∣ = | − 1 − L| = L + 1 ≥ 1 > ∣ 2 1 2 . =ε On the other hand, if L < 0 , then let x = δ/2 . Thus, |x − 0| =∣ δ 2 − 0 ∣= δ 2 <δ and ∣ ∣ ∣ ∣ ∣ Thus, for any value of L, lim x→0 δ 2 δ ∣ ∣ ∣ − L∣ = |1 − L| = |L| + 1 ≥ 1 > 2 ∣ 1 2 =ε . |x| ≠ L. x One-Sided Limits Just as we first gained an intuitive understanding of limits and then moved on to a more rigorous definition of a limit, we now revisit one-sided limits. To do this, we modify the epsilon-delta definition of a limit to give formal epsilon-delta definitions for limits from the right and left at a point. These definitions only require slight modifications from the definition of the limit. In the definition of the limit from the right, the inequality 0 < x − a < δ replaces 0 < |x − a| < δ , which ensures that we only consider values of x that are greater than (to the right of) a . Similarly, in the definition of the limit from the left, the inequality −δ < x − a < 0 replaces 0 < |x − a| < δ , which ensures that we only consider values of x that are less than (to the left of) a. Definition: One-Sided Limits (Formal) Limit from the Right: Let f (x) be defined over an open interval of the form (a, b) where a 0 , there exists a δ > 0 , such that if 0 < x − a < δ , then |f (x) − L| < ε . Limit from the Left: Let f (x) be defined over an open interval of the form (b, c) where b < c . Then, lim f (x) = L (1.5.3) − x→c 10/21/2020 1.5.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10281 if for every ε > 0 ,there exists a δ > 0 such that if −δ < x − c < 0 , then |f (x) − L| < ε . Example 1.5.7 : Proving a Statement about a Limit From the Right Prove that − − − − − lim √ x − 4 = 0. + x→4 Solution Let ε > 0 . − − − − − − − − − − Choose δ = ε . Since we ultimately want ∣√x − 4 − 0 ∣< ε , we manipulate this inequality to get √x − 4 < ε or, equivalently, 0 < x − 4 < ε , making δ = ε a clear choice. We may also determine δ geometrically, as shown in Figure 1.5.4. 2 2 2 Figure 1.5.4 : This graph shows how we find δ for the proof in Example. Assume 0 < x −4 < δ − − − − − lim √x − 4 = 0 . Thus, 0 < x −4 < ε 2 . Hence, − − − − − 0 < √x − 4 < ε . Finally, − − − − − ∣ ∣√x − 4 − 0 ∣ ∣ <ε . Therefore, . + x→4 Exercise 1.5.4 Find δ corresponding to ε for a proof that − − − − − lim √1 − x = 0 . − x→1 Hint Sketch the graph and use Example 1.5.7as a solving guide. Answer δ =ε 2 Infinite Limits We conclude the process of converting our intuitive ideas of various types of limits to rigorous formal definitions by pursuing a formal definition of infinite limits. To have lim f (x) = +∞ , we want the values of the function f (x) to get larger and larger x→a as x approaches a. Instead of the requirement that |f (x) − L| < ε for arbitrarily small ε when 0 < |x − a| < δ for small enough δ , we want f (x) > M for arbitrarily large positive M when 0 < |x − a| < δ for small enough δ . Figure 1.5.5 illustrates this idea by showing the value of δ for successively larger values of M . 10/21/2020 1.5.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10281 Figure 1.5.5 : These graphs plot values of δ for M to show that lim f (x) = +∞ . x→a Definition: Infinite Limits (Formal) Let f (x) be defined for all x ≠ a in an open interval containing a . Then, we have an infinite limit lim f (x) = +∞ (1.5.4) x→a if for every M >0 , there exists δ > 0 such that if 0 < |x − a| < δ , then f (x) > M . Let f (x) be defined for all x ≠ a in an open interval containing a . Then, we have a negative infinite limit lim f (x) = −∞ (1.5.5) x→a if for every M >0 , there exists δ > 0 such that if 0 < |x − a| < δ , then f (x) < −M . Example 1.5.8 : Proving a Statement about an Infinite Limit Prove that lim x→3 1 2 = ∞. (x − 3) Solution We use a very similar approach to our previous Problem-Solving Strategy. We first find an appropriate δ > 0 . Then we write our proof. Step 1: First we find an appropriate δ > 0 . 1. Let M be any real number such that M 2. Let f (x) = 1 (x − 3)2 >M >0 . . Then we solve for the expression x − 3 . Multiplying both sides of the inequality by the positive quantity (x − 3) and dividing both sides by the positive quantity M gives us: 2 1 2 > (x − 3 ) M Taking the square root of both sides, we have, −− − 1 √ − − 2 (Remember that √x = |x|. ) > |x − 3|. M Rewriting this statement gives us, 10/21/2020 −− − 1 0 < |x − 3| < √ M . From this we choose 1.5.11 CC-BY-NC-SA −− − 1 δ =√ M . https://math.libretexts.org/@go/page/10281 Step 2: Now we write a proof. −− − 1 3. Let δ = √ M −− − 1 and assume 0 < |x − 3| < δ = √ M . Thus, −− − 1 . M |x − 3| < √ Squaring both sides gives us, 2 (x − 3 ) 1 < . M Taking the reciprocal of both sides (and remembering that this will reverse the direction of the inequality), 1 > M. 2 (x − 3) Therefore, we have proven that 1 lim x→3 2 = ∞. (x − 3) A very similar proof will be needed for a limit that is equal to −∞ . Note that a one-sided limit approach will often need to be taken with this type of limit. For example, to prove: 1 lim + x→0 =∞ . x Key Concepts The intuitive notion of a limit may be converted into a rigorous mathematical definition known as the epsilon-delta definition of the limit. The epsilon-delta definition may be used to prove statements about limits. The epsilon-delta definition of a limit may be modified to define one-sided limits. A similar definition of an infinite limit can be used to prove statements about infinite limits. Glossary epsilon-delta definition of the limit lim f (x) = L if for every ε > 0 , there exists a δ > 0 such that if 0 < |x − a| < δ , then |f (x) − L| < ε x→a triangle inequality If a and b are any real numbers, then |a + b| ≤ |a| + |b| formal definition of an infinite limit lim f (x) = ∞ if for every M > 0 , there exists a δ > 0 such that if 0 < |x − a| < δ , then f (x) > M x→a lim f (x) = −∞ if for every M >0 , there exists a δ > 0 such that if 0 < |x − a| < δ , then f (x) < −M x→a Contributors and Attributions Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Paul Seeburger (Monroe Community College), added Example 1.5.8 and entries for infinite limits under Key Concepts and the Glossary. 10/21/2020 1.5.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/10281 1.5E: Exercises 10/21/2020 1.5E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10661 1.6: Continuity and the Intermediate Value Theorem Thinking out Loud Many functions have the property that their graphs can be traced with a pencil without lifting the pencil from the page. Normally, such functions are called continuous. Other functions have points at which a break in the graph occurs, but satisfy this property over intervals contained in their domains. They are continuous on these intervals and are said to have a discontinuity at a point where a break occurs. We begin our investigation of continuity by exploring what it means for a function to have continuity at a point. Intuitively, a function is continuous at a particular point if there is no break in its graph at that point. Continuity at a Point Before we look at a formal definition of what it means for a function to be continuous at a point, let’s consider various functions that fail to meet our intuitive notion of what it means to be continuous at a point. We then create a list of conditions that prevent such failures. Our first function of interest is shown in Figure. We see that the graph of f (x) has a hole at a. In fact, f (a) is undefined. At the very least, for f (x) to be continuous at a, we need the following condition: i. f (a) is defined. Figure 1.6.1 : The function f (x) is not continuous at a because f (a) is undefined. However, as we see in Figure, this condition alone is insufficient to guarantee continuity at the point a. Although f (a) is defined, the function has a gap at a. In this example, the gap exists because li m f (x) does not exist. We must add another condition for continuity at a—namely, x→a ii. lim x→a f (x) exists. Figure 1.6.2 : The function f (x) is not continuous at a because li m x→a f (x) does not exist. However, as we see in Figure, these two conditions by themselves do not guarantee continuity at a point. The function in this figure satisfies both of our first two conditions, but is still not continuous at a. We must add a third condition to our list: iii. lim x→a f (x) 9/23/2020 . = f (a) 1.6.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10282 Figure 1.6.3 : The function f (x) is not continuous at a because li m x→a . f (x) ≠ f (a) Now we put our list of conditions together and form a definition of continuity at a point. Definition A function f (x) is continuous at a point a if and only if the following three conditions are satisfied: i. f (a) is defined ii. li m f (x) exists iii. li m f (x) = f (a) x→a x→a A function is discontinuous at a point a if it fails to be continuous at a. App: Continuity at a point Limit: Continuity at a Point Example Example 1 Does the limit exist at x = 2? Check Is the function defined at x = 2? 9/23/2020 Check 1.6.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10282 Is the function continuous at x = 2? Check The following procedure can be used to analyze the continuity of a function at a point using this definition. Problem-Solving Strategy: Determining Continuity at a Point 1. Check to see if f (a) is defined. If f (a) is undefined, we need go no further. The function is not continuous at a. If f (a) is defined, continue to step 2. 2. Compute li m f (x). In some cases, we may need to do this by first computing li m f (x) and li m f (x). If li m f (x) does not exist (that is, it is not a real number), then the function is not continuous at a and the problem is solved. If li m f (x) exists, then continue to step 3. 3. Compare f (a) and li m f (x). If li m f (x) ≠ f (a), then the function is not continuous at a. If li m f (x) = f (a), then the function is continuous at a. − x→a + x→a x→a x→a x→a x→a x→a x→a The next three examples demonstrate how to apply this definition to determine whether a function is continuous at a given point. These examples illustrate situations in which each of the conditions for continuity in the definition succeeds or fails. Example 1.6.1A : Determining Continuity at a Point, Condition 1 Using the definition, determine whether the function f (x) = (x 2 − 4)/(x − 2) is continuous at x = 2 . Justify the conclusion. Solution Let’s begin by trying to calculate f (2). We can see that f (2) = 0/0, which is undefined. Therefore, f (x) = because f (2) is undefined. The graph of f (x) is shown in Figure. 2 x −4 x−2 is discontinuous at 2 Figure 1.6.4 : The function f (x) is discontinuous at 2 because f (2) is undefined. Example 1.6.1B : Determining Continuity at a Point, Condition 2 Using the definition, determine whether the function f (x) = { 2 −x +4 if x ≤ 3 4x − 8 if x > 3 is continuous at x = 3 . Justify the conclusion. Solution Let’s begin by trying to calculate f (3). 2 f (3) = −(3 ) + 4 = −5 Thus, f (3) is defined. Next, we calculate li m x→3 . . To do this, we must compute li m f (x) − x→3 f (x) and li m + x→3 : f (x) 2 li mx→3− f (x) = −(3 ) + 4 = −5 and li mx→3+ f (x) = 4(3) − 8 = 4 9/23/2020 . 1.6.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10282 Therefore, li m x→3 f (x) does not exist. Thus, f (x) is not continuous at 3. The graph of f (x) is shown in Figure. Figure 1.6.5 : The function f (x) is not continuous at 3 because li m x→3 f (x) does not exist. Exercise 1.6.2 ⎧ 2x + 1 Using the definition, determine whether the function f (x) = ⎨ 2 ⎩ −x + 4 if x < 1 if x = 1 is continuous at x =1 . If the function is not if x > 1 continuous at 1, indicate the condition for continuity at a point that fails to hold. Hint Check each condition of the definition. Answer f is not continuous at 1 because f (1) = 2 ≠ 3 = li m x→1 f (x) . Exercise 1.6.3 To send a letter to New Zealand from Canada costs $1.50 for the first ounce and $0.75 for each additional ounce. Let C (x) be the cost to mail x ounces. (a) Find lim C (x). + x→2 (b) Find llim − x→2 (c) Does lim x→2 C (x). C (x). exist? Why or why not? Answer C (x) = 1.50 + .75(x − 1) = .75 + .75x, limx→2 = C (x) limx→2+ C (x) = limx→2− C (x) = 2.25 Types of Discontinuities As we have seen in Example and Example, discontinuities take on several different appearances. We classify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or jump discontinuities. Intuitively, a removable discontinuity is a discontinuity for which there is a hole in the graph, a jump discontinuity is a noninfinite discontinuity for which the sections of the function do not meet up, and an infinite discontinuity is a discontinuity located at a vertical asymptote. Figure illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep in mind that not all discontinuities fit neatly into these categories. 9/23/2020 1.6.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10282 Figure 1.6.5 : Discontinuities are classified as (a) removable, (b) jump, or (c) infinite. These three discontinuities are formally defined as follows: Definition If f (x) is discontinuous at a, then 1. f has a removable discontinuity at a if li m , where L is a real number.) x→a f (x) exists. (Note: When we state that li mx→a f (x) exists, we mean that li mx→a f (x) = L 2. f has a jump discontinuity at a if li m When we state that li m f (x) and li m values ±∞.) x→a− x→a− x→a+ 3. f has an infinite discontinuity at a if li m − x→a and li m f (x) both exist, but li m f (x) ≠ li m f (x). (Note: both exist, we mean that both are real-valued and that neither take on the f (x) f (x) x→a+ f (x) = ±∞ x→a− or li m + x→a x→a+ . f (x) = ±∞ Example 1.6.6 : Classifying a Discontinuity 2 In Example, we showed that f (x) = x −4 x−2 is discontinuous at x = 2 . Classify this discontinuity as removable, jump, or infinite. Solution To classify the discontinuity at 2 we must evaluate li m x→2 : f (x) 2 li mx→2 f (x) = li mx→2 x −4 x−2 (x−2)(x+2) =lim x→2 =li m x→2 (x + 2) x−2 =4. Since f is discontinuous at 2 and li m x→2 f (x) exists, f has a removable discontinuity at x=2. Example 1.6.7 : Classifying a Discontinuity In Example, we showed that f (x) = { 2 −x +4 if x ≤ 3 4x − 8 if x > 3 is discontinuous at x = 3 . Classify this discontinuity as removable, jump, or infinite. Solution Earlier, we showed that f is discontinuous at 3 because li m f (x) does not exist. However, since li m f (x) = 4 both exist, we conclude that the function has a jump discontinuity at 3. x→3 li mx→3− f (x) = −5 and − x→3 Example 1.6.8 : Classifying a Discontinuity Determine whether jump, or infinite. f (x) = x+2 x+1 is continuous at −1. If the function is discontinuous at −1, classify the discontinuity as removable, Solution The function value f (−1) is undefined. Therefore, the function is not continuous at −1. To determine the type of discontinuity, we must determine the limit at −1. We see that li m = −∞ and li m = +∞ . Therefore, the function has an infinite discontinuity at −1. x+2 − x→−1 9/23/2020 x+2 + x+1 x→−1 x+1 1.6.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10282 Exercise 1.6.9 For 2 f (x) = { x if x ≠ 1 3 if x = 1 , decide whether f is continuous at 1. If f is not continuous at 1, classify the discontinuity as removable, jump, or infinite. Hint Follow the steps in Note. If the function is discontinuous at 1, look at li m f (x) x→1 Answer Discontinuous at 1; removable Continuity over an Interval Now that we have explored the concept of continuity at a point, we extend that idea to continuity over an interval. As we develop this idea for different types of intervals, it may be useful to keep in mind the intuitive idea that a function is continuous over an interval if we can use a pencil to trace the function between any two points in the interval without lifting the pencil from the paper. In preparation for defining continuity on an interval, we begin by looking at the definition of what it means for a function to be continuous from the right at a point and continuous from the left at a point. Continuity from the Right and from the Left A function f (x) is said to be continuous from the right at a if li m A function f (x) is said to be continuous from the left at a if li m x→a+ − x→a . f (x) = f (a) f (x) = f (a) A function is continuous over an open interval if it is continuous at every point in the interval. A function f (x) is continuous over a closed interval of the form [a, b] if it is continuous at every point in (a, b) and is continuous from the right at a and is continuous from the left at b. Analogously, a function f (x) is continuous over an interval of the form (a, b] if it is continuous over (a, b) and is continuous from the left at b. Continuity over other types of intervals are defined in a similar fashion. Requiring that li m f (x) = f (a) and li m f (x) = f (b) ensures that we can trace the graph of the function from the point (a, f (a)) to the point (b, f (b)) without lifting the pencil. If, for example, li m f (x) ≠ f (a), we would need to lift our pencil to jump from f (a) to the graph of the rest of the function over (a, b]. + − x→a x→b + x→a Example 1.6.10 : Continuity on an Interval State the interval(s) over which the function f (x) = x−1 2 x +2x is continuous. Solution Since f (x) = x−1 2 x +2x is a rational function, it is continuous at every point in its domain. The domain of f (x) is the set . Thus, f (x) is continuous over each of the intervals (−∞, −2), (−2, 0), and (0, +∞). (−∞, −2) ∪ (−2, 0) ∪ (0, +∞) Example 1.6.11 : Continuity over an Interval − −−− − State the interval(s) over which the function f (x) = √4 − x is continuous. 2 Solution − −−− − − −−− − 2 2 li mx→a √4 − x = √4 − a − −−− − 2 − √4 − x =0 x→2 From the limit laws, we know that − −−− − √4 − x li m = 0 exists and li m + 2 x→−2 for all values of a in (−2, 2). We also know that exists. Therefore, f (x) is continuous over the interval [−2, 2]. Exercise 1.6.12 − − − − − State the interval(s) over which the function f (x) = √x + 3 is continuous. Hint Use Example as a guide for solving. Answer [−3,+∞) 9/23/2020 1.6.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10282 The Note allows us to expand our ability to compute limits. In particular, this theorem ultimately allows us to demonstrate that trigonometric functions are continuous over their domains. Continuity of Polynomials and Rational Functions Theorem 1.6.1 :Continuity of Polynomials and Rational Functions Polynomials and rational functions are continuous at every point in their domains. Proof Previously, we showed that if p(x) li mx→a p(a) = q(x) q(a) p(x) and q(x) are polynomials, li mx→a p(x) = p(a) for every polynomial p(x) and as long as q(a) ≠ 0 . Therefore, polynomials and rational functions are continuous on their domains. We now apply Note to determine the points at which a given rational function is continuous. Example 1.6.4 :Continuity of a Rational Function For what values of x is f (x) = continuous? x+1 x−5 Solution The rational function f (x) = is continuous for every value of x except x = 5 . x+1 x−5 Exercise 1.6.5 For what values of x is f (x) = 3x 4 2 − 4x continuous? Hint Use Note Answer f (x) is continuous at every real number. Properties of Continuous Functions Properties of Continuous Functions Let f and g be continuous functions. Then 1. f ± g is a continuous function. 2. f g is a continuous function. 3. f g is a continuous function, whenever g(x) ≠ 0 . Composite Function Theorem If f (x) is continuous at L and li m x→a g(x) = L , then li mx→a f (g(x)) = f (li mx→a g(x)) = f (L). Note Let f and g be continuous functions. Then f ∘ g is a continuous function. Example 1.6.13 : Limit of a Absolute valued Function Evaluate li m x→2 . |x − 3| Solution Since absolute function is continuous and li m 9/23/2020 x→2 (x − 3) = 2 − 3 = −1 exists, li m x→2 1.6.7 CC-BY-NC-SA |x − 3| = | − 1| = 1 . https://math.libretexts.org/@go/page/10282 Exercise 1.6.10 Evaluate li m x→2 . |3 − x| Answer Since absolute function is continuous and li m x→2 exists,li m (3 − x) = 3 − 2 = 1 x→2 |3 − x| = |1| = 1 . Think out loud “True or False? At some point in your life, you were exactly π feet tall” As we continue our study of calculus, we revisit this theorem many times. The Intermediate Value Theorem Functions that are continuous over intervals of the form [a, b], where a and b are real numbers, exhibit many useful properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. The first of these theorems is the Intermediate Value Theorem. The Intermediate Value Theorem Let f be continuous over a closed, bounded interval [a, b]. If z is any real number between f (a) and f (b), then there is a number c in [a, b] satisfying f (c) = z in Figure. Figure 1.6.6 : There is a number c ∈ [a, b] that satisfies f (c) = z . Example 1.6.9 : Application of the Intermediate Value Theorem Show that f (x) = x − cosx has at least one zero. Solution Since f (x) = x − cosx is continuous over (−∞, +∞) , it is continuous over any closed interval of the form [a, b]. If you can find an interval [a, b] such that f (a) and f (b) have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number c in (a, b) that satisfies f (c) = 0 . Note that f (0) = 0 − cos(0) = −1 < 0 and f( π 2 ) = π 2 − cos π 2 = π 2 >0 . Using the Intermediate Value Theorem, we can see that there must be a real number c in Therefore, f (x) = x − cosx has at least one zero. [0, π/2] that satisfies f (c) = 0 . Example 1.6.14 : When Can You Apply the Intermediate Value Theorem? If f (x) is continuous over [0, 2], f (0) > 0 and f (2) > 0 , can we use the Intermediate Value Theorem to conclude that f(x) has no zeros in the interval [0, 2]? Explain. Solution 9/23/2020 1.6.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10282 No. The Intermediate Value Theorem only allows us to conclude that we can find a value between f (0) and f (2); it doesn’t allow us to conclude that we can’t find other values. To see this more clearly, consider the function f (x) = (x − 1) . It satisfies f (0) = 1 > 0, f (2) = 1 > 0 , and f (1) = 0 . 2 Example 1.6.15 : When Can You Apply the Intermediate Value Theorem? For f (x) = 1/x, f (−1) = −1 < 0 and f (1) = 1 > 0 . Can we conclude that f (x) has a zero in the interval [−1, 1]? Solution No. The function is not continuous over [−1, 1]. The Intermediate Value Theorem does not apply here. Exercise 1.6.16 Show that f (x) = x 3 2 −x − 3x + 1 has a zero over the interval [0, 1]. Hint Find f (0) and f (1). Apply the Intermediate Value Theorem. Answer f (0) = 1 > 0, f (1) = −2 < 0; f (x) is continuous over [0, 1]. It must have a zero on this interval. Key Concepts For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point. Discontinuities may be classified as removable, jump, or infinite. A function is continuous over an open interval if it is continuous at every point in the interval. It is continuous over a closed interval if it is continuous at every point in its interior and is continuous at its endpoints. The composite function theorem states: If f (x) is continuous at L and li m g(x) = L , then li m f (g(x)) = f (li m g(x)) = f (L) . x→a x→a x→a The Intermediate Value Theorem guarantees that if a function is continuous over a closed interval, then the function takes on every value between the values at its endpoints. Glossary continuity at a point A function f (x) is continuous at a point a if and only if the following three conditions are satisfied: (1) li m f (x) exists, and (3) limx → af (x) = f (a) f (a) is defined, (2) x→a continuity from the left A function is continuous from the left at b if li m f (x) = f (b) − x→b continuity from the right A function is continuous from the right at a if li m + x→a f (x) = f (a) continuity over an interval a function that can be traced with a pencil without lifting the pencil; a function is continuous over an open interval if it is continuous at every point in the interval; a function f (x) is continuous over a closed interval of the form [a, b] if it is continuous at every point in ( a, b), and it is continuous from the right at a and from the left at b discontinuity at a point A function is discontinuous at a point or has a discontinuity at a point if it is not continuous at the point infinite discontinuity An infinite discontinuity occurs at a point a if li m x→a− f (x) = ±∞ or li m x→a+ f (x) = ±∞ Intermediate Value Theorem 9/23/2020 1.6.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10282 Let f be continuous over a closed bounded interval [a, b] if z is any real number between f (a) and f (b), then there is a number c in [ a, b] satisfying f (c) = z jump discontinuity A jump discontinuity occurs at a point a if li m x→a− f (x) and li m x→a+ f (x) both exist, but li m removable discontinuity A removable discontinuity occurs at a point a if f (x) is discontinuous at a, but li m x→a f (x) x→a− f (x) ≠ li mx→a+ f (x) exists Source Calculus Applets using GeoGebra by Marc Renault is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah, Mount Royal University 9/23/2020 1.6.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10282 1.6E: Excercises Exercise 1.6E. 1 For exercises 1 - 8, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other. 1 1) f (x) = − √x Answer The function is defined for all x in the interval (0, ∞) . 2 2) f (x) = 2 x +1 x 3) f (x) = 2 x −x Answer Removable discontinuity at x = 0 ; infinite discontinuity at x = 1 . 4) g(t) = t −1 5) f (x) = +1 5 e x −2 Answer Infinite discontinuity at x = ln 2 6) f (x) = |x − 2| x −2 7) H (x) = tan 2x Answer Infinite discontinuities at x = 8) f (t) = (2k + 1)π 4 , for k = 0, ±1, ±2, ±3, … t +3 2 t + 5t + 6 Exercise 1.6E. 2 For exercises 1 - 6, decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it? 2 1) 2x − 5x + 3 x −1 at x = 1 Answer No. It is a removable discontinuity. 2) h(θ) = 10/21/2020 sin θ − cos θ tan θ at θ = π 1.6E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10662 2 ⎧ 6u ⎪ 3) g(u) = ⎨ +u −2 , if u ≠ 12 2u − 1 ⎩ ⎪ 7 , 2 , at u = 1 2 if u = 12 Answer Yes. It is continuous. 4) f (y) = sin(πy) tan(πy) 2 x 5) f (x) = { , at y = 1 x −e , if x < 0 x − 1, , at x = 0 if x ≥ 0 Answer Yes. It is continuous. 6) f (x) = { x sin(x), if x ≤ π x tan(x), if x > π , at x = π Exercise 1.6E. 3 In exercises 1 - 5, find the value(s) of k that makes each function continuous over the given interval. 1) f (x) = { 3x + 2, if x < k 2x − 3, if k ≤ x ≤ 8 Answer k = −5 2) f (θ) = { sin θ, if 0 ≤ θ < cos(θ + k), 2 ⎧ x if π 2 π 2 ≤θ ≤π + 3x + 2 , 3) f (x) = ⎨ if x ≠ −2 x +2 ⎩ k, if x = −2 Answer k = −1 4) f (x) = { 5) f (x) = { e kx , if 0 ≤ x < 4 x + 3, if 4 ≤ x ≤ 8 − − √kx , if 0 ≤ x ≤ 3 x + 1, if 3 < x ≤ 10 Answer k = 16 3 Exercise 1.6E. 4 In exercises 1 - 3, use the Intermediate Value Theorem (IVT). 1) Show that the equation x 3 2 +x − 2x − 1 = 0 has at least one solution in the interval [1, 3]. Answer Let f (x) = x 3 10/21/2020 2 +x − 2x − 1 . Then f (x) is continuous and f (1) = −1 and f (3) = 29. 1.6E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10662 Since −1 < 0 < 29 ,by the Intermediate Value Theorem there exist at least one real number such that f (k) = 0 . 2) Let 2 h(x) = { 3x − 4, 5 + 4x, h(0) < 10 if x ≤ 2 Over the interval if x > 2 , there is no value of [0, 4] x such that k in the interval h(x) = 10 [1, 3] , although and h(4) > 10 . Explain why this does not contradict the IVT. 3) A particle moving along a line for time t has a position function s(t) , which is continuous. Assume s(2) = 5 and s(5) = 2 . Another particle moves such that its position is given by h(t) = s(t) − t . Explain why there must be a value c for 2 < c < 5 such that h(c) = 0 . Answer Since both s and y = t are continuous everywhere, then h(t) = s(t) − t is continuous everywhere and, in particular, it is continuous over the closed interval [2, 5]. Also, h(2) = 3 > 0 and h(5) = −3 < 0 . Therefore, by the IVT, there is a value x = c such that h(c) = 0 . Exercise 1.6E. 5 1) [T] Use the statement “The cosine of t is equal to t cubed." a. Write a mathematical equation of the statement. b. Prove that the equation in part a. has at least one real solution. c. Use a calculator to find an interval of length 0.01 that contains a solution. 2) Apply the IVT to determine whether 2 = x has a solution in one of the intervals [1.25, 1.375] or [1.375, 1.5]. Briefly explain your response for each interval. x Answer The function endpoints. x f (x) = 2 3 −x 3 is continuous over the interval [1.25, 1.375] and has opposite signs at the Exercise 1.6E. 6 1) Consider the graph of the function y = f (x) shown in the following graph. a. Find all values for which the function is discontinuous. b. For each value in part a., state why the formal definition of continuity does not apply. c. Classify each discontinuity as either jump, removable, or infinite. 2) Let f (x) = { 3x, 3 x , if x > 1 if x < 1 . a. Sketch the graph of f . 10/21/2020 1.6E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10662 b. Is it possible to find a value explain. k such that f (1) = k , which makes f (x) continuous for all real numbers? Briefly Answer a. b. It is not possible to redefine f (1) since the discontinuity is a jump discontinuity. 4 3) Let f (x) = x 2 x −1 for x ≠ −1, 1. −1 a. Sketch the graph of f . b. Is it possible to find values k and k such that all real numbers? Briefly explain. 1 2 f (−1) = k and f (1) = k , and that makes f (x) continuous for 2 4) Sketch the graph of the function y = f (x) with properties i. through vii. i. The domain of f is (−∞, +∞). ii. f has an infinite discontinuity at x = −6 . iii. f (−6) = 3 iv. lim f (x) = − lim f (x) = 2 + x→−3 x→−3 v. f (−3) = 3 vi. f is left continuous but not right continuous at x = 3 . vii. lim x→−∞ f (x) = −∞ and lim f (x) = +∞ x→+∞ Answer Answers may vary; see the following example: 10/21/2020 1.6E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10662 5) Sketch the graph of the function y = f (x) with properties i. through iv. i. The domain of f is [0, 5]. ii. lim f (x) + x→1 and lim f (x) exist and are equal. − x→1 iii. f (x) is left continuous but not continuous at x = 2 , and right continuous but not continuous at x = 3 . iv. has a removable discontinuity at f (x) = −∞ and lim f (x) = 2 . f (x) lim − x→3 x =1 , a jump discontinuity at x =2 , and the following limits hold: + x→3 Exercise 1.6E. 7 In following exercises, suppose indicated property. 1) Discontinuous at x = 1 with y = f (x) is defined for all lim f (x) = −1 x→−1 x . For each description, sketch a graph with the and lim f (x) = 4 x→2 Answer Answers may vary; see the following example: 2) Discontinuous at x = 2 but continuous elsewhere with lim f (x) = x→0 10/21/2020 1 2 1.6E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10662 Exercise 1.6E. 8 Determine whether each of the given statements is true. Justify your response with an explanation or counterexample. 1) f (t) = 2 et − e−t is continuous everywhere. Answer False. It is continuous over (−∞, 0 ) ∪ (0, ∞). 2) If the left- and right-hand limits of f (x) as x → a exist and are equal, then f cannot be discontinuous at x = a . 3) If a function is not continuous at a point, then it is not defined at that point. Answer False. Consider f (x) = { x, if x ≠ 0 4, if x = 0 . 4) According to the IVT, cos x − sin x − x = 2 has a solution over the interval [−1, 1]. 5) If f (x) is continuous such that f (a) and f (b) have opposite signs, then f (x) = 0 has exactly one solution in [a, b]. Answer False. Consider f (x) = cos(x) on [−π, 2π ]. 2 6) The function f (x) = x − 4x + 3 2 x is continuous over the interval [0, 3]. −1 7) If f (x) is continuous everywhere and f (a), f (b) > 0 , then there is no root of f (x) in the interval [a, b]. Answer False. The IVT does not work in reverse! Consider (x − 1) over the interval [−2, 2]. 2 Exercise 1.6E. 9 [T] The following problems consider the scalar form of Coulomb’s law, which describes the electrostatic force between two point charges, such as electrons. It is given by the equation F (r) = k e | q1 q2 | 2 , where ke is Coulomb’s r constant, q are the magnitudes of the charges of the two particles, and r is the distance between the two particles. i 1) To simplify the calculation of a model with many interacting particles, after some threshold value approximate F as zero. r =R , we a. Explain the physical reasoning behind this assumption. b. What is the force equation? c. Evaluate the force F using both Coulomb’s law and our approximation, assuming two protons with a charge magnitude of 1.6022 × 10 coulombs (C), and the Coulomb constant k = 8.988 × 10 N m /C are 1 m apart. Also, assume R < 1 m. How much inaccuracy does our approximation generate? Is our approximation reasonable? −19 9 2 2 e d. Is there any finite value of R for which this system remains continuous at R? 2) Instead of making the force 0 at R , we let the force be 10 − 20 for r ≥ R . Assume two protons, which have a magnitude of charge 1.6022 × 10 C , and the Coulomb constant k = 8.988 × 10 N m / C . Is there a value R that can make this system continuous? If so, find it. −19 9 2 2 e 10/21/2020 1.6E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10662 Answer m R = 0.0001519 Exercise 1.6E. 10 The following problems consider a rocket launch from Earth’s surface. The force of gravity on the rocket is given by F (d) = −mk/d , where m is the mass of the rocket, d is the distance of the rocket from the center of Earth, and k is a constant. 2 1) [T] Determine the value and units of k given that the mass of the rocket on Earth is 3 million kg. (Hint: The distance from the center of Earth to its surface is 6378 km.) 2) [T] After a certain distance D approximate the force function by has passed, the gravitational effect of Earth becomes quite negligible, so we can ⎧ F (d) = ⎨ ⎩ mk − d 2 , 10, 000, if d < D . Find the necessary condition such that the force D if d ≥ D function remains continuous. Answer D = 63.78 km 3) As the rocket travels away from Earth’s surface, there is a distance D where the rocket sheds some of its mass, since it no longer needs the excess fuel storage. We can write this function as D such that this function is continuous, assuming m 1 ≠ m2 m1 k ⎧ ⎪− , ⎪ 2 d F (d) = ⎨ m2 k ⎪ ⎩− ⎪ , d ? 2 if d < D . Is there a value of if d ≥ D Exercise 1.6E. 11 In the following Exercises, prove each function is continuous everywhere. 1) f (θ) = sin θ Answer For all values of a , f (a) is defined, lim f (θ) exists, and θ→a lim f (θ) = f (a) . Therefore, f (θ) is continuous θ→a everywhere. 2) g(x) = |x| 3) Where is f (x) = { 0, if x is irrational 1, if x is rational continuous? Answer Nowhere Exercise 1.6E. 12 : Terms and Cocepts 1. In your own words, describe what it means for a function to be continuous. 2. In your own words, describe what the Intermediate Value Theorem states. 3. What is a “root” of a function? 4. T/F: If f is defined on an open interval containing c, and lim f (x) exists, then f is continuous at c. x→c 5. T/F: If f is continuous at c, then lim f (x) exist T/F: If f is continuous at c, then 10/21/2020 . lim f (x) = f (c) + x→c x→c 1.6E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10662 6. T/F: If f is continuous on [a, b], then lim f (x) = f (a) . − x→a 7. T/F: If f is continuous on [0, 1) and [1, 2), then f is continuous on [0, 2). 8. T/F: The sum of continuous functions is also continuous. Answer 8. T Exercise 1.6E. 13 Determine on which interval(s) the given function is continuous. 1. f (x) = x − 3x + 9 − −−− − 2. g(x) = √x − 4 −−− − −−− − 3. h(k) = √1 − k + √k + 1 − − − − − − − 4. f (t) = √5t − 30 5. g(t) = 2 2 2 1 √1−t2 6. g(x) = 7. f (x) = e 8. g(s) = ln s 9. h(t) = cos t − − − − − 10. f (k) = √1 − e 11. f (x) = sin(e + x 1 2 1+x x k x 2 ) Answer 11. every where. Exercise 1.6E. 14 : Countinuity For what values of x, is f continuous: 1. Given 2 ⎧ x ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − 5x + 6 if x < 3 3 −x f (x) = ⎨ (1.6E.1) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ −1 if x = 3 x −2 if x > 3 2. Given ⎧ −6 ⎪ ⎪ ⎪ x f (x) = ⎨ ⎪ ⎪ ⎩ ⎪ 2 3x if x ≤ −1 (1.6E.2) if x > −1 3. Given 10/21/2020 1.6E.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10662 ⎧ |x − 3| − 2 ⎪ ⎪ ⎪ ⎪ ⎪ x(x − 1) ⎪ ⎪ ⎪ ⎪ if x ≠ 0, x ≠ 1 f (x) = ⎨ (1.6E.3) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ 1 if x = 1 1 if x = 0 Answer R ∖ {3}, \(R ∖ {−1}, \(R ∖ {0, 1} Exercise 1.6E. 15 : Countinuity For what values of k , each of the following function is continuous every where: 1. 2 (x + k) ⎧ ⎪ if x ≥ 0 f (x) = ⎨ (1.6E.4) ⎩ ⎪ 2 (2 − x) if x < 0 2. (x + k) if x ≤ 2 − −−−−− − √ kx2 + 20 if x > 2 ⎧ ⎪ f (x) = ⎨ ⎩ ⎪ (1.6E.5) 3. 2x + k, x < 3 ⎧ ⎪ ⎪ ⎪ f (x) = ⎨ (1.6E.6) 2 ⎪ kx ⎪ ⎩ ⎪ +1 ,x ≥3 7 4. 2 ⎧ ⎪ x + kx + 1 if x ≤ 1, f (x) = ⎨ ⎩ ⎪ (1.6E.7) 2x + 3 if x > 1. Answer k = ±2, k = ±4 Exercise 1.6E. 16 : Countinuity Given 10/21/2020 1.6E.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10662 ⎧ sin(2x) ⎪ ⎪ ⎪ ⎪ ⎪ ax ⎪ ⎪ ⎪ f (x) = ⎨ 2 x ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ if x < 0, (1.6E.8) −5 if 0 ≤ x < 2 bx − 4 if x ≥ 2 1. For what value of a , f is continuous at x = 0 . 2. For what value of b , f is continuous at x = 2 . Answer a = −10, b = 3/2 Exercise 1.6E. 17 : Countinuity Find values of k and m, if possible, that will make the function f continuous everywhere. 2 x ⎧ ⎪ + 2, x > 1 f (x) = ⎨ m(x + 1) + k, ⎩ ⎪ 2 2x + x + 5, −1 < x ≤ 1 (1.6E.9) x ≤ −1. Answer k = 6, m = −3/2 Solution: when x = 1 : 2 limx→1+ f (x) = limx→1 x +2 = 3 and limx→1− f (x) = limx→1 m(x + 1) + k = 2m + k = f (1) , Since f is continuous at x = 1 , 2m + k = 3 . when x = −1 : limx→−1+ f (x) = limx→−1 m(x + 1) + k = k 2 limx→−1− f (x) = limx→−1 2 x and + x + 5 = 2 − 1 + 5 = 6 = f (1) , Since f is continuous at x = −1 , k = 6 . Therefore, 2m + 6 = 3 , implies m = −3/2. Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution Noncommercial (BY-NC) License. http://www.apexcalculus.com/ Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 1.6E.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10662 1.7: Limit of Trigonometric functions Trigonometric function an Introduction Radian Measure To use trigonometric functions, we first must understand how to measure the angles. Although we can use both radians and degrees, radians are a more natural measurement because they are related directly to the unit circle, a circle with radius 1. The radian measure of an angle is defined as follows. Given an angle θ , let s be the length of the corresponding arc on the unit circle (Figure). We say the angle corresponding to the arc of length 1 has radian measure 1. Figure 1.7.1 : The radian measure of an angle θ is the arc length s of the associated arc on the unit circle. Since an angle of 360° corresponds to the circumference of a circle, or an arc of length 2π, we conclude that an angle with a degree measure of 360° has a radian measure of 2π. Similarly, we see that 180° is equivalent to π radians. Table shows the relationship between common degree and radian values. The Six Basic Trigonometric Functions Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on any circle—not only on a unit circle—or to find an angle given a point on a circle. They also define the relationship among the sides and angles of a triangle. To define the trigonometric functions, first consider the unit circle centered at the origin and a point P = (x, y) on the unit circle. Let θ be an angle with an initial side that lies along the positive x-axis and with a terminal side that is the line segment OP . An angle in this position is said to be in standard position (Figure 1.7.2). We can then define the values of the six trigonometric functions for θ in terms of the coordinates x and y . Figure 1.7.2 : The angle θ is in standard position. The values of the trigonometric functions for θ are defined in terms of the coordinates x and y . Definition: Trigonometric functions Let P = (x, y) be a point on the unit circle centered at the origin O. Let θ be an angle with an initial side along the positive x-axis and a terminal side given by the line segment OP . The trigonometric functions are then defined as 1 cscθ = sin θ = y cosθ = x 10/21/2020 y 1 secθ = 1.7.1 CC-BY-NC-SA x https://math.libretexts.org/@go/page/10283 y x tan θ = cot θ = x y If x = 0, secθ and tan θ are undefined. If y = 0 , then cot θ and cscθ are undefined. We can see that for a point P = (x, y) on a circle of radius r with a corresponding angle θ , the coordinates x and y satisfy x cos θ = (1.7.1) r x = r cos θ (1.7.2) and y sin θ = (1.7.3) r y = r sin θ. (1.7.4) Figure 1.7.3.1 : Diagram demonstrating trigonometric functions in the unit circle., \). The values of the other trigonometric functions can be expressed in terms of x, y, and r (Figure 1.7.3). Figure 1.7.3.2 : For a point P = (x, y) on a circle of radius r, the coordinates x and y satisfy x = r cos θ and y = r sin θ . Table shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determine the values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometric functions are calculated easily from the values of sin θ and cos θ. Table 1.7.2: Values of sin θ and cos θ at Major Angles θ in the First Quadrant θ sin θ cos θ 0 0 1 π 1 – √3 6 2 2 π – √2 – √2 4 π 3 π 2 2 2 – √3 1 2 2 1 0 Example 1.7.2 : Evaluating Trigonometric Functions Evaluate each of the following expressions. a. 2π sin( 10/21/2020 ) 3 1.7.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10283 b. cos(− c. 5π ) 6 15π tan( ) 4 Solution: a) On the unit circle, the angle [Math Processing Error] corresponds to the point (− b) An angle θ = − c) An angle θ = 7π 4 6 15π 4 angle of 5π – √3 1 , ) 2 2 . Therefore, sin( corresponds to a revolution in the negative direction, as shown. Therefore, cos(− =2π+ 7π 5π ) 6 2π ) 3 =y = =x=− – √3 – √3 2 2 . . . Therefore, this angle corresponds to more than one revolution, as shown. Knowing the fact that an 4 corresponds to the point ( – √2 – √2 , 2 ) 2 , we can conclude that tan( 15π y ) = 4 = −1 x . Exercise 1.7.2 Evaluate cos(3π/4) and sin(−π/6). Hint Look at angles on the unit circle. Answer 10/21/2020 1.7.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10283 Trigonometric Identities A trigonometric identity is an equation involving trigonometric functions that is true for all angles θ for which the functions are defined. We can use the identities to help us solve or simplify equations. The main trigonometric identities are listed next. Rule: Trigonometric Identities Reciprocal identities sin θ tan θ = (1.7.5) cos θ cos θ cot θ = (1.7.6) sin θ 1 cscθ = (1.7.7) sin θ 1 secθ = (1.7.8) cos θ Pythagorean identities 2 sin 2 θ + cos 2 1 + tan 2 1 + cot θ =1 (1.7.9) 2 θ = sec θ (1.7.10) 2 θ = csc θ (1.7.11) Addition and subtraction formulas sin(α ± β) = sin α cos β ± cos αsinβ (1.7.12) cos(α ± β) = cos α cos β ∓ sin α sin β (1.7.13) Double-angle formulas sin(2θ) = 2 sin θ cos θ 2 cos(2θ) = 2 cos 2 θ − 1 = 1 − 2 sin (1.7.14) 2 θ = cos 2 θ − sin θ (1.7.15) Graphs and Periods of the Trigonometric Functions We have seen that as we travel around the unit circle, the values of the trigonometric functions repeat. We can see this pattern in the graphs of the functions. Let P = (x, y) be a point on the unit circle and let θ be the corresponding angle . Since the angle θ and θ + 2π correspond to the same point P , the values of the trigonometric functions at θ and at θ + 2π are the same. Consequently, the trigonometric functions are periodic functions. The period of a function f is defined to be the smallest positive value p such that f (x + p) = f (x) for all values x in the domain of f . The sine, cosine, secant, and cosecant functions have a period of 2π. Since the tangent and cotangent functions repeat on an interval of length π, their period is π (Figure 1.7.4). 10/21/2020 1.7.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10283 Figure 1.7.5 : The six trigonometric functions are periodic Limit of the Trigonometric Functions Consider the sine function f (x) = sin(x), where x is measured in radian. The sine function is continuous everywhere,as we see in the graph above:, there fore, lim sin(x) = sin(c). x→c Thingout Loud What is a natural domain of a function? This leads to the following theorem. Theorem 1.7.1 If a is any number in the natural domain of the corresponding trigonometric function, then 1. lim 2. lim 3. lim 4. lim 5. lim 6. lim x→a sin(x) = sin(a). x→a cos(x) = cos(a). x→a tan(x) = tan(a). x→a csc(x) = csc(a). x→a sec(x) = sec(a). x→a cot(x) = cot(a). Proof Since the trigonometric functions are continuous on their natural domain, the statements are valid. Example 1.7.1 : 2 Find lim x→0 x −1 sin ( . ) x −1 Solution: 10/21/2020 1.7.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10283 2 Since is sine 2 x limx→0 sin ( a continuous 2 −1 x −1 x ) = sin ( limx→0 function and x limx→0 ( −1 x −1 ) = sin(limx→0 (x + 1)) = sin(2) −1 x −1 ) = limx→0 (x + 1) = 2 , . The six basic trigonometric functions are periodic and do not approach a finite limit as x → ±∞. For example, sinx oscillates between 1and − 1 (Figure). The tangent function x has an infinite number of vertical asymptotes as x → ±∞ ; therefore, it does not approach a finite limit nor does it approach ±∞ as x → ±∞ as shown in Figure. Figure 1.7.19:The function f(x) = sinx oscillates between 1and − 1 as x → ±∞ Figure 1.7.20: The function f(x) = tanx does not approach a limit and does not approach ±∞ as x → ±∞ The Squeeze Theorem The techniques we have developed thus far work very well for algebraic functions, but we are still unable to evaluate limits of very basic trigonometric functions. The next theorem, called the squeeze theorem, proves very useful for establishing basic trigonometric limits. This theorem allows us to calculate limits by “squeezing” a function, with a limit at a point a that is unknown, between two functions having a common known limit at a. Figure illustrates this idea. 10/21/2020 1.7.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10283 Figure 1.7.4 : The Squeeze Theorem applies when f (x) ≤ g(x) ≤ h(x) and lim x→a f (x) = limx→a h(x) . The Squeeze Theorem Let f (x), g(x), and h(x) be defined for all x≠a over an open interval containing a. If f (x) ≤ g(x) ≤ h(x) (1.7.16) lim f (x) = L = lim h(x) (1.7.17) for all x ≠ a in an open interval containing a and x→a where L is a real number, then lim x→a x→a g(x) = L. Example 1.7.2 : Applying the Squeeze Theorem Apply the squeeze theorem to evaluate lim x→0 xcosx . Solution Because −1 ≤ cosx ≤ 1 for all x, we have −x ≤ xcosx ≤ x for x ≥ 0 and −x ≥ xcosx ≥ x for x ≤ 0 (if x is negative the direction of the inequalities changes when we multiply). Since lim (−x) = 0 = lim x , from the squeeze theorem, we obtain lim xcosx = 0 . The graphs of f (x) = −x, g(x) = xcosx , and h(x) = x are shown in Figure. x→0 x→0 x→0 Figure 1.7.5 : The graphs of f (x), g(x), and h(x) are shown around the point x = 0 . Exercise 1.7.3 Use the squeeze theorem to evaluate lim x→0 2 x sin 1 x . Hint Use the fact that −x 2 10/21/2020 2 2 ≤ x sin(1/x) ≤ x to help you find two functions such that x 2 1.7.7 CC-BY-NC-SA sin(1/x) is squeezed between them. https://math.libretexts.org/@go/page/10283 Answer 0 We now use the squeeze theorem to tackle several very important limits. Although this discussion is somewhat lengthy, these limits prove invaluable for the development of the material in both the next section and the next chapter. The first of these limits is lim sin θ . θ→0 Limits: An Important Trig Limit (GeoGebra) Limit: An Important Trig Limit Theorem 1.7.1 sin x lim x→0 = 1. x Proof We begin by considering the unit circle. Each point on the unit circle has coordinates (cos θ, sin θ) for some angle θ as shown in Figure 1.7.1. Using similar triangles, we can extend the line from the origin through the point to the point (1, tan θ), as shown. (Here we are assuming that 0 ≤ θ ≤ π/2 .Later we will show that we can also consider θ ≤ 0 .) Figure 1.7.1: The unit circle and related triangles. Figure 1.19 shows three regions have been constructed in the first quadrant, two triangles and a sector of a circle, which are also drawn below. The area of the large triangle is tan θ; the area of the sector is θ/2; the area of the triangle contained inside the sector is sin θ . It is then clear from the diagram that 1 2 1 2 10/21/2020 1.7.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10283 Multiply all terms by 2 sin θ , giving 1 θ ≥ cos θ ≥ 1. (1.7.18) sin θ Taking reciprocals reverses the inequalities, giving sin θ cos θ ≤ ≤ 1. (1.7.19) θ (These inequalities hold for all values of θ near 0, even negative values, since cos(−θ) = cos θ and sin(−θ) = − sin θ .) Now take limits. sin θ lim cos θ ≤ lim θ→0 ≤ lim 1 θ θ→0 (1.7.20) θ→0 sin θ cos 0 ≤ lim ≤1 (1.7.21) θ θ→0 sin θ 1 ≤ lim ≤1 Clearly this means that lim θ→0 sin θ θ (1.7.22) θ θ→0 =1 Example 1.7.12 : Evaluating an Important Trigonometric Limit Evaluate lim 1−cos θ θ→0 θ . Solution In the first step, we multiply by the conjugate so that we can use a trigonometric identity to convert the cosine in the numerator to a sine: limθ→0 1−cos θ θ = limθ→0 1−cos θ θ θ→0 θ(1+cos θ) 2 =lim sin θ θ→0 θ(1+cos θ) sin θ θ→0 =1 ⋅ 0 2 1+cos θ 2 1−co s θ =lim =lim 1+cos θ ⋅ =0 θ ⋅ sin θ 1+cos θ . Therefore, limθ→0 1−cos θ θ = 0. Exercise 1.7.10 Evaluate lim θ→0 1−cos θ sin θ . Answer Multiply numerator and denominator by 1 + cos θ . Answer 10/21/2020 1.7.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10283 0 Example 1.7.1 For each of the following functions f , valuate lim f (x) and a. . Determine the horizontal asymptote(s) for f . lim x→∞ f (x) x→−∞ sinx f (x) = x b. f (x) = tan −1 (x) Solution a. Since 1 ≤ sinx ≤ 1 for all x, we have −1 x ≤ sinx x ≤ 1 x for all x ≠ 0 . Also, since limx→∞ −1 = 0 = limx→∞ x 1 x , we can apply the squeeze theorem to conclude that limx→∞ sinx x = 0. Similarly, limx→−∞ sinx x = 0. Thus, f (x) = has a horizontal asymptote of y = 0 and f (x) approaches this horizontal asymptote as x → ±∞ as shown in the following graph. sinx x Figure 1.7.5 : This function crosses its horizontal asymptote multiple times. b. To evaluate lim tan (x) and lim (−π/2, π/2)as shown in the following graph. x→∞ −1 x→−∞ −1 tan (x) , we first consider the graph of y = tan(x) Figure 1.7.6 : The graph of tanx has vertical asymptotes at x = ± over the interval π 2 Since limx→(π/2)− tanx = ∞, it follows that −1 limx→∞ tan 10/21/2020 (x) = π 2 . 1.7.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10283 Similarly, since lim + x→(π/2) tanx = −∞, it follows that −1 limx→−∞ tan As a result, y = π 2 and y = − π 2 (x) = − are horizontal asymptotes of f (x) = tan −1 π 2 . (x) as shown in the following graph. Figure 1.7.7 : This function has two horizontal asymptotes. Key Concepts Radian measure is defined such that the angle associated with the arc of length 1 on the unit circle has radian measure 1. An angle with a degree measure of 180° has a radian measure of π rad. For acute angles θ ,the values of the trigonometric functions are defined as ratios of two sides of a right triangle in which one of the acute angles is θ . For a general angle θ , let (x, y) be a point on a circle of radius r corresponding to this angle θ . The trigonometric functions can be written as ratios involving x, y , and r. The trigonometric functions are periodic. The sine, cosine, secant, and cosecant functions have period 2π. The tangent and cotangent functions have period π. The squzze theorem limx→0 sin(x) x =1 . Source for the figure: Diagram demonstrating trigonometric functions in the unit circle, Daniel Kleitman. RES.18-003 Calculus for Beginners and Artists. Spring 2005. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NCSA. Contributors Michael Corral (Schoolcraft College). The content of this page is distributed under the terms of the GNU Free Documentation License, Version 1.2. Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution Noncommercial (BY-NC) License. http://www.apexcalculus.com/ Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 1.7.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10283 1.7E: Exercises Exercise 1.7E. 1 Evaluate the following: 1. lim cos x sin x x→π/4 2. lim csc x x→π/6 Answer Add texts here. Do not delete this text first. 10/21/2020 1.7E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10664 1.8: Limits and continuity of Inverse Trigonometric functions Inverse functions Recall that a function f is one-to-one (often written as 1 − 1 ) if it assigns distinct values of y to distinct values of x. In other words, if x ≠ x then f (x ) ≠ f (x ) . Equivalently, f is one-to-one if f (x ) = f (x ) implies x = x . There is a simple horizontal rule for determining whether a function y = f (x) is one-to-one: f is one-to-one if and only if every horizontal line intersects the graph of y = f (x) in the xy-coordinate plane at most once (see Figure 5.3.3). 1 1 2 1 2 1 2 2 Figure 5.3.3 Horizontal rule for one-to-one functions If a function f is one-to-one on its domain, then f has an inverse function, denoted by f range of f . The basic idea is that f −1 −1 , such that y = f (x) if and only if f −1 (y) = x . The domain of f −1 is the "undoes'' what f does, and vice versa. In other words, f −1 (f (x)) = x f (f −1 (y)) = y for all x in the domain of f , and for all y in the range of f . Theorem 1.8.1 If f is continuous and one to one, then \(f^{-1}\ is continuous on its domain. Inverse Trigonometric functions We know from their graphs that none of the trigonometric functions are one-to-one over their entire domains. However, we can restrict those functions to subsets of their domains where they are one-to-one. For example, y = sin x is one-to-one over the interval [− , ] , as we see in the graph below: For − ≤ x ≤ we have −1 ≤ sin x ≤ 1 , so we can define the inverse sine function y = sin whose domain is the interval [−1, 1] and whose range is the interval [− , ] . In other words: π π 2 2 −1 −1 sin π π 2 2 (sin y) = y −1 sin (sin for − x) = x π 2 ≤y ≤ π π 2 2 x (sometimes called the arc sine and denoted by y = arcsin (x π ) (1.8.1) 2 for −1 ≤ x ≤ 1 (1.8.2) Summary of Inverse Trigonometric functions Lets illustrate the summary of Trigonometric functions and Inverse Trigonometric functions in following table: Trigonometric function graph of the Trigonometric function Restricted domain and the range [− f (x) = sin(x) f (x) = cos(x) π 2 , π 2 Inverse Trigonometric function ] and [−1, 1] [0, π] f f −1 −1 (x) = sin −1 −1 (x) = cos graph of the Inverse Trigonometric function Properties x x −1 cos and [−1, 1] (cos y) −1 cos (cos 10/21/2020 1.8.1 CC-BY-NC-SA x) = y fo = x fo https://math.libretexts.org/@go/page/10556 (− f (x) = tan(x) π 2 π , 2 ) f and R −1 (x) = tan −1 tan x −1 (tan y) tan (tan −1 cot f (x) = cot(x) −1 (cot y) −1 cot (cot csc [0, π] , with x ≠ π x) −1 x) = y f = x fo = y fo = x fo (csc y) = y x) = x (sec y) = y = x csc (csc −1 2 and f (x) = sec(x) R sec −1 sec (sec −1 x) Below are examples: Example 1.8.1 : Find sin −1 (sin π 4 ) . Solution: Since − π ≤ 2 π π ≤ 4 2 , we know that sin −1 (sin π 4 π ) = 4 , by Equation 1.8.1. Example 1.8.2 : Find sin −1 (sin 5π ) 4 . Solution: Since 5π that − π > 4 π 2 , we can not use Equation ≤y ≤ 2 and sin π 2 1 y =− √2 . But we know that 1.8.1 5π sin 4 =− 1 √2 . Thus, −1 sin (sin 5π 4 −1 ) = sin (− 1 √2 ) is, by definition, the angle y such . That angle is y = − , since π 4 sin (− π 4 ) = − sin ( π 4 ) = − 1 √2 . Example 1.8.3 : Find cos −1 (cos π 3 . ) Solution: Since 0 ≤ π 3 ≤π , we know that cos −1 (cos π 3 π ) = 3 , by Equation 1.8.3. Example 1.8.4 : Find cos −1 (cos 4π 3 . ) Solution: Since 4π 3 >π 0 ≤y ≤π , we can not use Equation and cos 10/21/2020 y =− 1 2 . But we know that 1.8.3 . That angle is y = 2π 3 cos 4π 3 (i.e. 120 ). Thus, cos ∘ −1 =− (cos 1 2 . Thus, 4π 3 ) = −1 cos 2π 3 (cos 4π 3 −1 ) = cos (− 1 2 ) is, by definition, the angle y such that . 1.8.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10556 Example 1.8.5 : Find tan −1 . π (tan ) 4 Solution: Since − π 2 ≤ π 4 ≤ , we know that tan π −1 2 (tan π 4 π , by Equation 1.8.5. ) = 4 Example 1.8.6 : Find tan . −1 (tan π) Solution: Since π > , we can not use Equation 1.8.5. But we know that tan and tan y = 0 . That angle is y = 0 . Thus, tan (tan π) = 0 . π 2 π =0 . Thus, tan −1 −1 (tan π) = tan 0 is, by definition, the angle y such that − π 2 ≤y ≤ π 2 −1 Example1.8.7 : Find the exact value of cos −1 (sin (− 1 4 . )) Solution: Let θ = sin −1 (− 1 4 ) . We know that − π 2 ≤θ ≤ 2 cos π 2 , so since sin θ =− 1 4 <0 , θ must be in QIV. Hence cos 2 1 2 θ = 1 − sin θ = 1 − (− ) θ >0 . Thus, cos 15 ⇒ cos θ = 16 −1 (sin (− 1 4 . Thus, − − √15 = 4 Note that we took the positive square root above since cos θ >0 . 4 − − √15 . )) = 4 Example 1.8.8 : x Show that tan (sin −1 x) = − −−− − √1 − x2 for −1 < x < 1 . Solution: When x = 0 , the Equation holds trivially, since −1 tan (sin 0 0) = tan 0 = 0 = − − − − − 2 √1 − 0 . Now suppose that 0 < x < 1 . Let θ = sin x . Then θ is in QI and sin θ = x . Draw a right triangle with an angle θ such that the opposite leg has length x and the hypotenuse has length 1, as in Figure 5.3.10 (note that this is possible since 0 < x < 1 ). Then sin θ = = x . By the Pythagorean Theorem, the adjacent leg − −−− − has length √1 − x . Thus, tan θ = . −1 x 1 x 2 √1−x2 If −1 < x < 0 then θ = sin −1 x is in QIV. So we can draw the same triangle except that it would be "upside down'' and we would again have since the tangent and sine have the same sign (negative) in QIV. Thus, tan (sin −1 x x) = − −−− − √1 − x2 tan θ = x √1−x2 , for −1 < x < 1 . Example 1.8.9 : Prove the identity tan −1 −1 x + cot x = π 2 . Solution: Let θ = cot −1 x . Using relations, we have tan ( π 2 − θ) = − tan (θ − π 2 ) = cot θ = −1 cot (cot x) = x , by Equation 1.8.8. So since tan (tan x) = x for all x, this means that tan (tan x) = tan ( − θ) . Thus, tan (tan x) = tan ( − cot x) . Now, we know that 0 < cot x < π , so − < − cot x < , i.e. − cot x is in the restricted subset on which the tangent function is one-to-one. Hence, tan (tan x) = tan ( − cot x) implies that tan x = − cot x , which proves the identity. −1 −1 π 2 −1 −1 π −1 π π 2 2 π −1 π 2 −1 2 −1 π −1 2 −1 2 π −1 2 Continuity of Inverse Trigonometric functions 10/21/2020 1.8.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10556 Example 1.8.1 : Let f (x) = −1 3 sec (x) −1 4−tan (x) . Find the values(if any) for which f (x) is continuous. Exercise 1.8.1 Let f (x) = −1 3 sec (x) −1 8+2 tan . Find the values(if any) for which f (x) is continuous. (x) Answer Limit of Inverse Trigonometric functions Theorem 1.8.1 −1 li mx→∞ tan (x) = −1 li mx→−∞ tan li mx→∞ sec −1 π 2 . (x) = − π 2 . (x) = limx→∞ sec −1 (x) = π 2 . Example 1.8.1 : Find lim x→∞ −1 sin (2 tan . (x)) Exercise 1.8.1 Find lim x→−∞ −1 sin (2 tan . (x)) Answer Contributors Michael Corral (Schoolcraft College). The content of this page is distributed under the terms of the GNU Free Documentation License, Version 1.2. Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SANC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 1.8.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10556 1.8E: Exercises 10/21/2020 1.8E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10665 1.9: Limit of Exponential Functions and Logarithmic Functions Limit of Exponential Functions Definition A quantity grows linearly over time if it increases by a fixed amount with each time interval. A quantity decreases linearly over time if it decreases by a fixed amount with each time interval. Example 1.9.1 : If you start with $1000 and put $200 in a jar every month to save for a vacation, then every month the vacation savings grow by $200 and in x months you will have: Amount = 1000 + 200x Definition A quantity grows exponentially over time if it increases by a fixed percentage with each time interval. A quantity decays exponentially over time if it decreases by a fixed percentage with each time interval. Example 1.9.2 : If you start with a debt of $1000 and you are charged an annual interest rate of 24 percent (typical credit card interest rate) then how much will you owe after X months? 24 percent per year = 2 percent per month (this is how they convert it to a monthly interest rate) EXPONENTIAL FUNCTION For any real number x, an exponential function is a function with the form x f (x) = b where is any positive real number such that b ≠ 1 . The domain of f is all real numbers. The range of f is all positive real numbers. The y-intercept is (0, 1), and the horizontal asymptote is y = 0 . b CHARACTERISTICS OF THE EXPONENTIAL FUNCTION An exponential function with the form f (x) = b , b > 0 , b ≠ 1 ,has these characteristics: x one-to-one function horizontal asymptote: y = 0 domain: (– ∞, ∞) range: (0, ∞) x-intercept: none y-intercept: (0, 1) increasing if b > 1 decreasing if b < 1 10/21/2020 1.9.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10284 Rule: Laws of Exponents For any constants a > 0 ,b > 0 , and for all x and y, 1. b x y x+y ⋅b =b x 2. b x−y =b y b 3. (b ) 4. (ab) x y xy =b x x 5. a a x =( x x =a b x ) b b Example 1.9.2 : Using the Laws of Exponents Use the laws of exponents to simplify each of the following expressions. 2/3 a. (2x 3 ) −1/3 (4x 3 b. (x y −1 2 2 ) 2 ) −2 (xy ) Soution a. We can simplify as follows: 2/3 (2x 3 ) −1/3 (4x 2 3 ) 2/3 2 (x = 2 −1/3 4 (x 3 2 ) 2 2 8x = −2/3 = 2 16x ) 2/3 x x 8/3 = x . 2 b. We can simplify as follows: 3 (x y −1 2 2 2 ) −2 (x3 ) (y = (xy ) −1 2 2 −2 x − 2(y ) 6 ) −2 x y = −2 x 6 y −4 2 =x x y −2 y 4 8 =x y 2 . Exercise 1.9.2 : Use the laws of exponents to simplify (6x −3 2 −4 y )/(12 x 5 . y ) Hint: a b x /x a−b =x Answer: 3 x/(2 y ) The Number e 10/21/2020 1.9.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10284 A special type of exponential function appears frequently in real-world applications. To describe it, consider the following example of exponential growth, which arises from compounding interest in a savings account. Suppose a person invests P dollars in a savings account with an annual interest rate r, compounded annually. The amount of money after 1 year is A(1) = P + rP = P (1 + r) . The amount of money after 2 years is . 2 A(2) = A(1) + rA(1) = P (1 + r) + rP (1 + r) = P (1 + r) More generally, the amount after t years is . t A(t) = P (1 + r) If the money is compounded 2 times per year, the amount of money after half a year is 1 r A( r ) = P +( 2 )P = P (1 + ( 2 )) 2 . The amount of money after 1 year is 1 A(1) = A( r )+( 2 1 )A( 2 r ) = P (1 + 2 r r )+ (P (1 + 2 2 r )) = P (1 + 2 2 ) . 2 After t years, the amount of money in the account is r A(t) = P (1 + 2t ) 2 . More generally, if the money is compounded n times per year, the amount of money in the account after t years is given by the function r A(t) = P (1 + nt ) . n What happens as n → ∞? To answer this question, we let m = n/r and write r (1 + nt ) 1 = (1 + n and examine the behavior of (1 + 1/m) m 1 m (1 + ) , as m → ∞ , using a table of values (Table). Values of (1 + m mrt ) m 1 m ) m as m → ∞ 10 100 1000 10,000 100,000 1,000,000 2.5937 2.7048 2.71692 2.71815 2.718268 2.718280 m Looking at this table, it appears that (1 + 1/m) is approaching a number between 2.7 and 2.8 as m → ∞ . In fact, (1 + 1/m) does approach some number as m → ∞ . We call this number e . To six decimal places of accuracy, m m . e ≈ 2.718282 The letter e was first used to represent this number by the Swiss mathematician Leonhard Euler during the 1720s. Although Euler did not discover the number, he showed many important connections between e and logarithmic functions. We still use the notation e today to honor Euler’s work because it appears in many areas of mathematics and because we can use it in many practical applications. Returning to our savings account example, we can conclude that if a person puts P dollars in an account at an annual interest rate r, compounded continuously, then A(t) = P e . This function may be familiar. Since functions involving base e arise often in applications, we call the function f (x) = e the natural exponential function. Not only is this function interesting because of the definition of the number e , but also, as discussed next, its graph has an important property. rt x Since e > 1 , we know ex is increasing on (−∞, ∞). In Figure, we show a graph of f (x) = e along with a tangent line to the graph of at x = 0 . We give a precise definition of tangent line in the next chapter; but, informally, we say a tangent line to a graph of f at x = a is a line that passes through the point (a, f (a)) and has the same “slope” as f at that point . The function x 10/21/2020 1.9.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10284 is the only exponential function b with tangent line at x = 0 that has a slope of 1. As we see later in the text, having this property makes the natural exponential function the most simple exponential function to use in many instances. f (x) = e x x Figure 1.9.3 : The graph of f (x) = e has a tangent line with slope 1 at x = 0 . x Example 1.9.3 : Compounding Interest Suppose $500 is invested in an account at an annual interest rate of r = 5.5, compounded continuously. a. Let t denote the number of years after the initial investment and A(t) denote the amount of money in the account at time t . Find a formula for A(t) . b. Find the amount of money in the account after 10 years and after 20 years. Solution a. If P dollars are invested in an account at an annual interest rate r, compounded continuously, then A(t) = P e . Here P = $500 and r = 0.055. Therefore, A(t) = 500e . rt 0.055t b. After 10 years, the amount of money in the account is A(10) = 500 e 0.055⋅10 = 500 e 0.55 . ≈ $866.63 After 20 years, the amount of money in the account is A(20) = 500 e 0.055⋅20 = 500 e 1.1 . ≈ $1, 502.08 Exercise 1.9.3 If $750 is invested in an account at an annual interest rate of 4, compounded continuously, find a formula for the amount of money in the account after t years. Find the amount of money after 30 years. Hint A(t) = P e rt Answer (A(t)=750e^{0.04t}\). After 30 years, there will be approximately $2, 490.09. Limit and Continuity of Exponential functions Limit laws x limx→∞ b =∞ x limx→−∞ b x limx→∞ b =0 =0 x limx→−∞ b , if b > 1 . , if b > 1 . , if 0 < b < 1 . =∞ , if 0 < b < 1 . =∞ . Note limx→∞ e 10/21/2020 x 1.9.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10284 limx→−∞ e limx→∞ e x −x limx→−∞ e =0 . =0 . −x =∞ . 1 lim(1 + x ) x =e x→0 x lim x→0 e −1 x =1 Logarithmic Functions Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. These come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels. The exponential function f (x) = b is one-to-one, with domain (−∞, ∞) and range (0, ∞). Therefore, it has an inverse function, called the logarithmic function with base b . For any b > 0, b ≠ 1 , the logarithmic function with base b, denoted log , has domain (0, ∞) and range (−∞, ∞),and satisfies x b if and only if b y logb (x) = y =x . For example, log2 (8) = 3 since 2 3 1 log10 ( ) = −2 =8 , since 10 −2 100 logb (1) = 0 1 = 1 2 10 since b 0 Furthermore, since y = log b (x) =1 , = 100 for any base b > 0 . and y = b are inverse functions, x x logb (b ) = x (1.9.1) and logb (x) b = x. (1.9.2) The most commonly used logarithmic function is the function log . Since this function uses natural e as its base, it is called the natural logarithm. Here we use the notation ln(x) or ln x to mean log (x). For example, e e 3 3 ln(e) = loge (e) = 1, ln(e ) = loge (e ) = 3, ln(1) = loge (1) = 0. (1.9.3) Since the functions f (x) = e and g(x) = ln(x) are inverses of each other, x x ln(e ) = x and e ln x =x , and their graphs are symmetric about the line y = x (Figure). 10/21/2020 1.9.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10284 Figure 1.9.4 : The functions y = e and y = ln(x) are inverses of each other, so their graphs are symmetric about the line y = x. x In general, for any base b > 0 ,b ≠ 1 , the function g(x) = log (x) is symmetric about the line y = x with the function f (x) = b . Using this fact and the graphs of the exponential functions, we graph functions log for several values of b>1 (Figure). b x b Figure 1.9.5 : Graphs of y = log b (x) are depicted for b = 2, e, 10 . Before solving some equations involving exponential and logarithmic functions, let’s review the basic properties of logarithms. Properties of Logarithms If a, b, c > 0, b ≠ 1 , and r is any real number, then 1. log b (ac) 2. log b( 3. log = logb (a) + logb (c) a c r b (a ) = logb (a) − logb (c) ) = rlogb (a) (Product property) (Quotient property) (Power property) Example 1.9.4 : Solving Equations Involving Exponential Functions Solve each of the following equations for x. a. 5 b. e x x =2 + 6e −x =5 Solution a. Applying the natural logarithm function to both sides of the equation, we have x ln 5 = ln 2 . Using the power property of logarithms, x ln 5 = ln 2. Therefore, x = ln 2/ ln 5. b. Multiplying both sides of the equation by e ,we arrive at the equation x e 2x + 6 = 5e x . Rewriting this equation as e 2x − 5e x +6 = 0 , we can then rewrite it as a quadratic equation in e : x x 2 (e ) x − 5(e ) + 6 = 0. Now we can solve the quadratic equation. Factoring this equation, we obtain 10/21/2020 1.9.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10284 (e Therefore, the solutions satisfy x = ln 3, ln 2 . e x =3 and e x x =2 − 3)(e x − 2) = 0. . Taking the natural logarithm of both sides gives us the solutions Exercise 1.9.4 Solve e 2x /(3 + e 2x ) = 1/2 . Hint First solve the equation for e 2x Answer ln 3 x = 2 . Example 1.9.5 : Solving Equations Involving Logarithmic Functions Solve each of the following equations for x. a. 1 ln( ) =4 x b. log 0√− x + log 0x = 2 c. ln(2x) − 3 ln(x ) = 0 1 1 2 Solution a. By the definition of the natural logarithm function, 1 ln( ) =4 x if and only if e 4 1 = x . Therefore, the solution is x = 1/e . 4 b. Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as x x log1 0 log1 0x = log1 0x + 3/2 log1 0 x 3 = = log1 0x . 2 Therefore, the equation can be rewritten as 3 2 The solution is x = 10 4/3 [ = 10 ]10 3 log1 0x = 2 or log 1 0x 4 = 3 . . c. Using the power property of logarithmic functions, we can rewrite the equation as ln(2x) − ln(x 6 ) =0 . Using the quotient property, this becomes 2 ln( Therefore, 2/x 5 =1 x5 ) =0 – , which implies x = √2 . We should then check for any extraneous solutions. 5 Exercise 1.9.5 Solve ln(x 3 ) − 4 ln(x) = 1 . Hint First use the power property, then use the product property of logarithms. Answer 10/21/2020 1.9.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10284 1 x = e When evaluating a logarithmic function with a calculator, you may have noticed that the only options are log 0 or log, called the common logarithm, or \ln , which is the natural logarithm. However, exponential functions and logarithm functions can be expressed in terms of any desired base b . If you need to use a calculator to evaluate an expression with a different base, you can apply the change-of-base formulas first. Using this change of base, we typically write a given exponential or logarithmic function in terms of the natural exponential and natural logarithmic functions. 1 Rule: Change-of-Base Formulas Let a > 0, b > 0 , and a ≠ 1, b ≠ 1 . 1. a x xlo gb a =b for any real number x. If b = e , this equation reduces to a x 2. log ax = logb x logb a =e xlo ge a =e x ln a . for any real number x > 0 . ln x If b = e , this equation reduces to log ax = ln a . Proof For the first change-of-base formula, we begin by making use of the power property of logarithmic functions. We know that for any base b > 0, b ≠ 1 , log (a ) = xlog a . Therefore, x b b x lo gb ( a ) b In addition, we know that b and log x b (x) =b xlo gb a . are inverse functions. Therefore, x lo gb ( a ) b Combining these last two equalities, we conclude that a x x =a xlo gb a =b . . To prove the second property, we show that (logb a) ⋅ (loga x) = logb x. Let u = log know that b b a, u v = loga x v = a, a =x , and w = log x . We will show that u ⋅ v = w . By the definition of logarithmic functions, we , and b = x .From the previous equations, we see that b w uv b Therefore, b uv w =b u v = (b ) v =a w =x =b . . Since exponential functions are one-to-one, we can conclude that u ⋅ v = w . □ Example 1.9.6 : Changing Bases Use a calculating utility to evaluate log 37 with the change-of-base formula presented earlier. Solution Use the second equation with a = 3 and e = 3 : log 37 ln 7 = ≈ 1.77124 ln 3 . Exercise 1.9.6 Use the change-of-base formula and a calculating utility to evaluate log 46 . Hint Use the change of base to rewrite this expression in terms of expressions involving the natural logarithm function. Answer 10/21/2020 1.9.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10284 1.29248 Example 1.9.7 : The Richter Scale for Earthquakes Figure 1.9.6 : (credit: modification of work by Robb Hannawacker, NPS) In 1935, Charles Richter developed a scale (now known as the Richter scale) to measure the magnitude of an earthquake. The scale is a base-10 logarithmic scale, and it can be described as follows: Consider one earthquake with magnitude R on the Richter scale and a second earthquake with magnitude R on the Richter scale. Suppose R > R , which means the earthquake of magnitude R is stronger, but how much stronger is it than the other earthquake? A way of measuring the intensity of an earthquake is by using a seismograph to measure the amplitude of the earthquake waves. If A is the amplitude measured for the first earthquake and A is the amplitude measured for the second earthquake, then the amplitudes and magnitudes of the two earthquakes satisfy the following equation: 1 2 1 2 1 1 2 A1 R1 − R2 = log10 ( ) A2 . Consider an earthquake that measures 8 on the Richter scale and an earthquake that measures 7 on the Richter scale. Then, A1 8 − 7 = log10 ( ) A2 . Therefore, A1 log10 ( ) =1 A2 , which implies A /A = 10 or A = 10A . Since A is 10 times the size of A , we say that the first earthquake is 10 times as intense as the second earthquake. On the other hand, if one earthquake measures 8 on the Richter scale and another measures 6, then the relative intensity of the two earthquakes satisfies the equation 1 2 1 2 1 2 A1 log10 ( Therefore, A 1 = 100 A2 ) = 8 −6 = 2 A2 . .That is, the first earthquake is 100 times more intense than the second earthquake. How can we use logarithmic functions to compare the relative severity of the magnitude 9 earthquake in Japan in 2011 with the magnitude 7.3 earthquake in Haiti in 2010? Solution To compare the Japan and Haiti earthquakes, we can use an equation presented earlier: A1 9 − 7.3 = log10 ( ) . A2 Therefore, A /A = 10 the earthquake in Haiti. 1.7 1 2 , and we conclude that the earthquake in Japan was approximately 50 times more intense than Exercise 1.9.7 Compare the relative severity of a magnitude 8.4 earthquake with a magnitude 7.4 earthquake. Hint R1 − R2 = log10 (A1/A2) . Answer The magnitude 8.4 earthquake is roughly 10 times as severe as the magnitude 7.4 earthquake. 10/21/2020 1.9.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10284 EXAMPLE 1.9.8: DETERMINING END BEHAVIOR FOR A TRANSCENDENTAL FUNCTION x (2+3e ) Find the limits as x → ∞ and x → −∞ for f(x) = ) (7−5ex and describe the end behavior of f. Solution To find the limit as x → ∞, divide the numerator and denominator by e : x x 2 + 3e lim f(x) = lim x→∞ x→∞ x 7 − 5e x (2/e )+3 = limx→∞ As shown in Figure, e x → ∞ x (7/e )−5. as x → ∞. Therefore, 2 lim x→∞ 3 We conclude that lim (x) = − 3 5 as x → ∞. conclude that 2 7 x→∞ , and the graph of To find the limit as 2 lim f(x) = 7 x→∞ y = e = 0 = lim f x . e approaches the horizontal asymptote 5 x→∞f y = − 7 x as x → −∞ . x → −∞ , use the fact that x e → 0 as x → −∞ to , and therefore the graph of approaches the horizontal asymptote EXERCISE 1.9.8 Find the limits as x → ∞ and x → −∞ for f(x) = x (3e −4) x (5e +2). Hint x lim e = ∞ and im x x→∞ e = 0. x→∞ Answer 3 lim f(x) = x→∞ , 5 lim f(x) = −2 x→−∞ Key Concepts The exponential function y = b is increasing if b > 1 and decreasing if 0 < b < 1 . Its domain is (−∞, ∞) and its range is (0, ∞). The logarithmic function y = log (x) is the inverse of y = b . Its domain is (0, ∞) and its range is (−∞, ∞). The natural exponential function is y = e and the natural logarithmic function is y = ln x = log x . Given an exponential function or logarithmic function in base a , we can make a change of base to convert this function to any base b > 0 , b ≠ 1 . We typically convert to base e . x x b x e Glossary 10/21/2020 1.9.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10284 base the number b in the exponential function f (x) = b and the logarithmic function f (x) = log x bx exponent the value x in the expression b x natural exponential function the function f (x) = e x natural logarithm the function ln x = log ex number e as m gets larger, the quantity (1 + (1/m) of e is approximately 2.718282 m gets closer to some real number; we define that real number to be e; the value Contributors Contributed by OpenStax, Mathematics at OpenStax CNX Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 1.9.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10284 1.9E: Exercises Exercise 1.9E. 1 lim ln x x→0 Answer Add texts here. Do not delete this text first. 10/21/2020 1.9E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10666 1E: Review Exercises The following problems are collected from past exams. Exercise 1E. 1: Limits Find 1. 3 2 lim x + 6x − 16 x→−2 2. lim π 2 x→0 3. − −−−−−−−− − √3 x2 + 4x − 1 lim 3 −x x→−∞ 2 4. lim x +9 2 x x→4 −1 2 x 5. lim x→4 2 x + x − 20 2 6. lim x→0 − 16 x + 2x 2 x − 2x 2 7. lim x→1 8. lim 1 −x x2 + 5x − 6 3 x 9. x −4 lim |x − 4| − x→4 10. − 27 x −3 x→3 − −−− − √x2 − 4 lim 2x x→∞ 11. lim − − − − − √x − 5 x x→5 Answer: – 2 0, π , √3, 5/3, 8/9, 2, −2/7, 27, −1, 1/2, 0 (1E.1) Solution: 1. 3 lim x 2 + 6x − 16 x→−2 Input x = -2, tthen, 3 2 lim x + 6x − 16 = (−2) 3 2 + 6(2 ) − 16 = −8 + 24 − 16 = 0 x→−2 2. lim π 2 x→0 π 2 is just a number!! Therefore: =π 2 3. lim − −−−−−−−− − √3 x2 + 4x − 1 3 −x x→−∞ Because x is approaching −∞ , consider only the highest degree polynomials of the numerator and denominator: = − − − √3x2 lim x→−∞ 10/21/2020 x 1E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10391 − − Divide the numerator and denominator by √x : 2 √3x2 = √x2 lim x x→−∞ x Simplify and solve: = – √3 lim 1 x→−∞ = – √3 2 x 4. lim 2 x→4 x +9 −1 Because the only domain restrictions for this function are x=1 and x=-1, you may input x=4 and solve: 2 = (4 ) 2 (4 ) = 25 = 5 +9 −1 15 3 2 x 5. lim x→4 2 x − 16 + x − 20 Due to the fact that one of the domain restrictions for this function is x=4, both the numerator and denominator must be factored in order to answer: = lim x→4 (x − 4)(x + 4) (x − 4)(x + 5) Now the (x-4) from the numerator and the (x-4) from the denominator maybe be divided to equal 1: = lim x→4 x +4 x +5 Now, because the limit is no longer a domain restriction, 4 may be input for x: = = 4 +4 4 +5 8 9 2 6. lim x→0 x + 2x 2 x − 2x Due to the fact that one of the domain restrictions for this function is x=0, both the numerator and denominator must be factored in order to answer: = lim x→0 x(x + 2) x(1 − 2x) Now the x from the numerator and the x from the denominator maybe be divided to equal 1: = lim x→0 x +2 1 − 2x Now, because the limit is no longer a domain restriction, 0 may be input for x: = (0) + 2 1 − 2(0) = 2 1 =2 10/21/2020 1E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10391 2 1 −x 7. lim 2 x→1 x + 5x − 6 Due to the fact that one of the domain restrictions for this function is x=1, both the numerator and denominator must be factored in order to answer: (1 − x)(1 + x) = lim x→1 (x + 6)(x − 1) Looking at this, it doesn't seem as though any of the terms divide to equal 1, however you can make one of the terms in the numerator the same by factoring out a -1: −(−1 + x)(1 + x) = lim x→1 (x + 6)(x − 1) −(x − 1)(1 + x) = lim x→1 (x + 6)(x − 1) Now the (x-1) from the numerator and the (x-1) from the denominator maybe be divided to equal 1: −(1 + x) = lim x→1 (x + 6) Now, because the limit is no longer a domain restriction, 1 may be input for x: = −(1 + (1)) ((1) + 6) = −2 7 3 x 8. lim − 27 x −3 x→3 Due to the fact that one of the domain restrictions for this function is x=3, the numerator must be factored in order to answer: 2 2 (x − 3)(x = lim + 3x + 3 x −3 x→3 Now the (x-3) from the numerator and the (x-3) from the denominator maybe be divided to equal 1: = lim x 2 2 + 3x + 3 x→3 Now, because the limit is no longer a domain restriction, 3 may be input for x: = (3) 2 2 + 3(3) + 3 = 9 + 9 + 9 = 27 9. x −4 lim − x→4 |x − 4| Due to absolute value properties |x − 4| = x if x ≥ 0 or −x if x < 0 . Because x − 4 < 0 when x approaches 4 that means that |x − 4| = −x + 4 : − = x −4 lim − x→4 −x + 4 To make the numerator and denominator look similar, a -1 may be factored out of the numerator: = −(−x + 4) lim − x→4 −x + 4 Now the (x-3) from the numerator and the (x-3) from the denominator maybe be divided to equal 1: = lim −1 − x→4 10/21/2020 1E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10391 Because there are no variables left, the limit is simply equal to -1 − −−− − √x2 − 4 10. lim 2x x→∞ Because the limit is approaching ∞ , only the highest degree polynomials from the numerator and the denominator need to be considered: = − − √x2 lim 2x x→∞ This can now be simplified: = x lim 2x x→∞ = 1 lim 2 x→∞ Because there are no variables left, the limit is simply equal to 1/2 − − − − − √x − 5 11. lim x x→5 Because the only domain restriction is when x=0, 5 may simply be inputted for x: = − − − − − − √(5) − 5 5 This may now be simplified: = – √0 5 =0 Exercise 1E. 2: Trigonometry Limits 2 2 sin(t ) + t 1. lim t→0 2t sin(t) sin(2t) 2. lim t cos(t) t→0 3. lim + x→0 2 x csc 2 x √t sin(√t) 4. lim t→0 −4t + sin(t) 2 tan ( 5. lim x→0 x 3 ) 2 4x sin(π − x) 6. lim x→π π −x Answer −1 1, 2, 1, 1 , 3 ,1 (1E.2) 36 Solutions: Add texts here. Do not delete this text first. 2 1. lim 2 sin(t ) + t t→0 2t sin(t) Factor out t from the numerator and multiply the denominator by : 2 t t 10/21/2020 1E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10391 2 sin( t ) 2 t ( + 1) 2 t = lim t→0 sin(t) 2 2t t sin(x) Use the trig limit identity lim t→0 =1 : =1 : x 2 t (1 + 1) = lim t→0 2 2t 2 2t = lim t→0 2 2t =1 2. lim t→0 sin(2t) t cos(t) Multiply the limit by 2 2 : sin(2t) 2 = lim t cos(t) 2 t→0 = 2 lim t→0 sin(2t) 2t cos(t) sin(x) Use the trig limit identity lim t→0 x 1 = 2 lim t→0 cos(t) Input 0 for t: 1 =2 cos(0) =2 1 1 =2 3. lim + x→0 2 x csc 2 x Use trig property csc(x) = 1 2 : sin (x) = lim + x→0 1 2 x 2 sin (x) sin(x) Use the trig limit identity lim t→0 = lim + x→0 x =1 (Notice how the inverse is also equal to 1): =1 again: x 1 sin(x) Use the trig limit identity lim sin(x) t→0 x = lim + x→0 1 =1 4. lim √t sin(√t) t→0 −4t + sin(t) Multiply the numerator by 10/21/2020 √t √t and the factor the denominator by t: 1E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10391 √t (√t sin(√t)) √t =lim t→0 sin(t) t(−4 + ) t sin(x) Use the trig limit identity lim =1 t→0 x : 1t =lim t→0 t(−4 + 1) Simplify: 1 =lim t→0 −3 Therefore: = −1 3 2 tan ( 5. lim x→0 x 3 ) 2 4x Use trig identity tan(x) = sin 3 x 2 cos 1 3 limx→0 4 2 x Multiply the denominator by 2 = limx→0 4 3x(x) 3 3 and simplify the fraction: x sin 1 3 3 x cos2 3 sin(x) Use the trig limit identity lim t→0 = 1 12 limx→0 3 3 3 : x 3 2 cos x 3 sin(x) t→0 1 36 : 3 Use the trig limit identity lim = =1 x 2 3x : 3 (x) cos sin 1 12 =1 x x sin limx→0 Multiply the denominator by = : x 2 = sin(x) cos(x) x 1 limx→0 2 cos x 3 Input 0 for x and solve: = = = 1 1 36 1 36 cos2 (0) 1 1 1 36 6. lim x→π sin(π − x) π −x Let t = π − x and find lim t→0 10/21/2020 : 1E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10391 = lim sin(t)) t→0 t Use the trig limit identity lim sin(x) =1 t→0 : x = lim t→0 1 Change parameters back to x: = lim x→π 1 =1 Exercise 1E. 3: Limits Find the following limits: 1. 3x + 4 lim 2. |x + 2| lim (1E.4) x +2 − x→−2 3. 3 −x lim x→−2 (1E.5) x +2 − − − − − lim √ 2 − x − x 4. 5. (1E.3) −−−−−−− − √ x2 − x − 1 x→−∞ (1E.6) x→2 2x − 1 lim (1E.7) −−−−−−−− − √ 3 x2 + x + 6 x→−∞ − − − − − − − 2 lim √ −x + x 6. (1E.8) + x→1 7. −1 (3 + h ) −1 −3 lim (1E.9) h h→0 8. |3x − 1| − |3x + 1| lim (1E.10) x x→0 9. 1 lim ( t→0 10. 1 − − − − t√ t + 1 − ) (1E.11) t sin x lim x→0 (1E.12) tan x Answer −2 −3, 1, DN E, −1, – √3 −1 , 0, −1 , −6, 9 ,1 (1E.13) 2 Exercise 1E. 4: Asymptotes Find all vertical asymptoe(s) (if any) and all horizontal asymptote(s) (if any) for the function 2 − 5x + 6 2 − 4x + 3 x f (x) = x (1E.14) Justify your work using limits. Answer 10/21/2020 1E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10391 The vertical asymptote is x = 1 and the horizontal asymptote is y = 1 . Exercise 1E. 5 1. lim ln(1+x) x x→0 2. lim sin x x→0 x , where x is measured in degrees not radians. Answer Coming soon. Contributors Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution Noncommercial (BY-NC) License. http://www.apexcalculus.com/ Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 1E.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10391 CHAPTER OVERVIEW 2: DERIVATIVES 2.0: TANGENT LINES AND RATES OF CHANGE 2.0E: EXERCISES 2.1: DERIVATIVE AS A FUNCTION 2.1E: EXERCISES 2.2: TECHNIQUES OF DIFFERENTIATION 2.2E: EXERCISES 2.3: DERIVATIVE AS A RATE OF CHANGE 2.3E: EXERCISES 2.4: DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 2.4E: EXERCISES 2.5: CHAIN RULE 2.5E: EXERCISES 2.6: IMPLICIT DIFFERENTIATION 2.6E: EXERCISES 2.7: DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS In this section, we explore the relationship between the derivative of a function and the derivative of its inverse. For functions whose derivatives we already know, we can use this relationship to find derivatives of inverses without having to use the limit definition of the derivative. In particular, we will apply the formula for derivatives of inverse functions to trigonometric functions. This formula may also be used to extend the power rule to rational exponents. 2.7E: EXERCISES 2.8: DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS 2.8E: EXERCISES 2.9: L'HÔPITAL'S RULE 2.9E: EXERCISES 2E: EXERCISES 1 10/21/2020 2.0: Tangent lines and Rates of change Thing out Loud A common amusement park ride lifts riders to a height then allows them to freefall a certain distance before safely stopping them. Suppose such a ride drops riders from a height of 150 feet. Students of physics may recall that the height (in feet) of the riders, t seconds after freefall (and ignoring air resistance, etc.) can be accurately modeled by f (t) = −16t + 150 . 2 − − − Using this formula, it is easy to verify that, without intervention, the riders will hit the ground at t = 2.5√1.5 ≈ 3.06 seconds. Suppose the designers of the ride decide to begin slowing the riders' fall after 2 seconds (corresponding to a height of 86 ft.). How fast will the riders be traveling at that time? Answer: We have been given a position function, but what we want to compute is a velocity at a specific point in time, i.e., we want an instantaneous velocity. We do not currently know how to calculate this. However, we do know from common experience how to calculate an average velocity. (If we travel 60 miles in 2 hours, we know we had an average velocity of 30 mph.) We looked at this concept in Section 1.1 when we introduced the difference quotient. We have change in distance "rise'' = change in time = average velocity. (2.0.1) run We can approximate the instantaneous velocity at t = 2 by considering the average velocity over some time period containing t = 2 . If we make the time interval small, we will get a good approximation. (This fact is commonly used. For instance, high speed cameras are used to track fast moving objects. Distances are measured over a fixed number of frames to generate an accurate approximation of the velocity.) Consider the interval from t = 2 to t = 3 (just before the riders hit the ground). On that interval, the average velocity is f (3) − f (2) f (3) − f (2) = = −80 ft/s, 3 −2 (2.0.2) 1 where the minus sign indicates that the riders are moving down. By narrowing the interval we consider, we will likely get a better approximation of the instantaneous velocity. On [2, 2.5] we have f (2.5) − f (2) f (2.5) − f (2) = = −72 ft/s. 2.5 − 2 (2.0.3) 0.5 We can do this for smaller and smaller intervals of time. For instance, over a time span of 1/10 have th f (2.1) − f (2) 2.1 − 2 th , we [2, 2.1] f (2.1) − f (2) = Over a time span of 1/100 of a second, i.e., on = −65.6 ft/s. (2.0.4) 0.1 of a second, on [2, 2.01], the average velocity is f (2.01) − f (2) f (2.01) − f (2) = 2.01 − 2 = −64.16 ft/s. (2.0.5) 0.01 What we are really computing is the average velocity on the interval [2, 2 + h] for small values of h . That is, we are computing f (2 + h) − f (2) (2.0.6) h where h is small. What we really want is for h = 0 , but this, of course, returns the familiar "0/0'' indeterminate form. So we employ a limit, as we did in Section 1.1. We can approximate the value of this limit numerically with small values of h as seen in Figure 2.1. It looks as though the velocity is approaching −64 ft/s. Computing the limit directly gives 9/30/2020 2.0.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10287 2 f (2 + h) − f (2) lim h→0 −16(2 + h ) 2 + 150 − (−16(2 ) + 150) = lim h h h→0 2 −64h − 16h = lim h→0 h = lim −64 − 16h h→0 = −64. Figure 2.1: Approximating the instantaneous velocity with average velocities over a small time period h . Graphically, we can view the average velocities we computed numerically as the slopes of secant lines on the graph of f going through the points (2, f (2)) and (2 + h, f (2 + h)) . In Figure 2.2, the secant line corresponding to h = 1 is shown in three contexts. Figure 2.2(a) shows a "zoomed out'' version of f with its secant line. In (b), we zoom in around the points of intersection between f and the secant line. Notice how well this secant line approximates f between those two points -- it is a common practice to approximate functions with straight lines. As h → 0 , these secant lines approach the tangent line, a line that goes through the point (2, f (2)) with the special slope of −64. In parts (c) and (d) of Figure 2.2, we zoom in around the point (2, 86). In (c) we see the secant line, which approximates f well, but not as well the tangent line shown in (d). Figure 2.2: Parts (a), (b) and (c) show the secant line to f (x) with h = 1 , zoomed in different amounts. Part (d) shows the tangent line to f at x = 2. Now that we have both a conceptual understanding of a limit and the practical ability to compute limits, we have established the foundation for our study of calculus, the branch of mathematics in which we compute derivatives and integrals. Most mathematicians and historians agree that calculus was developed independently by the Englishman Isaac Newton (1643–1727) and the German Gottfried 9/30/2020 2.0.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10287 Leibniz (1646–1716), whose images appear in Figure. When we credit Newton and Leibniz with developing calculus, we are really referring to the fact that Newton and Leibniz were the first to understand the relationship between the derivative and the integral. Both mathematicians benefited from the work of predecessors, such as Barrow, Fermat, and Cavalieri. The initial relationship between the two mathematicians appears to have been amicable; however, in later years a bitter controversy erupted over whose work took precedence. Although it seems likely that Newton did, indeed, arrive at the ideas behind calculus first, we are indebted to Leibniz for the notation that we commonly use today. Figure 2.0.1 : Newton and Leibniz are credited with developing calculus independently. App:Average and Instantaneous Rate of Change Limits: Average and Instantaneous Rate of Change Tangent Lines We begin our study of calculus by revisiting the notion of secant lines and tangent lines. Recall that we used the slope of a secant line to a function at a point (a, f (a)) to estimate the rate of change, or the rate at which one variable changes in relation to another variable. We can obtain the slope of the secant by choosing a value of x near a and drawing a line through the points (a, f (a)) and (x, f (x)), as shown in Figure. The slope of this line is given by an equation in the form of a difference quotient: f (x) − f (a) msec = 9/30/2020 (2.0.7) x −a 2.0.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10287 We can also calculate the slope of a secant line to a function at a value a by using this equation and replacing x with a + h , where h is a value close to a. We can then calculate the slope of the line through the points (a, f (a)) and (a + h, f (a + h)) . In this case, we find the secant line has a slope given by the following difference quotient with increment h : f (a + h) − f (a) msec = f (a + h) − f (a) = a+h −a (2.0.8) h Definition: (Newton's) difference quotient Let f be a function defined on an interval I containing a . If x ≠ a is in I , then f (x) − f (a) Q = (2.0.9) x −a is a difference quotient. Also, if h ≠ 0 is chosen so that a + h is in I , then f (a + h) − f (a) Q = (2.0.10) h is a difference quotient with increment h . These two expressions for calculating the slope of a secant line are illustrated in Figure. We will see that each of these two methods for finding the slope of a secant line is of value. Depending on the setting, we can choose one or the other. The primary consideration in our choice usually depends on ease of calculation. Figure 2.0.2 : We can calculate the slope of a secant line in either of two ways. In Figure(a) we see that, as the values of x approach a , the slopes of the secant lines provide better estimates of the rate of change of the function at a . Furthermore, the secant lines themselves approach the tangent line to the function at a , which represents the limit of the secant lines. Similarly, Figure(b) shows that as the values of h get closer to 0, the secant lines also approach the tangent line. The slope of the tangent line at a is the rate of change of the function at a , as shown in Figure(c). Figure 2.0.3 : The secant lines approach the tangent line (shown in green) as the second point approaches the first. In Figure we show the graph of f (x) = √− x and its tangent line at (1, 1) in a series of tighter intervals about x = 1 . As the intervals become narrower, the graph of the function and its tangent line appear to coincide, making the values on the tangent line a good approximation to the values of the function for choices of x close to 1. In fact, the graph of f (x) itself appears to be locally linear in the immediate vicinity of x = 1 . 9/30/2020 2.0.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10287 Figure 2.0.4 : For values of x close to 1, the graph of f (x) = √− x and its tangent line appear to coincide. Formally we may define the tangent line to the graph of a function as follows. Definition: tangent line Let f (x) (a, f (a)) be a function defined in an open interval containing a . The tangent line to having slope f (x) at a is the line passing through the point f (x)−f (a) mtan = li mx→a x−a provided this limit exists. Equivalently, we may define the tangent line to f (x) at a to be the line passing through the point (a, f (a)) having slope f (a+h)−f (a) mtan = li mh→0 h provided this limit exists. Just as we have used two different expressions to define the slope of a secant line, we use two different forms to define the slope of the tangent line. In this text we use both forms of the definition. As before, the choice of definition will depend on the setting. Now that we have formally defined a tangent line to a function at a point, we can use this definition to find equations of tangent lines. Example 2.0.1 : Finding a Tangent Line Find the equation of the line tangent to the graph of f (x) = x at x = 3. 2 Solution First find the slope of the tangent line. In this example, use Equation. f (x)−f (3) mtan = li mx→3 x−3 Apply the definition. 2 = limx→3 x −9 x−3 Substitute f (x) = x and f (3) = 9 2 (x−3)(x+3) = li mx→3 x−3 = li mx→3 (x + 3) = 6 Factor the numerator to evaluate the limit. Next, find a point on the tangent line. Since the line is tangent to the graph of f (x) at x = 3 , it passes through the point (3, f (3)). We have f (3) = 9 , so the tangent line passes through the point (3, 9). Using the point-slope equation of the line with the slope m = 6 and the point (3, 9), we obtain the line y − 9 = 6(x − 3) . Simplifying, we have y = 6x − 9 . The graph of f (x) = x and its tangent line at 3 are shown in Figure. 2 9/30/2020 2.0.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10287 Figure 2.0.5 : The tangent line to f (x) at x = 3 . Example 2.0.2 : The Slope of a Tangent Line Revisited Use Equation to find the slope of the line tangent to the graph of f (x) = x at x = 3 . 2 Solution The steps are very similar to Example. See Equation for the definition. f (3+h)−f (3) mtan = li mh→0 Apply the definition. h 2 (3+h) −9 = limh→0 h Substitute f (3 + h) = (3 + h) 2 = limh→0 2 9+6h+h −9 h and f (3) = 9 . Expand and simplify to evaluate the limit. h(6+h) = li mh→0 = li mh→0 (6 + h) = 6 h We obtained the same value for the slope of the tangent line by using the other definition, demonstrating that the formulas can be interchanged. Example 2.0.c : Finding the Equation of a Tangent Line Find the equation of the line tangent to the graph of f (x) = 1/x at x = 2 . Solution We can use Equation, but as we have seen, the results are the same if we use Equation. f (x)−f (2) mtan = li mx→2 x−2 1 = limx→2 x = limx→2 = limx→2 =− 1 4 − 1 2 x−2 1 = li mx→2 Apply the definition. x − Substitute f (x) = 2 ⋅ 2x 2x (2−x) (x−2)(2x) 2x and f (2) = 1 2 1 x−2 −1 1 x Multiply numerator and denominator by 2x to simplify fractions. Simplify. Simplify using 2−x x−2 = −1 ,for x ≠ 2 . Evaluate the limit. We now know that the slope of the tangent line is − . To find the equation of the tangent line, we also need a point on the line. We know that f (2) = . Since the tangent line passes through the point (2, ) we can use the point-slope equation of a line to find the equation of the tangent line. Thus the tangent line has the equation y = − x + 1 . The graphs of f (x) = and y = − x + 1 are shown in Figure. 1 4 9/30/2020 1 1 2 2 1 1 1 4 x 4 2.0.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10287 Figure 2.0.6 :The line is tangent to f (x) at x = 2 . Exercise 2.0.1 Find the slope of the line tangent to the graph of f (x) = √− x at x = 4 . Hint Use either Equation or Equation. Multiply the numerator and the denominator by a conjugate. Answer \frac{1}{4} The Derivative of a Function at a Point The type of limit we compute in order to find the slope of the line tangent to a function at a point occurs in many applications across many disciplines. These applications include velocity and acceleration in physics, marginal profit functions in business, and growth rates in biology. This limit occurs so frequently that we give this value a special name: the derivative. The process of finding a derivative is called differentiation. Definition Let f (x) be a function defined in an open interval containing a . The derivative of the function f (x) at a , denoted by f '(a), is defined by f (x)−f (a) f '(a) = limx→a x−a provided this limit exists. Alternatively, we may also define the derivative of f (x) at a as f (a+h)−f (a) f '(a) = li mh→0 . h Velocities and Rates of Change Now that we can evaluate a derivative, we can use it in velocity applications. Recall that if s(t) is the position of an object moving along a coordinate axis, the average velocity of the object over a time interval [a, t] if t > a or [t, a] if t < a is given by the difference quotient s(t)−s(a) vave = As the values of t approach a , the values of velocity at a , denoted v(a) , is given by vave t−a . approach the value we call the instantaneous velocity at s(t)−s(a) v(a) = s'(a) = limt→a t−a a . That is, instantaneous . To better understand the relationship between average velocity and instantaneous velocity, see Figure. In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time t = a whose position at time t is given by the function s(t) . The slope of the secant line (shown in green) is the average velocity of the object over the time interval [a, t]. 9/30/2020 2.0.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10287 Figure 2.0.7 : The slope of the secant line is the average velocity over the interval [a, t]. The slope of the tangent line is the instantaneous velocity. Example 2.0.8 : Estimating Velocity A lead weight on a spring is oscillating up and down. Its position at time t with respect to a fixed horizontal line is given by s(t) = sint (Figure). Use a table of values to estimate v(0) . Check the estimate by using Equation. Figure 2.0.8 : A lead weight suspended from a spring in vertical oscillatory motion. Solution We can estimate the instantaneous velocity at shown in Table. t =0 by computing a table of average velocities using values of sint−sin0 t t−0 = t approaching 0, as sint t −0.1 0.998334166 −0.01 0.9999833333 −0.001 0.999999833 0.001 0.999999833 0.01 0.9999833333 0.1 0.998334166 Average velocities using values of t approaching 0 From the table we see that the average velocity over the time interval [−0.1, 0] is 0.998334166, the average velocity over the time interval [−0.01, 0] is 0.9999833333, and so forth. Using this table of values, it appears that a good estimate is v(0) = 1 . By using Equation, we can see that v(0) = s'(0) = li mt→0 sint−sin0 t−0 = li mt→0 sint t =1 . Thus, in fact, v(0) = 1 . Exercise 2.0.5 A rock is dropped from a height of 64 feet. Its height above ground at time t seconds later is given by s(t) = −16t Find its instantaneous velocity 1 second after it is dropped, using Equation. 2 + 64, 0 ≤ t ≤ 2 . Hint v(t) = s'(t) . Follow the earlier examples of the derivative using Equation. Answer 9/30/2020 2.0.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10287 −32 ft/s As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are related concepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change of a function, or the rate of change of a function at any point along the function. Definition: instantaneous rate of change The instantaneous rate of change of a function f (x) at a value a is its derivative f '(a). Example 2.0.9 : Chapter Opener: Estimating Rate of Change of Velocity Figure 2.0.9 : (credit: modification of work by Codex41, Flickr) Reaching a top speed of 270.49 mph, the Hennessey Venom GT is one of the fastest cars in the world. In tests it went from 0 to 60 mph in 3.05 seconds, from 0 to 100 mph in 5.88 seconds, from 0 to 200 mph in 14.51 seconds, and from 0 to 229.9 mph in 19.96 seconds. Use this data to draw a conclusion about the rate of change of velocity (that is, its acceleration) as it approaches 229.9 mph. Does the rate at which the car is accelerating appear to be increasing, decreasing, or constant? Solution: First observe that 60 mph = summarize the information in a table. 88 ft/s, 100 mph ≈146.67 ft/s, 200 t v(t) 0 0 3.05 88 5.88 147.67 14.51 293.33 19.96 337.19 v(t) mph ≈293.33 ft/s, and 229.9 mph ≈337.19 ft/s. We can at different values of t Now compute the average acceleration of the car in feet per second on intervals of the form [t, 19.96] as t approaches 19.96, as shown in the following table. v(t)−v(19.96) t t−19.96 0.0 16.89 3.05 14.74 5.88 13.46 14.51 8.05 = v(t)−337.19 t−19.96 Average acceleration The rate at which the car is accelerating is decreasing as its velocity approaches 229.9 mph (337.19 ft/s). Example 2.0.10 : Rate of Change of Temperature A homeowner sets the thermostat so that the temperature in the house begins to drop from 70°F at 9 p.m., reaches a low of 60° during the night, and rises back to 70° by 7 a.m. the next morning. Suppose that the temperature in the house is given by T (t) = 0.4 t − 4t + 70 for 0 ≤ t ≤ 10 , where t is the number of hours past 9 p.m. Find the instantaneous rate of change of the temperature at midnight. 2 9/30/2020 2.0.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10287 Solution Since midnight is 3 hours past 9 p.m., we want to compute T '(3). Refer to Equation. T '(3) = limt→3 T (t)−T (3) t−3 Apply the definition. 2 = limt→3 0.4 t −4t+70−61.6 t−3 2 = limt→3 Substitute T (t) = 0.4t 2 t−3 0.4(t−3)(t−7) = −1.6 0.4(t−3)(t−7) = limt→3 t−3 = li mt→3 0.4(t − 7) and T (3) = 61.6. Simplify. 0.4 t −4t+8.4 = limt→3 − 4t + 70 t−3 Cancel. Evaluate the limit. The instantaneous rate of change of the temperature at midnight is −1.6°F per hour. Example 2.0.11 : Rate of Change of Profit A toy company can sell x electronic gaming systems at a price of p = −0.01x + 400 dollars per gaming system. The cost of manufacturing x systems is given by C (x) = 100x + 10, 000 dollars. Find the rate of change of profit when 10, 000 games are produced. Should the toy company increase or decrease production? Solution The profit P (x) earned by producing x gaming systems is R(x) − C (x) , where games. Since the company can sell x games at p = −0.01x + 400 per game, R(x) 2 R(x) = xp = x(−0.01x + 400) = −0.01 x is the revenue obtained from the sale of + 400x x . Consequently, 2 P (x) = −0.01 x . + 300x − 10, 000 Therefore, evaluating the rate of change of profit gives P (x)−P (10000) P '(10000) = limx→10000 x−10000 2 = limx→10000 −0.01 x +300x−10000−1990000 x−10000 2 = limx→10000 −0.01 x +300x−2000000 x−10000 . = 100 Since the rate of change of profit P '(10, 000) > 0 and P (10, 000) > 0, the company should increase production. Exercise 2.0.6 A coffee shop determines that the daily profit on scones obtained by charging s dollars per scone is P (s) = −20s + 150s − 10 . The coffee shop currently charges $3.25 per scone. Find P '(3.25), the rate of change of profit when the price is $3.25 and decide whether or not the coffee shop should consider raising or lowering its prices on scones. 2 Answer Use Example for a guide. Answer ; raise prices P '(3.25) = 20 > 0 Key Concepts The slope of the tangent line to a curve measures the instantaneous rate of change of a curve. We can calculate it by finding the limit of the difference quotient or the difference quotient with increment h . The derivative of a function f (x) at a value a is found using either of the definitions for the slope of the tangent line. Velocity is the rate of change of position. As such, the velocity v(t) at time t is the derivative of the position s(t) at time t . Average velocity is given by vave = 9/30/2020 s(t)−s(a) t−a . 2.0.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10287 Instantaneous velocity is given by s(t)−s(a) v(a) = s'(a) = limt→a t−a . Key Equations (Newton's) Difference quotient f (x)−f (a) Q = x−a Difference quotient with increment h f (a+h)−f (a) Q = a+h−a f (a+h)−f (a) = h Slope of tangent line f (x)−f (a) mtan = li mx→a x−a f (a+h)−f (a) mtan = li mh→0 h Derivative of f(x) at a f (x)−f (a) f '(a) = limx→a x−a f (a+h)−f (a) f '(a) = limh→0 h Average velocity s(t)−s(a) vave = t−a Instantaneous velocity s(t)−s(a) v(a) = s'(a) = limt→a t−a Glossary derivative the slope of the tangent line to a function at a point, calculated by taking the limit of the difference quotient, is the derivative difference quotient of a function f (x) at a is given by f (a+h)−f (a) h or f (x)−f (a) x−a differentiation the process of taking a derivative instantaneous rate of change the rate of change of a function at any point along the function a , also called f '(a), or the derivative of the function at a . Source Calculus Applets using GeoGebra by Marc Renault is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Contributors Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/ Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 9/30/2020 2.0.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10287 9/30/2020 2.0.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/10287 2.0E: Exercises Exercise 2.0.1 For the following exercises, use the equation to find the slope of the secant line between the values x and x for each function y = f (x). 1 1) f (x) = 4x + 7; x = 2, x2 = 5 2) f (x) = 8x − 3; x = −1, x2 = 3 1 1 3) f (x) = x 2 + 2x + 1; x1 = 3, x2 = 3.5 4) f (x) = −x + x + 2; x1 = 0.5, x2 = 1.5 5) f (x) = ; x1 = 1, x2 = 3 2 6) f (x) = 7) 2 4 3x−1 x−7 2x+1 ; x1 = −2, x2 = 0 − f (x) = √x ; x1 = 1, x2 = 16 − − − − − 8) f (x) = √x − 9 ; x 1 9) f (x) = x 1/3 = 10, x2 = 13 + 1; x1 = 0, x2 = 8 10) f (x) = 6x 2/3 1/3 + 2x ; x1 = 1, x2 = 27 Answers to odd numbered questions 1. 4 3. 8.5 5. − 3 4 7. 0.2 9. 0.25 Exercise 2.0.2 For the following exercises, use Equation to find the slope of the secant line between the values x and x for each function \ 1 2 For the following functions, a. use Equation to find the slope of the tangent line m tan = f '(a) , and b. find the equation of the tangent line to f at x = a . 1) f (x) = 3 − 4x, a = 2 2) f (x) = x 5 + 6, a = −1 3) f (x) = x 2 + x, a = 1 4) f (x) = 1 − x − x 2 5) f (x) = 7 x ,a =0 ,a =3 − − − − − 6) f (x) = √x + 8 , a = 1 7) f (x) = 2 − 3x 2 8) f (x) = 9) f (x) = −3 x−1 2 x+3 10) f (x) = 3 2 x , a = −2 ,a =4 , a = −4 ,a =3 Answers to odd numbered questions 1. a. −4 b. y = 3 − 4x 3. a. 3 b. y = 3x − 1 5. a. −7 9 b. y = −7 9 x+ 14 3 7. a. 12b. y = 12x + 14 9. a. −2b. y = −2x − 10 Exercise 2.0.3 For the following functions y = f (x), find f '(a) using Equation. 1) f (x) = 5x + 4, a = −1 10/21/2020 2.0E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10629 2) f (x) = −7x + 1, a = 3 3) f (x) = x 2 + 9x, a = 2 4) f (x) = 3x 2 5) − x + 2, a = 1 − f (x) = √x , a = 4 − − − − − 6) f (x) = √x − 2 , a = 6 7) f (x) = 8) f (x) = 9) f (x) = 1 ,a =2 x 1 , a = −1 x−3 1 3 x 10) f (x) = ,a =1 1 √x ,a =4 Answers to odd numbered questions 1. 5 3. 13 5. 1 4 7. − 1 4 9. −3 Exercise 2.0.4 For the following exercises, given the function y = f (x), a. find the slope of the secant line P Q for each point Q(x, f (x))with x value given in the table. b. Use the answers from a. to estimate the value of the slope of the tangent line at P . c. Use the answer from b. to find the equation of the tangent line to f at point P . 1) f (x) = x 2 + 3x + 4 , P (1, 8) (Round to 6 decimal places.) x SlopemPQ x SlopemPQ 1.1 (i) 0.9 (vii) 1.01 (ii) 0.99 (viii) 1.001 (iii) 0.999 (ix) 1.0001 (iv) 0.9999 (x) 1.00001 (v) 0.99999 (xi) 1.000001 (vi) 0.999999 (xii) x SlopemPQ x SlopemPQ 0.1 (i) −0.1 (vii) 0.01 (ii) −0.01 (viii) 0.001 (iii) −0.001 (ix) 0.0001 (iv) −0.0001 (x) 0.00001 (v) −0.00001 (xi) 0.000001 (vi) −0.000001 (xii) 2) f (x) = x+1 2 x −1 3) f (x) = 10e 10/21/2020 , P (0, −1) 0.5x , P (0, 10) (Round to 4 decimal places.) 2.0E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10629 x SlopemPQ −0.1 (i) −0.01 (ii) −0.001 (iii) −0.0001 (iv) −0.00001 (v) −0.000001 (vi) 4) f (x) = tan(x) , P (π, 0) x SlopemPQ 3.1 (i) 3.14 (ii) 3.141 (iii) 3.1415 (iv) 3.14159 (v) 3.141592 (vi) Answers to odd numbered questions 1. a. (i)5.100000, (ii)5.010000, (iii)5.001000, (iv)5.000100, (v)5.000010, (vi)5.000001, (vii)4.900000, (viii)4.990000, (ix)4.999000, (x)4.999900, (xi)4.999990, (x)4.999999 b. m =5 tan c. y = 5x + 3 3. a. (i)4.8771, (ii)4.9875(iii)4.9988, (iv)4.9999, (v)4.9999, (vi)4.9999 b. m tan =5 c. y = 5x + 10 Exercise 2.0.5 For the following position functions y = s(t) , an object is moving along a straight line, where t is in seconds and s is in meters. Find a. the simplified expression for the average velocity from t = 2 to t = 2 + h ; b. the average velocity between t = 2 and t = 2 + h , where (i)h = 0.1, (ii)h = 0.01, (iii)h = 0.001, and (iv)h = 0.0001; and c. use the answer from a. to estimate the instantaneous velocity at t = 2 second. 1) s(t) = 1 3 t +5 2) s(t) = t 2 − 2t 3) s(t) = 2t 3 4) s(t) = 16 2 +3 − t 4 t Answers to odd numbered questions 1. a. 1 3 ; b. (i)0.3 m/s, (ii)0.3 m/s, (iii)0.3 m/s, (iv)0.3 m/s; c. 0.3 = 13 m/s 3. a. 2(h 2 + 6h + 12) ; b. $(i)25.22 m/s, (ii)24.12 m/s, (iii)24.01 m/s, (iv)24 m/s; 10/21/2020 2.0E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10629 c. 24 m/s$ Exercise 2.0.6 1) Use the following graph to evaluate a. f '(1) and b. f '(6). 2) Use the following graph to evaluate a. f '(−3) and b. f '(1.5). Answer 1. a. 1.25 b. 0.5 Exercise 2.0.7 For the following exercises, use the limit definition of derivative to show that the derivative does not exist at x=a for each of the given functions. 1) f (x) = x ,x =0 2) f (x) = x ,x =0 1/3 2/3 3) f (x) = { 4) f (x) = 1 x <1 x x ≥1 ,x =1 |x| ,x =0 x Answers to odd numbered questions 1. li m 3. li m 1/3 x −0 − x→0 x−0 1−1 − x→1 x−1 = li mx→0− 1 2/3 =∞ x = 0 ≠ 1 = li mx→1+ , x−1 x−1 Exercise 2.0.8 The position in feet of a race car along a straight track after t seconds is modeled by the function s(t) = 8t 2 − 1 16 3 t . a. Find the average velocity of the vehicle over the following time intervals to four decimal places: i. [4, 4.1] ii. [4, 4.01] iii. [4, 4.001] iv. [4, 4.0001] b. Use a. to draw a conclusion about the instantaneous velocity of the vehicle at t = 4 seconds. Answer a. (i)61.7244f t/s, (ii)61.0725f t/s(iii)61.0072f t/s(iv)61.0007f t/s b. At 4 seconds the race car is traveling at a rate/velocity of 61 ft/s. 10/21/2020 2.0E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10629 Exercise 2.0.9 The distance in feet that a ball rolls down an incline is modeled by the function s(t) = 14t , 2 where t is seconds after the ball begins rolling. a. Find the average velocity of the ball over the following time intervals: i. [5, 5.1] ii. [5, 5.01] iii. [5, 5.001] iv. [5, 5.0001] b. Use the answers from a. to draw a conclusion about the instantaneous velocity of the ball at t = 5 seconds. Answer Under Construction Exercise 2.0.10 Two vehicles start out traveling side by side along a straight road. Their position functions, shown in the following graph, are given by s = g(t) , where s is measured in feet and t is measured in seconds. s = f (t) and a. Which vehicle has traveled farther at t = 2 seconds? b. What is the approximate velocity of each vehicle at t = 3 seconds? c. Which vehicle is traveling faster at t = 4 seconds? d. What is true about the positions of the vehicles at t = 4 seconds? Answer a. The vehicle represented by f (t), because it has traveled 2 feet, whereas g(t) has traveled 1 foot. b. The velocity of f (t) is constant at 1 ft/s, while the velocity of g(t) is approximately 2 ft/s. c. The vehicle represented by g(t), with a velocity of approximately 4 ft/s. d. Both have traveled 4 feet in 4 seconds. Exercise 2.0.11 The total cost C (x), in hundreds of dollars, to produce x jars of mayonnaise is given by C (x) = 0.000003x 3 . + 4x + 300 a. Calculate the average cost per jar over the following intervals: i. [100, 100.1] ii. [100, 100.01] iii. [100, 100.001] iv. [100, 100.0001] b. Use the answers from a. to estimate the average cost to produce 100 jars of mayonnaise. Answer Under Construction Exercise 2.0.12 For the function f (x) = x 3 10/21/2020 2 − 2x − 11x + 12 , do the following. 2.0E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10629 a. Use a graphing calculator to graph f in an appropriate viewing window. b. Use the ZOOM feature on the calculator to approximate the two values of x = a for which m tan = f '(a) = 0 . Answer a. b. a ≈ −1.361, 2.694 Exercise 2.0.13 For the function f (x) = x 2 1+x , do the following. a. Use a graphing calculator to graph f in an appropriate viewing window. b. Use the ZOOM feature on the calculator to approximate the values of x = a for which m tan = f '(a) = 0 . Answer Under Construction Exercise 2.0.14 Suppose that N (x) computes the number of gallons of gas used by a vehicle traveling x miles. Suppose the vehicle gets 30 mpg. a. Find a mathematical expression for N (x). b. What is N (100)? Explain the physical meaning. c. What is N '(100)? Explain the physical meaning. Answer: a. N (x) = x 30 b. ∼3.3 gallons. When the vehicle travels 100 miles, it has used 3.3 gallons of gas. c. 1 30 . The rate of gas consumption in gallons per mile that the vehicle is achieving after having traveled 100 miles. Exercise 2.0.15 For the function f (x) = x 4 2 − 5x +4 , do the following. a. Use a graphing calculator to graph f in an appropriate viewing window. b. Use the nDeriv function, which numerically finds the derivative, on a graphing calculator to estimate f '(−2), f '(−0.5), f '(1.7), and f '(2.718). Answer Under Construction Exercise 2.0.16 For the function f (x) = 2 x 2 x +1 , do the following. a. Use a graphing calculator to graph f in an appropriate viewing window. b. Use the nDeriv function on a graphing calculator to find f '(−4), f '(−2), f '(2), and f '(4). Answer a. 10/21/2020 2.0E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10629 b. −0.028, −0.16, 0.16, 0.028 Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SANC 4.0 license. Download for free at http://cnx.org. 10/21/2020 2.0E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10629 2.1: Derivative as a Function As we have seen, the derivative of a function at a given point gives us the rate of change or slope of the tangent line to the function at that point. If we differentiate a position function at a given time, we obtain the velocity at that time. It seems reasonable to conclude that knowing the derivative of the function at every point would produce valuable information about the behavior of the function. However, the process of finding the derivative at even a handful of values using the techniques of the preceding section would quickly become quite tedious. In this section we define the derivative function and learn a process for finding it. Derivative Functions The derivative function gives the derivative of a function at each point in the domain of the original function for which the derivative is defined. We can formally define a derivative function as follows. Definition: derivative function Let f be a function. The derivative function, denoted by f′, is the function whose domain consists of those values of x such that the following limit exists: f (x + h) − f (x) f '(x) = lim . (2.1.1) h h→0 A function f (x) is said to be differentiable at a if f (a) exists. More generally, a function is said to be differentiable on S if it is differentiable at every point in an open set S , and a differentiable function is one in which f '(x) exists on its domain. ′ Notations We use a variety of different notations to express the derivative of a function. For the function y = f (x), each of the following notations represents the derivative of f (x): dy d f '(x), , y', dx In place of f '(a) we may also use dy ∣ ∣ dx ∣x=a . (f (x)) dx . Use of the dy dx notation (called Leibniz notation) is quite common in engineering and physics. To understand this notation better, recall that the derivative of a function at a point is the limit of the slopes of secant lines as the secant lines approach the tangent line. The slopes of these secant lines are often expressed in the form Δy where Δy is the difference in the y values corresponding to the difference in the x values, which are Δx expressed as Δx (Figure). Thus the derivative, which can be thought of as the instantaneous rate of change of y with respect to x, is expressed as dy Δy = dx 9/30/2020 lim Δx→0 Δx . 2.1.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10613 Figure 2.1.1 : The derivative is expressed as $ \displaystyle\frac{dy}{dx}=\lim_{Δx→0}\frac{Δy}{Δx}$. In the next few examples we use Equation to find the derivative of a function. Example 2.1.1 : Finding the Derivative of a Square-Root Function Find the derivative of f (x) = √− x. Solution Start directly with the definition of the derivative function. − − − − − − − − − − − √x + h − √x f '(x) = lim h→0 f(x+h)−f(x) f '(x) = lim h→0 h − − − − − − √x + h − √x = lim h→0 − − − − − − √x + h + √x ⋅ = lim h→0 = − − − − − − h(√x + h + √x . − − − − − h = lim h→0 h − Multiply numerator and denominator by √x + h + √x without distributing in thedenominator. − − − − − − √x + h + √x h − Substitutef (x + h) = √x + h and f (x) = √x into Multiply the numerators and simplify. ) 1 Cancel the h . − − − − − − (√x + h + √x ) 1 Evaluate the limit − 2 √x Example 2.1.2 : Finding the Derivative of a Quadratic Function Find the derivative of the function f (x) = x 2 − 2x . Solution Follow the same procedure here, but without having to multiply by the conjugate. 2 ((x + h ) 2 − 2(x + h)) − (x f '(x) = lim h→0 2 lim h→0 x 2 2 − 2x − 2h − x 2xh − 2h + h 9/30/2020 Expand (x + h ) and f (x) = x 2 h→0 − 2x 2 h − 2(x + h) . Simplify h h(2x − 2 + h) = lim h→0 + 2x − 2(x + h) f(x+h)−f(x) h 2 = lim h→0 into f '(x) = lim h + 2xh + h Substitute f (x + h) = (x + h ) 2 − 2x) Factor out h from the numerator h 2.1.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10613 = lim h→0 (2x − 2 + h) Cancel the common factor of h = 2x − 2 Evaluate the limit Exercise 2.1.1 Find the derivative of f (x) = x . 2 Hint Use Equation and follow the example. Answer f '(x) = 2x Exercise 2.1.1 −−−− − Find the derivative of f (x) = √2x + 3 . Answer Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page. Example 2.1.3 : Consider f (x) = a) Find f ′ (x) 3 x . using the definition of a derivative as a limit. b) Find the equation of the tangent line to this function at x = 3 . Exercise 2.1.3 Use the limit definition of the derivative to find f ‘(x) where f (x) = your answer). Then find the equation of the tangent line to f (x) at x = 2 : 1 x+1 (You may using differentiation rules to check Answer Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page. Example 2.1.4 : Use the limit definition of the derivative to find f ‘(x) where f (x) = 1 √x . Exercise 2.1.5 Use the limit definition of the derivative to find f ‘(x) where f (x) = 1 √x+5 . Answer 1 √x+5 Graphing a Derivative 9/30/2020 2.1.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10613 We have already discussed how to graph a function, so given the equation of a function or the equation of a derivative function, we could graph it. Given both, we would expect to see a correspondence between the graphs of these two functions, since f '(x) gives the rate of change of a function f (x) (or slope of the tangent line to f (x)). − In Example we found that for f (x) = √− x , f '(x) = 1/2 √x. If we graph these functions on the same axes, as in Figure, we can use the graphs to understand the relationship between these two functions. First, we notice that f (x) is increasing over its entire domain, which means that the slopes of its tangent lines at all points are positive. Consequently, we expect f '(x) > 0 for all values of x in its domain. Furthermore, as x increases, the slopes of the tangent lines to f (x) are decreasing and we expect to see a corresponding decrease in f '(x). We also observe that f (0) is undefined and that lim f '(x) = +∞ , corresponding to a vertical tangent to f (x) at 0. + x→0 Figure 2.1.2 : The derivative f '(x) is positive everywhere because the function f (x) is increasing. In Example we found that for f (x) = x − 2x, f '(x) = 2x − 2 . The graphs of these functions are shown in Figure. Observe that f (x) is decreasing for x < 1 . For these same values of x, f '(x) < 0. For values of x > 1 , f (x) is increasing and f '(x) > 0. Also, f (x) has a horizontal tangent at x = 1 and f '(1) = 0 . 2 Figure 2.1.3 : The derivative f '(x) < 0 where the function f (x) is decreasing and f '(x) > 0 where f (x) is increasing. The derivative is zero where the function has a horizontal tangent Example 2.1.3 : Sketching a Derivative Using a Function Use the following graph of f (x) to sketch a graph of f '(x). 9/30/2020 2.1.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10613 Solution The solution is shown in the following graph. Observe that f (x) is increasing and f '(x) > 0 on (– 2, 3). Also, f (x) is decreasing and f '(x) < 0 on (−∞, −2) and on (3, +∞). Also note that f (x) has horizontal tangents at – 2 and 3, and f '(−2) = 0 and f '(3) = 0 . Exercise 2.1.2 Sketch the graph of f (x) = x 2 −4 . On what interval is the graph of f '(x) above the x-axis? Hint The graph of f '(x) is positive where f (x) is increasing. Answer (0,+∞) An Applet for Reconstructing a function from its first derivative 9/30/2020 2.1.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10613 Reset the graph Check accuracy Show results! Derivatives and Continuity Now that we can graph a derivative, let’s examine the behavior of the graphs. First, we consider the relationship between differentiability and continuity. We will see that if a function is differentiable at a point, it must be continuous there; however, a function that is continuous at a point need not be differentiable at that point. In fact, a function may be continuous at a point and fail to be differentiable at the point for one of several reasons. Theorem 2.1.1 Differentiability Implies Continuity Let f (x) be a function and a be an element in its domain. If f (x) is differentiable at a , then f is continuous at a . Proof If f (x) is differentiable at a , then f '(a) exists and f (x)−f (a) f '(a) = limx→a x−a We want to show that f (x) is continuous at a by showing that lim x→a . f (x) = f (a). Thus, limx→a f (x) = limx→a (f (x) − f (a) + f (a)) f (x)−f (a) = limx→a ( x−a ⋅ (x − a) + f (a)) Multiply and divide f (x) − f (a) by x − a . f (x)−f (a) = (limx→a x−a ) ⋅ (limx→a (x − a)) + limx→a f (a) ′ = f (a) ⋅ 0 + f (a) = f (a). 9/30/2020 2.1.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10613 Therefore, since f (a) is defined and lim x→a f (x) = f (a) , we conclude that f is continuous at a .□ Thing out Loud We have just proven that differentiability implies continuity, does continuity implies differentiability? Definition: Vertical Cusps Let f be a function and a be point in the domain. Then f is said to have a vertical tangent line x = a if f ‘(a) approaches to ∞ or −∞ . Example 2.1.1 : Fails to be differentiable at Corner To determine an answer to above question, we examine the function f (x) = |x|. This function is continuous everywhere; however, f '(0) is undefined. This observation leads us to believe that continuity does not imply differentiability. Let’s explore further. For f (x) = |x|, f (x)−f (0) f '(0) = limx→0 x−0 |x|−|0| = limx→0 |x| = limx→0 x−0 x . This limit does not exist because |x| limx→0− x = −1 and lim |x| + x→0 x =1 . See Figure. Figure 2.1.4 : The function f (x) = |x| is continuous at 0 but is not differentiable at 0. Exercise 2.1.1 Consider f (x) = |x − 2| . What can you say about the tangent line at x = 2 ? Answer Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page. Example 2.1.2 : Cusp-Vertical Tangent Lines Let’s consider some additional situations in which a continuous function fails to be differentiable. Consider the function − f (x) = √x : 3 9/30/2020 2.1.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10613 3 √x−0 f '(0) = limx→0 x−0 Thus f '(0) does not exist. A quick look at the graph of tangent line at 0 (Figure). = limx→0 3 − f (x) = √x 1 3 √x2 = +∞ . clarifies the situation. The function has a vertical Figure 2.1.5 : The function f (x) = √− x has a vertical tangent at x = 0 . It is continuous at 0 but is not differentiable at 0 . 3 Exercise 2.1.2 Consider f(x)=\sqrt_{5}{2-x} . What can you say about the tangent line at x = 2 ? Answer The function has a vertical tangent line at x = 2 . Example 2.1.3 : Neither Cusp nor corner The function f (x) = { xsin( 1 x ) if x ≠ 0 also has a derivative that exhibits interesting behavior at 0. We see that 0if x = 0 xsin(1/x)−0 f '(0) = limx→0 = limx→0 sin( x−0 1 x ) . This limit does not exist, essentially because the slopes of the secant lines continuously change direction as they approach zero (Figure). Figure 2.1.6 : The function f (x) = { xsin( 0 1 x ) if x ≠ 0 is not differentiable at 0. if x = 0 In summary: 1. We observe that if a function is not continuous, it cannot be differentiable, since every differentiable function must be continuous. However, if a function is continuous, it may still fail to be differentiable. 9/30/2020 2.1.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10613 2. We saw that f (x) = |x| failed to be differentiable at 0 because the limit of the slopes of the tangent lines on the left and right were not the same. Visually, this resulted in a sharp corner on the graph of the function at 0. From this we conclude that in order to be differentiable at a point, a function must be “smooth” at that point. 3. As we saw in the example of f (x) = √− x , a function fails to be differentiable at a point where there is a vertical tangent line. 3 4. As we saw with f (x) = { xsin( 1 x ) if x ≠ 0 0 a function may fail to be differentiable at a point in more complicated if x = 0 ways as well. Example 2.1.4 : A Piecewise Function that is Continuous and Differentiable A toy company wants to design a track for a toy car that starts out along a parabolic curve and then converts to a straight 1 line (Figure). The function that describes the track is to have the form f (x) = { 10 − 2 x 1 4 + bx + c x+ 5 2 if x < −10 where x if x ≥ −10 and f (x) are in inches. For the car to move smoothly along the track, the function f (x) must be both continuous and differentiable at −10. Find values of b and c that make f (x) both continuous and differentiable. Figure 2.1.7 : For the car to move smoothly along the track, the function must be both continuous and differentiable. Solution For the function to be continuous at x = −10 , lim − x→10 limx→−10− f (x) = 1 10 f (x) = f (−10) 2 (−10 ) . Thus, since − 10b + c = 10 − 10b + c and f (−10) = 5 , we must have 10 − 10b + c = 5 . Equivalently, we have c = 10b − 5 . For the function to be differentiable at −10, f (x)−f (−10) f '(10) = limx→−10 x+10 must exist. Since f (x) is defined using different rules on the right and the left, we must evaluate this limit from the right and the left and then set them equal to each other: 1 limx→−10− f (x)−f (−10) x+10 = limx→−10− 1 10 = limx→−10− 10 2 x +bx+c−5 x+10 x2+bx+(10b−5)−5 x+10 Substitute c = 10b − 5 . 2 = limx→−10− x −100+10bx+100b 10(x+10) (x+10)(x−10+10b) = limx→−10− = b −2 10(x+10) Factor by grouping . We also have f (x)−f (−10) limx→−10+ x+10 − = limx→−10+ 9/30/2020 1 4 x+ 5 −5 2 x+10 2.1.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10613 −(x+10) = limx→−10+ =− 1 4 4(x+10) . This gives us b − 2 = − . Thus b = 1 7 4 4 and c = 10( 7 4 25 )−5 = 2 . Exercise 2.1.3 Find values of a and b that make f (x) = { ax + b if x < 3 2 x both continuous and differentiable at 3. if x ≥ 3 Hint Use Example as a guide. Answer a =6 and b = −9 Higher-Order Derivatives The derivative of a function is itself a function, so we can find the derivative of a derivative. For example, the derivative of a position function is the rate of change of position, or velocity. The derivative of velocity is the rate of change of velocity, which is acceleration. The new function obtained by differentiating the derivative is called the second derivative. Furthermore, we can continue to take derivatives to obtain the third derivative, fourth derivative, and so on. Collectively, these are referred to as higher-order derivatives. The notation for the higher-order derivatives of y = f (x) can be expressed in any of the following forms: f ′′ y (x), f ′′ (x), y d 2 ′′′ ′′′ y dx2 It is interesting to note that the notation for d dx ( d dx ( dy dx )) = d ( dx d 2 y dx2 ) = d 3 y dx3 d 2 y 2 dx , (x), f (x), y d 3 y dy 3 , (4) (4) d 4 y dy 4 (x), … , f (x), … , y ,…, d n dy y n (n) (n) (x) (x) . may be viewed as an attempt to express d dx dy ( dx ) more compactly. Analogously, . Example 2.1.5 : Finding a Second Derivative For f (x) = 2x 2 − 3x + 1 , find f ′′ . (x) Solution First find f '(x). Substitute f (x) = 2x 2 2 2 (2(x+h ) −3(x+h)+1)−(2x −3x+1) f '(x) = lim h→0 h and − 3(x + h) + 1 into f(x+h)−f(x) f '(x) = lim h→0 2 = lim h→0 − 3x + 1 2 f (x + h) = 2(x + h ) . h Simplify the numerator. 4xh+h −3h h = lim h→0 (4x + h − 3) Factor out the h in the numerator and cancel with the h in the denominator. = 4x − 3 Take the limit. Next, find f f ′′ ′′ (x) by taking the derivative of f '(x) = 4x − 3. f'(x+h)−f'(x) (x) = lim h→0 h (4(x+h)−3)−(4x−3) = lim h→0 = lim h→0 4 9/30/2020 h Use f '(x) = lim f(x+h)−f(x) h→0 h with f '(x) in place of f (x). Substitute f '(x + h) = 4(x + h) − 3 and f '(x) = 4x − 3. Simplify. 2.1.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10613 Take the limit. = 4 Exercise 2.1.4 Find f ′′ (x) for f (x) = x . 2 Hint We found f '(x) = 2x in a previous checkpoint. Use Equation to find the derivative of f '(x) Answer f ′′ (x) = 2 Example 2.1.6 : Finding Acceleration The position of a particle along a coordinate axis at time t (in seconds) is given by the function that describes its acceleration at time t . 2 s(t) = 3 t − 4t + 1 (in meters). Find Solution Since v(t) = s'(t) and a(t) = v'(t) = s ′′ (t) , we begin by finding the derivative of s(t) : s(t+h)−s(t) s'(t) = limh→0 h 2 2 3(t+h) −4(t+h)+1−(3 t −4t+1) = limh→0 h = 6t − 4. Next, s'(t+h)−s'(t) ′′ s (t) = limh→0 h = limh→0 6(t+h)−4−(6t−4) h = 6. Thus, a = 6m/s . 2 Exercise 2.1.5 For s(t) = t , find a(t). 3 Hint Use Example as a guide. Answer a(t) = 6t Key Concepts The derivative of a function f (x) is the function whose value at x is f '(x). The graph of a derivative of a function f (x) is related to the graph of f (x). Where f (x) has a tangent line with positive slope, f '(x) > 0. Where f (x) has a tangent line with negative slope, f′(x)<0. Where f (x) has a horizontal tangent line, f '(x) = 0. If a function is differentiable at a point, then it is continuous at that point. A function is not differentiable at a point if it is not continuous at the point, if it has a vertical tangent line at the point, or if the graph has a sharp corner or cusp. Higher-order derivatives are derivatives of derivatives, from the second derivative to the nth derivative. Key Equations 9/30/2020 2.1.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10613 The derivative function f (x+h)−f (x) f '(x) = limh→0 h Glossary derivative function gives the derivative of a function at each point in the domain of the original function for which the derivative is defined differentiable at a a function for which f '(a) exists is differentiable at a differentiable on S a function for which f '(x) exists for each x in the open set S is differentiable on S differentiable function a function for which f '(x) exists is a differentiable function higher-order derivative a derivative of a derivative, from the second derivative to the nth derivative, is called a higher-order derivative Source Calculus Applets using GeoGebra by Marc Renault is licensed under a Creative Commons Attribution-NonCommercialShareAlike 3.0 Unported License. Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 9/30/2020 2.1.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/10613 2.1E: Exercises Exercise 2.1E . 1 For the following functions, use the definition of a derivative to find f '(x). You may use derivative rules (will be learned in the next section to check if your answer is correct. 1) f (x) = 6 2) f (x) = 2 − 3x 3) f (x) = 2x +1 7 4) f (x) = 4x 2 5) f (x) = 5x − x 2 − − 6) f (x) = √2x − − − − − 7) f (x) = √x − 6 8) f (x) = 9 x 9) f (x) = x + 10) f (x) = 1 x 1 √x Answer Under Construction Exercise 2.1E . 2 For the following exercises, use the graph of y = f (x) to sketch the graph of its derivative f '(x).. 1) 2) 10/21/2020 2.1E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10628 3) 4) Answers to even numbered questions 10/21/2020 2.1E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10628 2. 4. Exercise 2.1E . 3 For the following exercises, the given limit represents the derivative of a function y = f (x) at x = a . Find f (x) and a . 2/3 (1+h) 1) lim h→0 2) lim h→0 3) lim h→0 2 [3(2+h) +2]−14 h cos(π+h)+1 h 4 (2+h) −16 4) lim h→0 5) lim h→0 6) lim −1 h h 2 [2(3+h) −(3+h)]−15 h h e −1 h→0 h Answers to even numbered questions 2. f (x) = 3x 2 10/21/2020 + 2, a = 2 2.1E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10628 4. f (x) = x ,a =2 6. f (x) = e ,a =0 4 x Exercise 2.1E . 4 For the following functions, a. sketch the graph and b. use the definition of a derivative to show that the function is not differentiable at x = 1 1) f (x) = { − 2 √x 0 ≤x ≤1 3x − 1 x >1 2) f (x) = { 3 x <1 3x x ≥1 3) f (x) = { −x 4) f (x) = { 2x, 2 +2 x x ≤1 x >1 2 x ≤1 x >1 x Answers to even numbered questions 2a. b. li m 3−3 − h→1 h ≠ li mh→1+ 3h h 4a. 10/21/2020 2.1E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10628 2 b. li m 2h − h→1 h ≠ li mh→1+ x+ h −2x h . Exercise 2.1E . 5 For the following graphs, a. determine for which values of x = a the li m x→a f (x) exists but f is not continuous at x = a , and b. determine for which values of x = a the function is continuous but not differentiable at x = a . 1) 2) 10/21/2020 2.1E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10628 Answer 2. a. x = 1, b. x = 2 Exercise 2.1E . 6 Use the graph to evaluate a. f '(−0.5), b. f '(0), c. f '(1), d. f '(2),and e. f '(3), if it exists. Answer Under Construction Exercise 2.1E . 7 For the following exercises, describe what the two expressions represent in terms of each of the given situations. Be sure to include units. a. f (x+h)−f (x) h b. f '(x) = lim f (x+h)−f (x) h→0 h 1) P (x) denotes the population of a city at time x in years. 2) C (x) denotes the total amount of money (in thousands of dollars) spent on concessions by x customers at an amusement park. 3) R(x) denotes the total cost (in thousands of dollars) of manufacturing x clock radios 4) g(x) denotes the grade (in percentage points) received on a test, given x hours of studying. 10/21/2020 2.1E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10628 5) B(x)denotes the cost (in dollars) of a sociology textbook at university bookstores in the United States in x years since 1990. 6) p(x) denotes atmospheric pressure at an altitude of x feet. Answers to even numbered questions 2a. Average rate at which customers spent on concessions in thousands per customer. b. Rate (in thousands per customer) at which x customers spent money on concessions in thousands per customer. 4a. Average grade received on the test with an average study time between two values. b. Rate (in percentage points per hour) at which the grade on the test increased or decreased for a given average study time of x hours. 6a. Average change of atmospheric pressure between two different altitudes. b. Rate (torr per foot) at which atmospheric pressure is increasing or decreasing at x feet. Exercise 2.1E . 8 Sketch the graph of a function y = f (x) with all of the following properties: a. f '(x) > 0 for −2 ≤ x < 1 b. f '(2) = 0 c. f '(x) > 0 for x > 2 d. f (2) = 2 and f (0) = 1 e. li m x→−∞ f (x) = 0 and li m x→∞ f (x) = ∞ f. f '(1) does not exist. Answer Under Construction Exercise 2.1E . 9 Suppose temperature T in degrees Fahrenheit at a height x in feet above the ground is given by y = T (x). a. Give a physical interpretation, with units, of T '(x). b. If we know that T '(1000) = −0.1, explain the physical meaning. Answer a. The rate (in degrees per foot) at which temperature is increasing or decreasing for a given height x. b. The rate of change of temperature as altitude changes at 1000feet is −0.1 degrees per foot. Exercise 2.1E . 10 Suppose the total profit of a company is y = P (x) thousand dollars when x units of an item are sold. a. What does P (b)−P (a) b−a for 0 < a < b measure, and what are the units? b. What does P '(x) measure, and what are the units? c. Suppose that P '(30) = 5 , what is the approximate change in profit if the number of items sold increases from 30 to 31? Answer 10/21/2020 2.1E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10628 Under Construction Exercise 2.1E . 11 The graph in the following figure models the number of people initial outbreak in a town with a population of 50,000 citizens. N (t) who have come down with the flu t weeks after its a. Describe what N '(t) represents and how it behaves as t increases. b. What does the derivative tell us about how this town is affected by the flu outbreak? Answer a. The rate at which the number of people who have come down with the flu is changing t weeks after the initial outbreak. b. The rate is increasing sharply up to the third week, at which point it slows down and then becomes constant. Exercise 2.1E . 12 For the following exercises, use the following table, which shows the height mission t seconds after launch. h T ime(seconds) H eight(meters) 0 0 1 2 2 4 3 13 4 25 5 32 of the Saturn V rocket for the Apollo 11 1) What is the physical meaning of h'(t)? What are the units? 10/21/2020 2.1E.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10628 2) Construct a table of values for h'(t) and graph both h(t) and h'(t) on the same graph. (Hint: for interior points, estimate both the left limit and right limit and average them.) 3) The best linear fit to the data is given by H (t) = 7.229t − 4.905, where H is the height of the rocket (in meters) and t is the time elapsed since takeoff. From this equation, determine H '(t). Graph H (t with the given data and, on a separate coordinate plane, graph H '(t). 4) The best quadratic fit to the data is given by G(t) = 1.429t + 0.0857t − 0.1429, where G is the height of the rocket (in meters) and t is the time elapsed since takeoff. From this equation, determine G'(t). Graph G(t) with the given data and, on a separate coordinate plane, graph G'(t). 2 5) The best cubic fit to the data is given by F (t) = 0.2037t + 2.956t − 2.705t + 0.4683, where F is the height of the rocket (in m) and t is the time elapsed since take off. From this equation, determine F '(t). Graph F (t) with the given data and, on a separate coordinate plane, graph F '(t). Does the linear, quadratic, or cubic function fit the data best? 3 2 6) Using the best linear, quadratic, and cubic fits to the data, determine what physical meanings of H (t), G (t)andF (t), and what are their units? ′′ ′′ H ′′ ′′ (t), G (t)andF ′′ (t) are. What are the ′′ Answers to even numbered questions 2. T ime(seconds) h'(t)(m/s) 0 2 1 2 2 5.5 3 10.5 4 9.5 5 7 4. G'(t) = 2.858t + 0.0857 10/21/2020 2.1E.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10628 6. H (t) = 0, G (t) = 2.858andf per second squared (m/s2). ′′ ′′ ′′ (t) = 1.222t + 5.912 represent the acceleration of the rocket, with units of meters Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 2.1E.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10628 2.2: Techniques of differentiation Finding derivatives of functions by using the definition of the derivative can be a lengthy and, for certain functions, a rather challenging process. For example, previously we found that d dx − (√x ) = 1 (2.2.1) − 2 √x by using a process that involved multiplying an expression by a conjugate prior to evaluating a limit. The process that we could use to evaluate d dx 3 − (√x ) (2.2.2) using the definition, while similar, is more complicated. In this section, we develop rules for finding derivatives that allow us to bypass this process. We begin with the basics. Basic Rules Theorem 2.2.1 : Basic rules 1. Let c be a constant. If f (x) = c , then f '(x) = 0. Alternatively, we may express this rule as $$ \dfrac{d}{dx}(c)=0.$$ 2. Power Rule : Let n be a positive integer. If f (x) = x ,then f '(x) = nx n (n−1) d n x . Alternatively, we may express this rule as (n−1) = nx . (2.2.3) dx 3. Let f (x) and g(x) be differentiable functions and k be a constant. Then each of the following equations holds. Sum Rule. The derivative of the sum of a function f and a function g is the same as the sum of the derivative of f and the derivative of g . d d (f (x) + g(x)) = dx d (f (x)) + dx (g(x)); (2.2.4) dx that is, if j(x) = f (x) + g(x), then j'(x) = f '(x) + g'(x). (2.2.5) Difference Rule. The derivative of the difference of a function f and a function g is the same as the difference of the derivative of f and the derivative of g : d d (f (x) − g(x)) = dx d (f (x)) − dx (g(x)); (2.2.6) dx that is, if j(x) = f (x) − g(x), then j'(x) = f '(x) − g'(x). (2.2.7) Constant Multiple Rule. The derivative of a constant c multiplied by a function f is the same as the constant multiplied by the derivative: d d (kf (x)) = k dx (f (x)); (2.2.8) dx that is, f orj(x) = kf (x), j'(x) = kf '(x). (2.2.9) Proof 9/30/2020 2.2.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10288 2. For f (x) = x where n is a positive integer, we have n n (x + h ) f '(x) = limh→0 Since (x + h) n n =x n−1 + nx n n−2 h + ( )x 2 n −x . h 2 h n n−3 3 + ( )x h 3 n−1 n + … + nx h +h , we see that n (x + h ) n n−1 −x = nx n n−2 h + ( )x 2 2 h n n−3 + ( )x 3 3 h n−1 + … + nx h n +h . Next, divide both sides by h: n (x + h ) n−1 n nx −x n n−2 h + ( )x 2 = 2 h n n−3 + ( )x 3 3 h n−1 + … + nx h n +h . h h Thus, n (x + h ) n −x n−1 = nx n n−2 + ( )x 2 h n n−3 h + ( )x 3 2 h n−2 + … + nx h n−1 +h . Finally, n−1 f '(x) = limh→0 (nx n n−2 + ( )x 2 n n−3 h + ( )x 3 n−1 = nx 2 h n−1 + … + nx h n +h ) . 3. We provide only the proof of the sum rule here. The rest follow in a similar manner. For differentiable functions f (x) and g(x) , we set j(x) = f (x) + g(x) . Using the limit definition of the derivative we have j(x + h) − j(x) j'(x) = lim . By substituting j(x + h) = f (x + h) + g(x + h) (2.2.10) h h→0 and j(x) = f (x) + g(x), we obtain (f (x + h) + g(x + h)) − (f (x) + g(x)) j'(x) = lim . (2.2.11) h h→0 Rearranging and regrouping the terms, we have (f (x + h) − f (x) j'(x) = lim h→0 g(x + h) − g(x) + ). h (2.2.12) h We now apply the sum law for limits and the definition of the derivative to obtain (f (x + h) − f (x) j'(x) = lim h→0 (g(x + h) − g(x) ) + lim h ) = f '(x) + g'(x). (2.2.13) h h→0 Example 2.2.1 : Find the derivative of 2 f (x) = π . (2.2.14) Solution Since π is a constant, f '(x) = 0. 2 9/30/2020 2.2.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10288 Example 2.2.1 : Applying Basic Derivative Rules Find the derivative of 5 f (x) = 2 x + 7. (2.2.15) Solution We begin by applying the rule for differentiating the sum of two functions, followed by the rules for differentiating constant multiples of functions and the rule for differentiating powers. To better understand the sequence in which the differentiation rules are applied, we use Leibniz notation throughout the solution: d f '(x) = 5 (2 x + 7) dx d 5 = d (2 x ) + dx (7) Apply the sum rule. (7) Apply the constant multiple rule. dx d 5 =2 d (x ) + dx dx 4 = 2(5 x ) + 0 4 = 10x Apply the power rule and the constant rule. Simplify. Exercise 2.2.1 Find the derivative of 3 f (x) = 2 x 2 − 6x + 3. (2.2.16) Hint Use the preceding example as a guide. Answer 2 f '(x) = 6 x − 12x. Example 2.2.2 : Finding the Equation of a Tangent Line Find the equation of the line tangent to the graph of 2 f (x) = x − 4x + 6 (2.2.17) at x = 1 Solution To find the equation of the tangent line, we need a point and a slope. To find the point, compute 2 f (1) = 1 − 4(1) + 6 = 3. (2.2.18) This gives us the point (1, 3). Since the slope of the tangent line at 1 is f '(1), we must first find f '(x). Using the definition of a derivative, we have f '(x) = 2x − 4 (2.2.19) so the slope of the tangent line is f '(1) = −2 . Using the point-slope formula, we see that the equation of the tangent line is y − 3 = −2(x − 1). (2.2.20) Putting the equation of the line in slope-intercept form, we obtain y = −2x + 5. 9/30/2020 2.2.3 CC-BY-NC-SA (2.2.21) https://math.libretexts.org/@go/page/10288 Exercise 2.2.2 Find the equation of the line tangent to the graph of f (x) = 3x 2 at x = 2 . Use the point-slope form. − 11 Hint Use the preceding example as a guide. Answer y = 12x − 23 The Product Rule Now that we have examined the basic rules, we can begin looking at some of the more advanced rules. The first one examines the derivative of the product of two functions. Although it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the product rule does not follow this pattern. To see why we cannot use this pattern, consider the function f (x) = x , whose derivative is f '(x) = 2x and not 2 d d (x) ⋅ dx (x) = 1 ⋅ 1 = 1. dx Theorem 2.2.2 : Product Rule Let f (x) and g(x) be differentiable functions. Then d d (f (x)g(x)) = dx d (f (x)) ⋅ g(x) + dx (g(x)) ⋅ f (x). (2.2.22) dx That is, if j(x) = f (x)g(x), then j'(x) = f '(x)g(x) + g'(x)f (x). (2.2.23) This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function. PROOF: We begin by assuming that f (x) and g(x) are differentiable functions. At a key point in this proof we need to use the fact that, since g(x) is differentiable, it is also continuous. In particular, we use the fact that since g(x) is continuous, limh→0 g(x + h) = g(x). By applying the limit definition of the derivative to (x) = f (x)g(x), we obtain f (x + h)g(x + h) − f (x)g(x) j'(x) = lim h→0 . (2.2.24) h By adding and subtracting f (x)g(x + h) in the numerator, we have f (x + h)g(x + h) − f (x)g(x + h) + f (x)g(x + h) − f (x)g(x) j'(x) = lim . (2.2.25) h h→0 After breaking apart this quotient and applying the sum law for limits, the derivative becomes (f (x + h)g(x + h) − f (x)g(x + h) j'(x) = lim (f (x)g(x + h) − f (x)g(x) ) + lim h h→0 . (2.2.26) h h→0 Rearranging, we obtain (f (x + h) − f (x) j'(x) = lim h→0 By using the continuity of arrive at the product rule, 9/30/2020 g(x) g(x + h) − g(x) ⋅ g(x + h)) + lim( h ⋅ f (x)). , the definition of the derivatives of f (x) 2.2.4 CC-BY-NC-SA (2.2.27) h h→0 and g(x) , and applying the limit laws, we https://math.libretexts.org/@go/page/10288 j'(x) = f '(x)g(x) + g'(x)f (x). (2.2.28) Note: Visually Example 2.2.3 : Applying the Product Rule to Constant Functions For j(x) = f (x)g(x) , use the product rule to find j'(2) if f (2) = 3, f '(2) = −4, g(2) = 1 , and g'(2) = 6 . Solution Since j(x) = f (x)g(x), j'(x) = f '(x)g(x) + g'(x)f (x), and hence j'(2) = f '(2)g(2) + g'(2)f (2) = (−4)(1) + (6)(3) = 14. (2.2.29) Example 2.2.4 : Applying the Product Rule to Binomials For j(x) = (x + 2)(3x then differentiating. 2 3 − 5x), find j'(x) by applying the product rule. Check the result by first finding the product and Solution If we set f (x) = x 2 +2 and g(x) = 3x 3 − 5x 3 j'(x) = f '(x)g(x) + g'(x)f (x) = (2x)(3 x , then f '(x) = 2x and g'(x) = 9x 2 2 − 5x) + (9 x 2 − 5)(x −5 . Thus, + 2). Simplifying, we have 4 j'(x) = 15 x To check, we see that j(x) = 3x 5 3 +x − 10x 2 + 3x − 10. (2.2.30) and, consequently, j'(x) = 15x 4 2 + 3x − 10. Exercise 2.2.3 Use the product rule to obtain the derivative of 5 2 j(x) = 2 x (4 x + x). (2.2.31) Hint Set f (x) = 2x and g(x) = 4x 5 2 +x and use the preceding example as a guide. Answer 4 2 j'(x) = 10 x (4 x 5 6 + x) + (8x + 1)(2 x ) = 56 x 5 + 12 x . (2.2.32) The Quotient Rule Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that 9/30/2020 2.2.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10288 d 3 (x ) d 2 (x ) = 2x, not dx d dx 2 3x 2 = = 3x . (2.2.33) 1 (x) dx The Quotient Rule Let f (x) and g(x) be differentiable functions. Then d ( dx d (f (x)) ⋅ g(x) − f (x) d ) = (g(x)) ⋅ f (x) dx dx . 2 g(x) (2.2.34) (g(x)) That is, if f (x) j(x) = (2.2.35) g(x) then f '(x)g(x) − g'(x)f (x) j'(x) = . (g(x))2 (2.2.36) The proof of the quotient rule is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this new rule for finding derivatives in the next example. Example 2.2.5 : Applying the Quotient Rule Use the quotient rule to find the derivative of 2 5x k(x) = . (2.2.37) 4x + 3 Solution Let f (x) = 5x and g(x) = 4x + 3 . Thus, f '(x) = 10x and g'(x) = 4 . Substituting into the quotient rule, we have 2 2 f '(x)g(x) − g'(x)f (x) k'(x) = 10x(4x + 3) − 4(5 x ) = 2 2 (g(x)) . (2.2.38) (4x + 3) Simplifying, we obtain 2 20 x + 30x k'(x) = (2.2.39) 2 (4x + 3) Exercise 2.2.4 Find the derivative of h(x) = 3x + 1 . 4x − 3 Answer Apply the quotient rule with f (x) = 3x + 1 and g(x) = 4x − 3 . Answer 13 k'(x) = − 2 . (2.2.40) (4x − 3) It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form x where k is a negative integer. k 9/30/2020 2.2.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10288 Extended Power Rule If k is a negative integer, then d k k−1 (x ) = kx . (2.2.41) dx Proof: If k is a negative integer, we may set integer n ,x −n 1 = n−1 , so that n is a positive integer with k = −n . Since for each positive , we may now apply the quotient rule by setting f (x) = 1 and g(x) = x . In this case, f '(x) = 0 n xn and g'(x) = nx n = −k . Thus, n d −n (x n−1 0(x ) − 1(nx ) ) = . n d (2.2.42) 2 (x ) Simplifying, we see that d n−1 −n (x −nx (n−1)−2n )\)$ = $$ = −nx −n−1 $\( = −nx . (2.2.43) x2 n d Finally, observe that since k = −n , by substituting we have d k k−1 (x ) = kx . (2.2.44) dx □ Example 2.2.6 : Using the Extended Power Rule Find d −4 (x ) dx . Solution By applying the extended power rule with k = −4 , we obtain d −4 (x −4−1 ) = −4 x −5 = −4 x . (2.2.45) dx Example 2.2.7 : Using the Extended Power Rule and the Constant Multiple Rule Use the extended power rule and the constant multiple rule to find f (x) = 6 2 . x Solution It may seem tempting to use the quotient rule to find this derivative, and it would certainly not be incorrect to do so. However, it is far easier to differentiate this function by first rewriting it as f (x) = 6x . −2 d f '(x) = 6 ( dx d =6 d 2 ) = −2 (x ) dx −3 = 6(−2 x −3 = −12x −2 (6 x ) dx x ) Rewrite 6 2 as 6x . −2 x Apply the constant multiple rule. Use the extended power rule to differentiate x . −2 Simplify. Exercise 2.2.5 Find the derivative of g(x) = 1 7 using the extended power rule. x 9/30/2020 2.2.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10288 Hint 1 \Rewrite g(x) = −7 =x x7 . Use the extended power rule with k = −7 . Answer −8 g'(x) = −7x . Combining Differentiation Rules As we have seen throughout the examples in this section, it seldom happens that we are called on to apply just one differentiation rule to find the derivative of a given function. At this point, by combining the differentiation rules, we may find the derivatives of any polynomial or rational function. Later on we will encounter more complex combinations of differentiation rules. A good rule of thumb to use when applying several rules is to apply the rules in reverse of the order in which we would evaluate the function. Example 2.2.8 : Combining Differentiation Rules For k(x) = 3h(x) + x 2 g(x) , find k'(x). Solution: Finding this derivative requires the sum rule, the constant multiple rule, and the product rule. d k'(x) = d 2 (3h(x)) + dx d d = 3 d (3h(x) + x g(x)) = dx (h(x)) + ( dx d 2 (x )g(x) + dx Apply the sum rule. 2 (x g(x)) dx Apply the constant multiple rule todifferentiate productrule to differentiate x g(x) . 2 (g(x))x ) 3h(x) and the 2 dx 2 = 3h'(x) + 2xg(x) + g'(x)x Example 2.2.9 : Extending the Product Rule For k(x) = f (x)g(x)h(x), express k'(x) in terms of f (x), g(x), h(x), and their derivatives. Solution We can think of the function k(x) = (f (x)g(x)) ⋅ h(x) . Thus, d k'(x) = as the product of the function k(x) d (f (x)g(x)) ⋅ h(x) + dx (h(x)) ⋅ (f (x)g(x)). f (x)g(x) and the function . That is, h(x) Apply the product rule to the productoff(x)g(x)andh(x). dx = (f '(x)g(x) + g'(x)f (x)h)(x) + h'(x)f (x)g(x) Apply the product rule to f (x)g(x)\) = f '(x)g(x)h(x) + f (x)g'(x)h(x) + f (x)g(x)h'(x). Simplify. Example 2.2.10 : Combining the Quotient Rule and the Product Rule 3 For h(x) = 2 x k(x) 3x + 2 , find h'(x). Solution This procedure is typical for finding the derivative of a rational function. d d 3 3 (2 x k(x)) ⋅ (3x + 2) − dx h'(x) = (3x + 2) ⋅ (2 x k(x)) dx Apply the quotient rule 2 (3x + 2) 2 3 3 (6 x k(x) + k'(x) ⋅ 2 x )(3x + 2) − 3(2 x k(x)) = 2 Apply the product rule to find (3x + 2) d (3x + 2) = 3 dx 9/30/2020 d 3 .Use (2 x k(x)) dx . 2.2.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10288 3 3 2 4 3 −6 x k(x) + 18 x k(x) + 12 x k(x) + 6 x k'(x) + 4 x k'(x) = (3x + 2)2 Simplify Exercise 2.2.6 Find d (3f (x) − 2g(x)). dx Hint Apply the difference rule and the constant multiple rule. Answer 3f '(x) − 2g'(x). Example 2.2.11 : Determining Where a Function Has a Horizontal Tangent Determine the values of x for which f (x) = x 3 2 − 7x + 8x + 1 has a horizontal tangent line. Solution To find the values of x for which f (x) has a horizontal tangent line, we must solve f '(x) = 0. Since 2 f '(x) = 3 x we must solve (3x − 2)(x − 4) = 0 − 14x + 8 = (3x − 2)(x − 4), (2.2.46) . Thus we see that the function has horizontal tangent lines at 2 x = and x =4 as 3 shown in the following graph. Figure 2.2.2 : This function has horizontal tangent lines at x = 2/3 and x = 4 . Example 2.2.12 : Finding a Velocity The position of an object on a coordinate axis at time t is given by t s(t) = 2 t . What is the initial velocity of the +1 object? Solution Since the initial velocity is v(0) = s'(0), begin by finding s'(t) by applying the quotient rule: 2 1(t2 + 1) − 2t(t) s'(t) = 2 (t 2 + 1) 1 −t = t 2 . ( 2 + 1) After evaluating, we see that v(0) = 1. Exercise 2.2.7 9/30/2020 2.2.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10288 Find the values of x for which the line tangent to the graph of f (x) = 4x 2 − 3x + 2 has a tangent line parallel to the line y = 2x + 3. Hint Solve f '(x) = 2 . Answer 5 8 Formula One Grandstands Formula One car races can be very exciting to watch and attract a lot of spectators. Formula One track designers have to ensure sufficient grandstand space is available around the track to accommodate these viewers. However, car racing can be dangerous, and safety considerations are paramount. The grandstands must be placed where spectators will not be in danger should a driver lose control of a car (Figure). Figure 2.2.3 : The grandstand next to a straightaway of the Circuit de Barcelona-Catalunya race track, located where the spectators are not in danger. Safety is especially a concern on turns. If a driver does not slow down enough before entering the turn, the car may slide off the racetrack. Normally, this just results in a wider turn, which slows the driver down. But if the driver loses control completely, the car may fly off the track entirely, on a path tangent to the curve of the racetrack. Suppose you are designing a new Formula One track. One section of the track can be modeled by the function f (x) = x + 3x + x (Figure). The current plan calls for grandstands to be built along the first straightaway and around a portion of the first curve. The plans call for the front corner of the grandstand to be located at the point (−1.9, 2.8). We want to determine whether this location puts the spectators in danger if a driver loses control of the car. 3 9/30/2020 2.2.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10288 Figure 2.2.4 : (a) One section of the racetrack can be modeled by the function f (x) = x of the grandstand is located at (−1.9, 2.8). 3 + 3x + x . (b) The front corner 1. Physicists have determined that drivers are most likely to lose control of their cars as they are coming into a turn, at the point where the slope of the tangent line is 1. Find the (x, y) coordinates of this point near the turn. 2. Find the equation of the tangent line to the curve at this point. 3. To determine whether the spectators are in danger in this scenario, find the x-coordinate of the point where the tangent line crosses the line y = 2.8 . Is this point safely to the right of the grandstand? Or are the spectators in danger? 4. What if a driver loses control earlier than the physicists project? Suppose a driver loses control at the point ( −2.5, 0.625). What is the slope of the tangent line at this point? 5. If a driver loses control as described in part 4, are the spectators safe? 6. Should you proceed with the current design for the grandstand, or should the grandstands be moved? Key Concepts The derivative of a constant function is zero. The derivative of a power function is a function in which the power on x becomes the coefficient of the term and the power on x in the derivative decreases by 1. The derivative of a constant c multiplied by a function f is the same as the constant multiplied by the derivative. The derivative of the sum of a function f and a function g is the same as the sum of the derivative of f and the derivative of g. The derivative of the difference of a function f and a function g is the same as the difference of the derivative of f and the derivative of g. The derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function. The derivative of the quotient of two functions is the derivative of the first function times the second function minus the derivative of the second function times the first function, all divided by the square of the second function. We used the limit definition of the derivative to develop formulas that allow us to find derivatives without resorting to the definition of the derivative. These formulas can be used singly or in combination with each other. Glossary constant multiple rule the derivative of a constant c multiplied by a function f is the same as the constant multiplied by the derivative: d (cf (x)) = cf '(x) dx 9/30/2020 2.2.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10288 constant rule the derivative of a constant function is zero: d (c) = 0 ,where c is a constant dx difference rule the derivative of the difference of a function f and a function g is the same as the difference of the derivative of f and the derivative of g: d (f (x) − g(x)) = f '(x) − g'(x) dx power rule the derivative of a power function is a function in which the power on x becomes the coefficient of the term and the power on x in the derivative decreases by 1: If n is an integer, then d n x n−1 = nx dx product rule the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function: d (f (x)g(x)) = f '(x)g(x) + g'(x)f (x) dx quotient rule the derivative of the quotient of two functions is the derivative of the first function times the second function minus the derivative of the second function times the first function, all divided by the square of the second function: f (x) d ( dx f '(x)g(x) − g'(x)f (x) = g(x)) 2 (g(x)) sum rule the derivative of the sum of a function f and a function g is the same as the sum of the derivative of f and the derivative of g: d (f (x) + g(x)) = f '(x) + g'(x) dx Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 9/30/2020 2.2.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/10288 2.2E: Exercises Exercise 2.2E . 1 For the following exercises, assume that each of the functions h(x). 1) h(x) = 4f (x) + 2) h(x) = x 3 3) h(x) = 4) h(x) = f (x) and g(x) are both differentiable functions for all x. Find the derivative of g(x) 7 f (x) f (x)g(x) 2 3f (x) g(x)+2 Answers to even numbered questions 2. h'(x) = 3x 2 4. h'(x) = 3 f (x) + x f '(x) 3f '(x)(g(x)+2)−3f (x)g'(x) 2 (g(x)+2) Exercise 2.2E . 2 For the following exercises, assume that f(x) and g(x) are both differentiable functions with values as given in the following table. Use the following table to calculate the following derivatives. x 1 2 3 4 f (x) 3 5 −2 0 g(x) 2 3 −4 6 f '(x) −1 7 8 −3 g'(x) 4 1 2 9 1) Find h'(1) if h(x) = xf (x) + 4g(x) . 2) Find h'(2) if h(x) = f (x) g(x) . 3) Find h'(3) if h(x) = 2x + f (x)g(x) . 4) Find h'(4) if h(x) = 1 x g(x) + f (x) . Answers to even numbered questions 2. 16 9 4. Undefined Exercise 2.2E . 3 For the following exercises, use the following figure to find the indicated derivatives, if they exist. 10/21/2020 2.2E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10625 1) Let h(x) = f (x) + g(x) . Find a. h'(1), b. h'(3), and c. h'(4). 2) Let h(x) = f (x)g(x). Find a. h'(1), b. h'(3), and c. h'(4). 3) Let h(x) = f (x) g(x) Find . a. h'(1), b. h'(3), and c. h'(4). Answers to even numbered questions 2a. 2 b. does not exist c. 2.5 Exercise 2.2E . 4 For the following exercises, a. evaluate f '(a), and b. graph the function f (x) and the tangent line at x = a. 1) f (x) = 2x 3 2) f (x) = 1 2 x 3) f (x) = x −x , a = 1 2 4) f (x) = 2 + 3x − x , a = 2 1 x 12 −x 2/3 −x + 3x + 2, a = 0 , a = −1 Answers to odd numbered questions 1a. 23 b. y = 23x − 28 10/21/2020 2.2E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10625 3a. 3 b. y = 3x + 2 Exercise 2.2E . 5 Find the derivative of the following functions and simplify your answer: 1) f (x) = x 7 + 10 2) f (x) = 5x −x +1 3) f (x) = 4x − 7x 4) f (x) = 8x + 9x 3 2 4 5) f (x) = x 4 2 −1 + 2x 6) f (x) = 3x(18x 4 + 13 x+1 ) 7) (f(x)=(x+2)(2x^2−3)\) 8) f (x) = x 2 9) f (x) = ( 2 2 x 3 + 5 3 x ) 2 x +2 x −4 3 10) \(f(x)=\frac{4x^3−2x+1}{x^2} 10/21/2020 2.2E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10625 2 11) f (x) = x +4 2 x −4 12) f (x) = x+9 2 x −7x+1 2 3x 13) y(x) = 2 2x 2 2 2x . +x −3 3x 14) y(x) = +5 +5 +x −3 Answers to even numbered questions 2. f '(x) = 15x −1 4. f '(x) = 32x + 18x 2 3 6.f '(x) = 270x 4 39 + 2 (x+1) 8. f '(x) = −5 x2 4 2 10. f '(x) = 4 x +2 x −2x 12. f '(x) = −x −18x+64 4 x 2 2 ( x2 −7x+1 ) Exercise 2.2E . 6 For the following exercises, find the equation of the tangent line T (x) to the graph of the given function at the indicated point. Use a graphing calculator to graph the function and the tangent line. 1) y = 3x 2 2) + 4x + 1 − y = 2 √x + 1 3) y = 4) y = 2x x−1 2 x − at (0, 1) at (4, 5) at (−1, 1) at (1, −1) 3 2 x Answers to even numbered questions 2. T (x) = 1 2 x +3 , 4. T (x) = 4x − 5 Exercise 2.2E . 7 1) Find the equation of the tangent line to the graph of f (x) = 2x 3 2) Find the equation of the tangent line to the graph of f (x) = x 2 2 + 4x + 4 x − 5x − 3 − 10 3) Find the equation of the tangent line to the graph of f (x) = (3x − x 2 at x = −1. at x = 8 . 2 )(3 − x − x ) at x = 1 . 4) Find the point on the graph of f (x) = x such that the tangent line at that point has an x intercept of 6. 3 5) Find the equation of the line passing through the point P (3, 3) and tangent to the graph of f (x) = 6 x−1 . Answers to odd numbered questions 1. y = −7x − 3 3. y = −5x + 7 5. y = − 10/21/2020 3 2 x+ 15 2 2.2E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10625 Exercise 2.2E . 8 Determine all points on the graph of f (x) = x 3 2 +x −x −1 for which the slope of the tangent line is a. horizontal b. −1. Answer Under Construction Exercise 2.2E . 9 Find a quadratic polynomial such that f (1) = 5, f '(1) = 3 and f ′′ (1) = −6. Answer 2 y = −3 x + 9x − 1 Exercise 2.2E . 10 A car driving along a freeway with traffic has traveled s(t) = t 3 2 − 6t + 9t meters in t seconds. a. Determine the time in seconds when the velocity of the car is 0. b. Determine the acceleration of the car when the velocity is 0. Answer Under Construction Exercise 2.2E . 11 A herring swimming along a straight line has traveled s(t) = 2 t 2 t +2 feet in t seconds. Determine the velocity of the herring when it has traveled 3 seconds. Answer 12 121 or 0.0992 ft/s Exercise 2.2E . 12 The population in millions of arctic flounder in the Atlantic Ocean is modeled by the function measured in years. P (t) = 8t+3 2 0.2 t +1 , where t is a. Determine the initial flounder population. b. Determine P '(10) and briefly interpret the result. Answer Under Construction Exercise 2.2E . 12 The concentration of antibiotic in the bloodstream t hours after being injected is given by the function where C is measured in milligrams per liter of blood. 2 C (t) = 2 t +t 3 t +50 , a. Find the rate of change of C (t). 10/21/2020 2.2E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10625 b. Determine the rate of change for t = 8, 12, 24,and 36. c. Briefly describe what seems to be occurring as the number of hours increases. Answer 4 a. 3 −2 t −2 t +200t+50 3 b. −0.02395 2 mg/L-hr, −0.01344 mg/L-hr, −0.003566 mg/L-hr, −0.001579 mg/L-hr c. The rate at ( t +50 ) which the concentration of drug in the bloodstream decreases is slowing to 0 as time increases.mplate active on the page. Exercise 2.2E . 13 3 x +2x+3 A book publisher has a cost function given by C (x) = , where x is the number of copies of a book in thousands and C is the cost per book, measured in dollars. Evaluate C '(2) and explain its meaning. 2 x Answer Under Construction Exercise 2.2E . 14 According to Newton’s law of universal gravitation, the force F between two bodies of constant mass m and m is given by the formula F = , where G is the gravitational constant and d is the distance between the bodies. 1 2 Gm1 m2 d 2 a. Suppose that G, m , and m are constants. Find the rate of change of force F with respect to distance d . 1 2 b. Find the rate of change of force F with gravitational constant meters apart, each with a mass of 1000 kilograms. −11 G = 6.67 × 10 2 N m /kg 2 , on two bodies 10 Answer a. F ′ (d) = −2Gm1 m2 d3 b. −1.33 × 10 −7 ' N /m Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 2.2E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10625 2.3: Derivative as a rate of Change In this section we look at some applications of the derivative by focusing on the interpretation of the derivative as the rate of change of a function. These applications include acceleration and velocity in physics, population growth rates in biology, and marginal functions in economics. Amount of Change Formula One application for derivatives is to estimate an unknown value of a function at a point by using a known value of a function at some given point together with its rate of change at the given point. If f (x) is a function defined on an interval [a, a + h] , then the amount of change of f (x) over the interval is the change in the y values of the function over that interval and is given by f (a + h) − f (a). (2.3.1) The average rate of change of the function f over that same interval is the ratio of the amount of change over that interval to the corresponding change in the x values. It is given by f (a + h) − f (a) . (2.3.2) h As we already know, the instantaneous rate of change of f (x) at a is its derivative f (a + h) − f (a) f '(a) = li mh→0 For small enough values of h , f '(a) ≈ f (a+h)−f (a) h . (2.3.3) h . We can then solve for f (a + h) to get the amount of change formula: f (a + h) ≈ f (a) + f '(a)h. (2.3.4) We can use this formula if we know only f (a)and f '(a) and wish to estimate the value of f (a + h) . For example, we may use the current population of a city and the rate at which it is growing to estimate its population in the near future. As we can see in Figure, we are approximating f (a + h) by the y coordinate at a+h on the line tangent to f (x) at x = a . Observe that the accuracy of this estimate depends on the value of h as well as the value of f '(a). Figure 2.3.1 : The new value of a changed quantity equals the original value plus the rate of change times the interval of change: f (a + h) ≈ f (a) + f '(a)h. Example 2.3.1 : Estimating the Value of a Function If f (3) = 2 and f '(3) = 5, estimate f (3.2). Solution Begin by finding h . We have h = 3.2 − 3 = 0.2. Thus, 10/21/2020 2.3.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10645 f (3.2) = f (3 + 0.2) ≈ f (3) + (0.2)f '(3) = 2 + 0.2(5) = 3. Exercise 2.3.1 Given f (10) = −5 and f '(10) = 6, estimate f (10.1). Hint Use the same process as in the preceding example. Answer −4.4 Motion along a Line Another use for the derivative is to analyze motion along a line. We have described velocity as the rate of change of position. If we take the derivative of the velocity, we can find the acceleration, or the rate of change of velocity. It is also important to introduce the idea of speed, which is the magnitude of velocity. Thus, we can state the following mathematical definitions. Definition Let s(t) be a function giving the position of an object at time t. The velocity of the object at time t is given by v(t) = s'(t) . The speed of the object at time t is given by |v(t)| . The acceleration of the object at t is given by a(t) = v'(t) = s ′′ (t) . Example 2.3.2 : Comparing Instantaneous Velocity and Average Velocity A ball is dropped from a height of 64 feet. Its height above ground (in feet) t seconds later is given by s(t) = −16t 2 + 64 . a. What is the instantaneous velocity of the ball when it hits the ground? b. What is the average velocity during its fall? Solution The first thing to do is determine how long it takes the ball to reach the ground. To do this, set −16 t + 64 = 0 , we get t = 2 , so it take 2 seconds for the ball to reach the ground. s(t) = 0 . Solving 2 a. The instantaneous velocity of the ball as it strikes the ground is v(t) = −64f t/s . b. The average velocity of the ball during its fall is s(2)−s(0) vave = 2−0 = 0−64 2 v(2) . Since v(t) = s'(t) = −32t , we obtain = −32f t/s. Example 2.3.3 : Interpreting the Relationship between v(t) and a(t) 10/21/2020 2.3.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10645 A particle moves along a coordinate axis in the positive direction to the right. Its position at time s(t) = t − 4t + 2 . Find v(1) and a(1) and use these values to answer the following questions. t is given by 3 a. Is the particle moving from left to right or from right to left at time t = 1 ? b. Is the particle speeding up or slowing down at time t = 1 ? Solution Begin by finding v(t) and a(t) . and a(t) = v'(t) = s ′′ . (t) = 6t Evaluating these functions at t = 1) , we obtain v(1) = −1 and a(1) = 6 . a. Because v(1) < 0 , the particle is moving from right to left. b. Because v(1) < 0 and a(1) > 0 , velocity and acceleration are acting in opposite directions. In other words, the particle is being accelerated in the direction opposite the direction in which it is traveling, causing |v(t)| to decrease. The particle is slowing down. Example 2.3.4 : Position and Velocity The position of a particle moving along a coordinate axis is given by s(t) = t 3 2 − 9t + 24t + 4, t ≥ 0. a. Find v(t) . b. At what time(s) is the particle at rest? c. On what time intervals is the particle moving from left to right? From right to left? d. Use the information obtained to sketch the path of the particle along a coordinate axis. Solution a. The velocity is the derivative of the position function: 2 v(t) = s'(t) = 3 t − 18t + 24. b. The particle is at rest when v(t) = 0 , so set 3t − 18t + 24 = 0 . Factoring the left-hand side of the equation produces 3(t − 2)(t − 4) = 0 . Solving, we find that the particle is at rest at t = 2 and t = 4 . 2 c. The particle is moving from left to right when v(t) > 0 and from right to left when v(t) < 0 . Figure gives the analysis of the sign of v(t) for t≥0, but it does not represent the axis along which the particle is moving. Figure 2.3.2 :The sign of v(t) determines the direction of the particle. Since 3t Since 3t 2 − 18t + 24 > 0 2 − 18t + 24 < 0 on [0, 2) ∪ (2, +∞), the particle is moving from left to right on these intervals. on (2, 4), the particle is moving from right to left on this interval. d. Before we can sketch the graph of the particle, we need to know its position at the time it starts moving (t = 0) and at the times that it changes direction (t = 2, 4) . We have s(0) = 4, s(2) = 24 , and s(4) = 20 . This means that the particle begins on the coordinate axis at 4 and changes direction at 0 and 20 on the coordinate axis. The path of the particle is shown on a coordinate axis in Figure. Figure 2.3.3 : The path of the particle can be determined by analyzing v(t) . Exercise 2.3.2 A particle moves along a coordinate axis. Its position at time t is given by s(t) = t right to left or from left to right at time t = 3 ? 2 10/21/2020 2.3.3 CC-BY-NC-SA − 5t + 1 . Is the particle moving from https://math.libretexts.org/@go/page/10645 Hint Find v(3) and look at the sign. Answer left to right Population Change In addition to analyzing velocity, speed, acceleration, and position, we can use derivatives to analyze various types of populations, including those as diverse as bacteria colonies and cities. We can use a current population, together with a growth rate, to estimate the size of a population in the future. The population growth rate is the rate of change of a population and consequently can be represented by the derivative of the size of the population. Definition If P (t) is the number of entities present in a population, then the population growth rate of P (t) is defined to be P '(t). Example 2.3.5 : Estimating a Population The population of a city is tripling every 5 years. If its current population is 10,000, what will be its approximate population 2 years from now? Solution Let P (t) be the population (in thousands) t years from now. Thus, we know that P (0) = 10 and based on the information, we anticipate P (5) = 30. Now estimate P '(0), the current growth rate, using P (5)−P (0) P '(0) ≈ 5−0 = 30−10 5 =4 . By applying Equation to P (t), we can estimate the population 2 years from now by writing P (2) ≈ P (0) + (2)P '(0) ≈ 10 + 2(4) = 18 ; thus, in 2 years the population will be 18,000. Exercise 2.3.3 The current population of a mosquito colony is known to be 3,000; that is, size of the population in 3 days, where t is measured in days. . If P (0) = 3, 000 , estimate the P '(0) = 100 Hint Use P (3) ≈ P (0) + 3P '(0) Answer 3,300 Changes in Cost and Revenue In addition to analyzing motion along a line and population growth, derivatives are useful in analyzing changes in cost, revenue, and profit. The concept of a marginal function is common in the fields of business and economics and implies the use of derivatives. The marginal cost is the derivative of the cost function. The marginal revenue is the derivative of the revenue function. The marginal profit is the derivative of the profit function, which is based on the cost function and the revenue function. Definition If C (x) is the cost of producing x items, then the marginal cost M C (x) is M C (x) = C '(x). If R(x) is the revenue obtained from selling x items, then the marginal revenue M R(x) is M R(x) = R'(x) . 10/21/2020 2.3.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10645 If P (x) = R(x) − C (x) is the profit obtained from selling x items, then the marginal profit M P (x) is defined to be M P (x) = P '(x) = M R(x) − M C (x) = R'(x) − C '(x) . We can roughly approximate C (x + h) − C (x) M C (x) = C '(x) = lim (2.3.5) h h→0 by choosing an appropriate value for h . Since x represents objects, a reasonable and small value for h is 1. Thus, by substituting h = 1 , we get the approximation M C (x) = C '(x) ≈ C (x + 1) − C (x) . Consequently, C '(x) for a given value of x can be thought of as the change in cost associated with producing one additional item. In a similar way, M R(x) = R'(x) approximates the revenue obtained by selling one additional item, and M P (x) = P '(x) approximates the profit obtained by producing and selling one additional item. Example 2.3.6 : Applying Marginal Revenue Assume that the number of barbeque dinners that can be sold, x, can be related to the price charged, p, by the equation p(x) = 9 − 0.03x, 0 ≤ x ≤ 300 . In this case, the revenue in dollars obtained by selling x barbeque dinners is given by 2 R(x) = xp(x) = x(9 − 0.03x) = −0.03 x + 9xf or0 ≤ x ≤ 300 . Use the marginal revenue function to estimate the revenue obtained from selling the 101st barbeque dinner. Compare this to the actual revenue obtained from the sale of this dinner. Solution First, find the marginal revenue function: M R(x) = R'(x) = −0.06x + 9. Next, use to approximate R(101) − R(100), the revenue obtained from the sale of the 101st dinner. Since R'(100) = 3 , the revenue obtained from the sale of the 101st dinner is approximately $3. R'(100) The actual revenue obtained from the sale of the 101st dinner is . R(101) − R(100) = 602.97 − 600 = 2.97, or$2.97 The marginal revenue is a fairly good estimate in this case and has the advantage of being easy to compute. Exercise 2.3.4 Suppose that the profit obtained from the sale of x fish-fry dinners is given by P (x) = −0.03x marginal profit function to estimate the profit from the sale of the 101st fish-fry dinner. 2 + 8x − 50 . Use the Hint Use P '(100) to approximate P (101) − P (100). Answer $2 Key Concepts Using f (a + h) ≈ f (a) + f '(a)h , it is possible to estimate f (a + h) given f '(a) and f (a). The rate of change of position is velocity, and the rate of change of velocity is acceleration. Speed is the absolute value, or magnitude, of velocity. The population growth rate and the present population can be used to predict the size of a future population. Marginal cost, marginal revenue, and marginal profit functions can be used to predict, respectively, the cost of producing one more item, the revenue obtained by selling one more item, and the profit obtained by producing and selling one more item. Glossary 10/21/2020 2.3.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10645 acceleration is the rate of change of the velocity, that is, the derivative of velocity amount of change the amount of a function f (x) over an interval [x, x + h]isf (x + h) − f (x) average rate of change is a function f (x) over an interval [x, x + h] is f (x+h)−f (a) b−a marginal cost is the derivative of the cost function, or the approximate cost of producing one more item marginal revenue is the derivative of the revenue function, or the approximate revenue obtained by selling one more item marginal profit is the derivative of the profit function, or the approximate profit obtained by producing and selling one more item population growth rate is the derivative of the population with respect to time speed is the absolute value of velocity, that is, |v(t)| is the speed of an object at time t whose velocity is given by v(t) Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 2.3.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10645 2.3E: Exercises Exercise 2.3E . 1 For the following exercises, the given functions represent the position of a particle traveling along a horizontal line. a. Find the velocity and acceleration functions. b. Determine the time intervals when the object is slowing down or speeding up. 1) s(t) = 2t − 3t 2) s(t) = 2t − 15 t 3 3 2 − 12t + 8 2 + 36t − 10 Answer 1a. v(t) = 6t 2 − 30t + 36, a(t) = 12t − 30 b. speeds up (2, 2.5) ∪ (3, ∞), slows down (0, 2) ∪ (2.5, 3) Exercise 2.3E . 2 A rocket is fired vertically upward from the ground. The distance s in feet that the rocket travels from the ground after t seconds is given by s(t) = −16t + 560t . 2 a. Find the velocity of the rocket 3 seconds after being fired. b. Find the acceleration of the rocket 3 seconds after being fired. Answer a. 464f t/s 2 b.−32f t/s 2 Exercise 2.3E . 3 A ball is thrown downward with a speed of 8 ft/s from the top of a 64-foot-tall building. After t seconds, its height above the ground is given by s(t) = −16t − 8t + 64. 2 a. Determine how long it takes for the ball to hit the ground. b. Determine the velocity of the ball when it hits the ground. Answer Under Construction Exercise 2.3E . 4 The position function s(t) = t − 3t − 4 represents the position of the back of a car backing out of a driveway and then driving in a straight line, where s is in feet and t is in seconds. In this case, s(t) = 0 represents the time at which the back of the car is at the garage door, so s(0) = −4 is the starting position of the car, 4 feet inside the garage. 2 a. Determine the velocity of the car when s(t) = 0 . b. Determine the velocity of the car when s(t) = 14 . Answer a. 5f t/s 10/21/2020 2.3E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10646 b. 9f t/s Exercise 2.3E . 5 7) The position of a hummingbird flying along a straight line in t seconds is given by s(t) = 3t 3 − 7t meters. a. Determine the velocity of the bird at t = 1 sec. b. Determine the acceleration of the bird at t = 1 sec. c. Determine the acceleration of the bird when the velocity equals 0. Answer Under Construction Exercise 2.3E . 6 A potato is launched vertically upward with an initial velocity of 100 ft/s from a potato gun at the top of an 85-foot-tall building. The distance in feet that the potato travels from the ground after t seconds is given by s(t) = −16 t + 100t + 85 . 2 a. Find the velocity of the potato after 0.5s and 5.75s. b. Find the speed of the potato at 0.5 s and 5.75 s. c. Determine when the potato reaches its maximum height. d. Find the acceleration of the potato at 0.5 s and 1.5 s. e. Determine how long the potato is in the air. f. Determine the velocity of the potato upon hitting the ground. Answer a. 84 ft/s, −84 ft/s b. 84 ft/s c. 25 8 s d. −32f t/s in both cases 2 e. 1 8 − − − (25 + √965)s − − − f. −4√965f t/s Exercise 2.3E . 7 The position function s(t) = t is measured in hours. 3 − 8t gives the position in miles of a freight train where east is the positive direction and t a. Determine the direction the train is traveling when s(t) = 0 . b. Determine the direction the train is traveling when a(t) = 0 . c. Determine the time intervals when the train is slowing down or speeding up. Answer Under Construction Exercise 2.3E . 8 10/21/2020 2.3E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10646 The following graph shows the position y = s(t) of an object moving along a straight line. a. Use the graph of the position function to determine the time intervals when the velocity is positive, negative, or zero. b. Sketch the graph of the velocity function. c. Use the graph of the velocity function to determine the time intervals when the acceleration is positive, negative, or zero. d. Determine the time intervals when the object is speeding up or slowing down. Answer a. Velocity is positive on (0, 1.5) ∪ (6, 7), negative on (1.5, 2) ∪ (5, 6), and zero on (2, 5) Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 2.3E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10646 2.4: Derivatives of Trigonometric functions One of the most important types of motion in physics is simple harmonic motion, which is associated with such systems as an object with mass oscillating on a spring. Simple harmonic motion can be described by using either sine or cosine functions. In this section we expand our knowledge of derivative formulas to include derivatives of these and other trigonometric functions. We begin with the derivatives of the sine and cosine functions and then use them to obtain formulas for the derivatives of the remaining four trigonometric functions. Being able to calculate the derivatives of the sine and cosine functions will enable us to find the velocity and acceleration of simple harmonic motion. Derivatives of the Sine and Cosine Functions We begin our exploration of the derivative for the sine function by using the formula to make a reasonable guess at its derivative. Recall that for a function f (x), f (x + h) − f (x) f '(x) = lim . (2.4.1) h h→0 Consequently, for values of h very close to 0, f (x + h) − f (x) f '(x) ≈ . (2.4.2) h We see that by using h = 0.01, d sin(x + 0.01) − sin x (sin x) ≈ dx (2.4.3) 0.01 By setting sin(x + 0.01) − sin x D(x) = (2.4.4) 0.01 and using a graphing utility, we can get a graph of an approximation to the derivative of sin x (Figure). Figure 2.4.1 : The graph of the function D(x) looks a lot like a cosine curve. Upon inspection, the graph of D(x) appears to be very close to the graph of the cosine function. Indeed, we will show that d (sin x) = cos x. (2.4.5) dx If we were to follow the same steps to approximate the derivative of the cosine function, we would find that d (cos x) = − sin x. (2.4.6) dx 10/21/2020 2.4.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10289 Theorem 2.4.1 Derivatives of the Sine and Cosine Functions The derivative of the sine function is the cosine and the derivative of the cosine function is the negative sine. d (sin x) = cos x (2.4.7) (cos x) = − sin x (2.4.8) dx d dx Proof Because the proofs for proof for Limits: d (sin x) = cos x dx d (sin x) = cos x. dx and d Before beginning, recall two important trigonometric limits we learned in Introduction to sinh limh→0 The graphs of y = (sinh) and y = h use similar techniques, we provide only the (cos x) = − sin x dx =1 h (cosh − 1) h and lim coshh − 1 h→0 =0 h . are shown in Figure. Figure 2.4.2: These graphs show two important limits needed to establish the derivative formulas for the sine and cosine functions. We also recall the following trigonometric identity for the sine of the sum of two angles: sin(x + h) = sin x cosh + cos x sinh. (2.4.9) Now that we have gathered all the necessary equations and identities, we proceed with the proof. sin(x + h) − sin x d dx sin x = lim h→0 Apply the definition of the derivative. h sin xcosh + cosxsinh − sin x = lim h→0 (sin xcosh − sin x = lim h→0 Use trig identity for the sine of the sum of two angles. h cosxsinh + h cosxh − 1 = limh → 0(sin x( Regroup. ) h h cosxsinh )+ h )) Factor out sin x and cosx . = sin x(0) + cosx(1) Apply trig limit formulas. = cosx Simplify. 10/21/2020 2.4.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10289 Figure shows the relationship between the graph of f (x) = sin x and its derivative f '(x) = cos x. Notice that at the points where f (x) = sin x has a horizontal tangent, its derivative f '(x) = cos x takes on the value zero. We also see that where f (x) = sin x is increasing, f '(x) = cos x > 0 and where f (x) = sin x is decreasing, f '(x) = cos x < 0. Example 2.4.1 : Differentiating a Function Containing sin x Find the derivative of f (x) = 5x 3 sin x . Solution Using the product rule, we have d ′ d 3 f (x) = (5 x ) ⋅ sin x + 3 (sin x) ⋅ 5 x dx 2 = 15 x 3 ⋅ sin x + cos x ⋅ 5 x . dx After simplifying, we obtain 2 f '(x) = 15 x 3 sin x + 5 x cos x. Example 2.4.1 : Find the derivative of f (x) = sin x x . Caution: Note that f (x) is not 1. lim x→0 sin x =1 x . Solution Using the quotient rule, we have f ‘(x) = xcos(x) − sin(x) 2 . x Exercise 2.4.1 Find the derivative of f (x) = sin x cos x. (2.4.10) Hint Don’t forget to use the product rule. Answer 2 f '(x) = cos 10/21/2020 2 x − sin x 2.4.3 CC-BY-NC-SA (2.4.11) https://math.libretexts.org/@go/page/10289 Example 2.4.2 : Finding the Derivative of a Function Containing cos x Find the derivative of g(x) = cos x 2 . 4x Solution By applying the quotient rule, we have 2 (− sin x)4 x g'(x) = − 8x(cos x) 2 . 2 (2.4.12) (4x ) Simplifying, we obtain 2 −4 x sin x − 8x cos x g'(x) = −x sin x − 2 cos x = 4 . 3 16x (2.4.13) 4x Exercise 2.4.2 Find the derivative of x f (x) = . (2.4.14) cos x Hint Use the quotient rule. Answer cos x + x sin x (2.4.15) 2 cos x Example 2.4.3 : An Application to Velocity A particle moves along a coordinate axis in such a way that its position at time 0 ≤ t ≤ 2π. At what times is the particle at rest? t is given by s(t) = 2sint − t for Solution To determine when the particle is at rest, set s'(t) = v(t) = 0. Begin by finding s'(t). We obtain s'(t) = 2 cos t − 1, (2.4.16) 2 cos t − 1 = 0 (2.4.17) so we must solve for 0 ≤ t ≤ 2π. The solutions to this equation are t = π 3 and t = 5π 3 (2.4.18) . Thus the particle is at rest at times t = π 3 and t = 5π 3 .6) Exercise 2.4.3 A particle moves along a coordinate axis. Its position at time times is the particle at rest? t is given by – s(t) = √3t + 2cost for 0 ≤ t ≤ 2π. At what Hint Use the previous example as a guide. Answer 10/21/2020 2.4.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10289 π 2π t = ,t = (2.4.19) 3 3 Derivatives of Other Trigonometric Functions Since the remaining four trigonometric functions may be expressed as quotients involving sine, cosine, or both, we can use the quotient rule to find formulas for their derivatives. Example 2.4.4 : The Derivative of the Tangent Function Find the derivative of f (x) = tan x. Solution Start by expressing tan x as the quotient of sin x and cos x: sin x f (x) = tan x = cos x . Now apply the quotient rule to obtain cos x cos x − (− sin x) sin x f '(x) = 2 . (cos x) Simplifying, we obtain 2 cos f '(x) = 2 x + si n x 2 cos Recognizing that cos 2 2 x + sin x = 1, . (2.4.20) x by the Pythagorean theorem, we now have 1 f '(x) = Finally, use the identity secx = 1 cos x (2.4.21) cos2 x to obtain 2 f '(x) = sec x . Exercise 2.4.4 Find the derivative of f (x) = cot x. (2.4.22) Hint Rewrite cot x as cos x sin x and use the quotient rule. Answer 2 f '(x) = −csc x (2.4.23) The derivatives of the remaining trigonometric functions may be obtained by using similar techniques. We provide these formulas in the following theorem. Derivatives of tan x , cot x , secx , and cscx The derivatives of the remaining trigonometric functions are as follows: d (tan x) = sec 2 x (2.4.24) dx d (cot x) = − csc 2 x (2.4.25) dx 10/21/2020 2.4.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10289 d (secx) = sec x tan x (2.4.26) (cscx) = − csc x cot x. (2.4.27) dx d dx Example 2.4.5 : Finding the Equation of a Tangent Line π Find the equation of a line tangent to the graph of f (x) = cot x at x = . 4 Solution To find the equation of the tangent line, we need a point and a slope at that point. To find the point, compute π f( π ) = cot 4 Thus the tangent line passes through the point evaluating it at π 4 π ( =1 4 . . Next, find the slope by finding the derivative of , 1) 4 f (x) = cot x and : 2 f '(x) = −csc x and f '( π π 2 ) = −csc ( 4 ) = −2 4 . Using the point-slope equation of the line, we obtain π y − 1 = −2(x − ) 4 or equivalently, π y = −2x + 1 + 2 . Example 2.4.6 : Finding the Derivative of Trigonometric Functions Find the derivative of f (x) = cscx + x tan x. Solution To find this derivative, we must use both the sum rule and the product rule. Using the sum rule, we find d f '(x) = In the first term, d (cscx) = −cscx cot x, dx d (cscx) + dx (x tan x) dx . and by applying the product rule to the second term we obtain d 2 (x tan x) = (1)(tan x) + (sec x)(x) dx . Therefore, we have 2 f '(x) = −cscx cot x + tan x + xsec x . Exercise 2.4.5 Find the derivative of f (x) = 2 tan x − 3 cot x. Hint Use the rule for differentiating a constant multiple and the rule for differentiating a difference of two functions. 2 2 2 2 f '(x) = 2sec x + 3csc x Answer f '(x) = 2sec x + 3csc x 10/21/2020 2.4.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10289 Exercise 2.4.6 π Find the slope of the line tangent to the graph of f (x) = tan x at x = . 6 Hint Evaluate the derivative at x = π 6 . Answer 4 3 Higher-Order Derivatives The higher-order derivatives of sin x and cos x follow a repeating pattern. By following the pattern, we can find any higherorder derivative of sin x and cos x. Example 2.4.7 : Finding Higher-Order Derivatives of y = sin x Find the first four derivatives of y = sin x. Solution Each step in the chain is straightforward: y = sin x (2.4.28) dy = cos x (2.4.29) dx 2 d y = − sin x (2.4.30) = − cos x (2.4.31) 2 dx 3 d y 3 dx 4 d y 4 = sin x (2.4.32) dx Analysis Once we recognize the pattern of derivatives, we can find any higher-order derivative by determining the step in the pattern to which it corresponds. For example, every fourth derivative of sin x equals sin x, so d 4 4 d (sin x) = dx d 5 5 dx 8 d 9 9 dx 12 (sin x) = dx d (sin x) = 8 dx d (sin x) = 12 d (sin x) = … = 13 13 dx 4n 4n (sin x) = sin x (2.4.33) dx d 4n+1 (sin x) = … = 4n+1 (sin x) = cos x. (2.4.34) dx Exercise 2.4.7 4 For y = cos x , find d y 4 . dx Hint See the previous example. Answer cos x 10/21/2020 2.4.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10289 Example 2.4.8 : Using the Pattern for Higher-Order Derivatives of y = sin x Find d 74 74 . (sin x) dx Solution We can see right away that for the 74th derivative of sin x,74 = 4(18) + 2 , so d 74 74 d 72+2 (sin x) = dx d 72+2 2 (sin x) = dx 2 (sin x) = − sin x. (2.4.35) dx Exercise 2.4.8 For y = sin x , find d 59 59 (sin x). (2.4.36) dx Hint d 59 59 d 4⋅14+3 (sin x) = dx 4⋅14+3 (sin x) (2.4.37) dx Answer − cos x Example 2.4.9 : An Application to Acceleration A particle moves along a coordinate axis in such a way that its position at time t is given by s(t) = 2 − sint . Find v(π/4) and a(π/4). Compare these values and decide whether the particle is speeding up or slowing down. Solution First find v(t) = s'(t) v(t) = s'(t) = − cos t. (2.4.38) Thus, π v( 1 ) =− 4 . – √2 Next, find a(t) = v'(t) . Thus, a(t) = v'(t) = sint and we have π a( 4 Since π v( ) =− 4 1 – <0 √2 and π a( ) = 4 1 – >0 √2 1 ) = – √2 . , we see that velocity and acceleration are acting in opposite directions; that is, the object is being accelerated in the direction opposite to the direction in which it is travelling. Consequently, the particle is slowing down. Exercise 2.4.9 A block attached to a spring is moving vertically. Its position at time t is given by s(t) = 2sint . Find v( Compare these values and decide whether the block is speeding up or slowing down. 5π ) 6 and 5π a( ) 6 . Hint Use Example 17 as a guide. 10/21/2020 2.4.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10289 Answer 5π v( – ) = −√3 < 0 6 and a( 5π ) = −1 < 0 6 . The block is speeding up. Key Concepts We can find the derivatives of sin x and cos x by using the definition of derivative and the limit formulas found earlier. The results are d d sin x = cos x dx cos x = − sin x . dx With these two formulas, we can determine the derivatives of all six basic trigonometric functions. Key Equations Derivative of sine function d (sin x) = cos x dx Derivative of cosine function d (cos x) = − sin x dx Derivative of tangent function d 2 (tan x) = sec x dx Derivative of cotangent function d 2 (cot x) = −csc x dx Derivative of secant function d (secx) = secx tan x dx Derivative of cosecant function d (cscx) = −cscx cot x dx Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 2.4.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10289 2.4E: Exercises Exercise 2.4E. 1 For the following exercises, find 1) y = x 2 2 5 x cotx 4) y = x − x 3 5) y = for the given functions: − secx + 1 2) y = 3cscx + 3) y = x dy dx sinx secx x 6) y = sinxtanx 7) y = (x + cosx)(1 − sinx) 8) y = 9) y = tanx 1−secx 1−cotx 1+cotx 10) y = cosx(1 + cscx) Answers to odd numbered questions 1. 3. 5. 7. 9. dy dx dy dx dy dx = 2x − secxtanx 2 2 = 2xcotx − x csc x xsecxtanx−secx = 2 x dy dx = (1 − sinx)(1 − sinx) − cosx(x + cosx) 2 dy dx 2csc x = 2 (1+cotx) Exercise 2.4E. 2 For the following exercises, find the equation of the tangent line to each of the given functions at the indicated values of x. 1) f (x) = −sinx, x = 0 2) f (x) = cscx, x = 3) f (x) = 1 + cosx, 4) f (x) = secx, 5) f (x) = x 2 π 2 x = x = 3π 2 π 4 − tanx, x = 0 6) f (x) = 5cot(x), x = π 4 Answers to odd numbered questions 1. y = −x 10/21/2020 2.4E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10649 3. y = x + 2−3π 5. y = x + 2−3π 2 2 Exercise 2.4E. 3 For the following exercises, find d 2 y dx2 for the given functions. 1) y = xsinx − cosx 2) y = sinxcosx 3) y = x − 4) y = 1 x 1 2 sinx + tanx 5) y = 2cscx 6) y = sec 2 x Answers to odd numbered questions 10/21/2020 2.4E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10649 1. 3cosx − xsinx 3. 1 2 sinx 5. csc(x)(3csc 2 2 (x) − 1 + cot (x)) Exercise 2.4E. 4 Find all x values on the graph of f (x) = −3sinxcosx where the tangent line is horizontal. Answer (2n+1)π 4 ,where n is an integer. Exercise 2.4E. 5 Find all x values on the graph of f (x) = x − 2cosx for 0 < x < 2π where the tangent line has slope 2. Answer Under Construction Exercise 2.4E. 6 Let f (x) = cotx. Determine the points on the graph of line y = −2x . f for 0 < x < 2π where the tangent line(s) is (are) parallel to the Answer ( π 4 , 1), ( 3π 4 , −1) Exercise 2.4E. 7 A mass on a spring bounces up and down in simple harmonic motion, modeled by the function s(t) = −6cost where s is measured in inches and t is measured in seconds. Find the rate at which the spring is oscillating at t = 5 s. Answer Under Construction Exercise 2.4E. 8 Let the position of a swinging pendulum in simple harmonic motion be given by s(t) = acost + bsint . Find the constants a and b such that when the velocity is 3 cm/s, s = 0 and t = 0 . Answer a = 0, b = 3 Exercise 2.4E. 9 After a diver jumps off a diving board, the edge of the board oscillates with position given by t seconds after the jump. s(t) = −5cost cm at a. Sketch one period of the position function for t ≥ 0 . b. Find the velocity function. c. Sketch one period of the velocity function for t ≥ 0 . 10/21/2020 2.4E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10649 d. Determine the times when the velocity is 0 over one period. e. Find the acceleration function. f. Sketch one period of the acceleration function for t ≥ 0 . Answer Under Construction Exercise 2.4E. 10 The number of hamburgers sold at a fast-food restaurant in Pasadena, California, is given by y = 10 + 5sinx where y is the number of hamburgers sold and x represents the number of hours after the restaurant opened at 11 a.m. until 11 p.m., when the store closes. Find y and determine the intervals where the number of burgers being sold is increasing. ′ Answer , increasing on (0, y' = 5cos(x) π 2 ), ( 3π 2 , 5π 2 , and ( ) 7π 2 , 12) Exercise 2.4E. 11 The amount of rainfall per month in Phoenix, Arizona, can be approximated by y(t) = 0.5 + 0.3cost , where t is months since January. Find y'and use a calculator to determine the intervals where the amount of rain falling is decreasing. Answer Under Construction Exercise 2.4E. 12 For the following exercises, use the quotient rule to derive the given equations. 1) 2) 3) d 2 dx d dx d dx (cotx) = −csc x (secx) = secxtanx (cscx) = −cscxcotx Answer Under Construction Exercise 2.4E. 13 For the following exercises, find the requested higher-order derivative for the given functions. 1. 2. 3. 4. 5. d 3 y 3 dx d 2 y 2 dx d 4 y dx4 d 2 y 2 dx d 3 y 3 dx of y = 3cosx of y = 3sinx + x 2 cosx of y = 5cosx of y = secx + cotx of y = x 10 − secx Answers to odd numbered questions 1. 3sinx 3. 5cosx 5. 720x 7 10/21/2020 3 3 − 5tan(x)sec (x) − tan (x)sec(x) 2.4E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10649 Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 2.4E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10649 2.5: Chain Rule We have seen the techniques for differentiating basic functions (x , sin x, cos x, etc. ) as well as sums, differences, products, quotients, and constant multiples of these functions. However, these techniques do not allow us to differentiate compositions − −−−− − of functions, such as h(x) = sin(x ) or k(x) = √3x + 1 . In this section, we study the rule for finding the derivative of the composition of two or more functions. n 3 2 Deriving the Chain Rule When we have a function that is a composition of two or more functions, we could use all of the techniques we have already learned to differentiate it. However, using all of those techniques to break down a function into simpler parts that we are able to differentiate can get cumbersome. Instead, we use the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. To put this rule into context, let’s take a look at an example: h(x) = sin(x ) . We can think of the derivative of this function with respect to x as the rate of change of sin(x ) relative to the change in x. Consequently, we want to know how sin(x ) changes as x changes. We can think of this event as a chain reaction: As x changes, x changes, which leads to a change in sin(x ). This chain reaction gives us hints as to what is involved in computing the derivative of sin(x ). First of all, a change in x forcing a change in x suggests that somehow the derivative of x is involved. In addition, the change in x forcing a change in sin(x ) suggests that the derivative of sin(u) with respect to u, where u = x , is also part of the final derivative. 3 3 3 3 3 3 3 3 3 3 3 We can take a more formal look at the derivative of h(x) = sin(x a specific value a in the domain of h(x) = sin(x ) . 3 ) by setting up the limit that would give us the derivative at 3 3 3 sin(x ) − sin(a ) h'(a) = lim (2.5.1) x −a x→a This expression does not seem particularly helpful; however, we can modify it by multiplying and dividing by the expression x − a \) to obtain 3 3 3 3 3 sin(x ) − sin(a ) h'(a) = lim x3 − a3 x→a x 3 −a ⋅ . (2.5.2) x −a From the definition of the derivative, we can see that the second factor is the derivative of x at x = a. That is, 3 3 x 3 −a lim x→a d 3 = 2 (x ) = 3 a . x −a (2.5.3) dx However, it might be a little more challenging to recognize that the first term is also a derivative. We can see this by letting u = x and observing that as x → a, u → a : 3 3 3 3 3 sin(x ) − sin(a ) lim x→a 3 x sin u − sin(a ) = lim 3 −a 3 u→a (2.5.4) 3 u −a . d = du (sin u )u=a3 (2.5.5) = cos(a3) Thus, h'(a) = cos(a 3 In other words, if 2 ) ⋅ 3a (2.5.6) . . Thus, if we think of h(x) = sin(x ) as the composition (f ∘ g)(x) = f (g(x)) where f (x) = sin x and g(x) = x , then the derivative of h(x) = sin(x ) is the product of the derivative of g(x) = x and the derivative of the function f (x) = sin x evaluated at the function g(x) = x . At this point, we anticipate that for h(x) = sin(g(x)) , it is quite likely that h'(x) = cos(g(x))g'(x). As we determined above, this is the case for h(x) = sin(x ) . 3 h(x) = sin(x ) , then 3 2 3 h'(x) = cos(x ) ⋅ 3 x 3 3 3 3 3 Now that we have derived a special case of the chain rule, we state the general case and then apply it in a general form to other composite functions. An informal proof is provided at the end of the section. Rule: The Chain Rule 10/21/2020 2.5.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10636 Let f and g be functions. For all x in the domain of g for which g is differentiable at x and f is differentiable at derivative of the composite function g(x) , the h(x) = (f ∘ g)(x) = f (g(x)) (2.5.7) h'(x) = f '(g(x))g'(x). (2.5.8) is given by Alternatively, if y is a function of u, and u is a function of x, then dy dy du = dx ⋅ du dx . Problem-Solving Strategy: Applying the Chain Rule 1. To differentiate h(x) = f (g(x)) , begin by identifying f (x) and g(x). 2. Find f (x) and evaluate it at g(x) to obtain f '(g(x)). 3. Find g'(x). 4. Write h'(x) = f '(g(x)) ⋅ g'(x). ′ Note: When applying the chain rule to the composition of two or more functions, keep in mind that we work our way from the outside function in. It is also useful to remember that the derivative of the composition of two functions can be thought of as having two parts; the derivative of the composition of three functions has three parts; and so on. Also, remember that we never evaluate a derivative at a derivative. Example 2.5.1A : Find d − − |x|, dx by writing |x| = √x . 2 Solution: d d |x| = dx − − √x2 = dx 1 2 √x2 2x = x |x| , x ≠ 0. Example 2.5.1B : Find d | sin(x)|, dx Solution: d sin(x) | sin(x)| = dx | sin(x)| cos(x), x ≠ 0 Exercise 2.5.1 Find d | cos(x)|, dx Answer -\frac{\cos(x)}{|\cos(x)|} \sin(x), x \ne \pi/2\). The Chain and Power Rules Combined We can now apply the chain rule to composite functions, but note that we often need to use it with other rules. For example, to find derivatives of functions of the form h(x) = (g(x)) , we need to use the chain rule combined with the power rule. To do so, we can think of h(x) = (g(x)) as f (g(x)) where f (x) = x . Then f '(x) = nx . Thus, f '(g(x)) = n(g(x)) . This leads us to the derivative of a power function using the chain rule, n n n n−1 n−1 h'(x) = n(g(x)) 10/21/2020 n−1 g'(x) 2.5.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10636 Rule: Power Rule for Composition of Functions For all values of x for which the derivative is defined, if n h(x) = (g(x)) . Then n−1 h'(x) = n(g(x)) g'(x) . Example 2.5.1 : Using the Chain and Power Rules 1 Find the derivative of h(x) = 2 (3 x 2 . + 1) Solution First, rewrite h(x) = 1 2 (3 x 2 = (3 x 2 −2 + 1) . + 1) Applying the power rule with g(x) = 3x 2 , we have +1 2 h'(x) = −2(3 x −3 + 1) (6x) . Rewriting back to the original form gives us −12x h'(x) = 2 (3 x 3 + 1) Exercise 2.5.1 Find the derivative of h(x) = (2x 3 4 + 2x − 1 ) . Hint Use Equation with g(x) = 2x 3 + 2x − 1 Answer 3 h'(x) = 4(2 x 3 3 + 2x − 1 ) (6x + 2) = 8(3x + 1)(2 x 3 + 2x − 1 ) Example 2.5.2 : Using the Chain and Power Rules with a Trigonometric Function Find the derivative of h(x) = si n 3 x . Solution First recall that si n 3 3 x = (sin x ) , so we can rewrite h(x) = si n 3 as h(x) = (sin x) . 3 x Applying the power rule with g(x) = sin x , we obtain 2 . 2 h'(x) = 3(sin x ) cos x = 3si n x cos x Example 2.5.3 : inding the Equation of a Tangent Line 1 Find the equation of a line tangent to the graph of h(x) = 2 at x = 2 . (3x − 5) Solution Because we are finding an equation of a line, we need a point. The x-coordinate of the point is 2. To find the y-coordinate, substitute 2 into h(x). Since h(2) = 1 (3(2) − 5)2 =1 , the point is (2, 1). For the slope, we need h'(2). To find h'(x), first we rewrite h(x) = (3x − 5) and apply the power rule to obtain −2 −3 h'(x) = −2(3x − 5 ) 10/21/2020 −3 (3) = −6(3x − 5 ) 2.5.3 CC-BY-NC-SA . https://math.libretexts.org/@go/page/10636 By substituting, we have h'(2) = −6(3(2) − 5) −3 = −6. Therefore, the line has equation y − 1 = −6(x − 2) . Rewriting, the equation of the line is y = −6x + 13 . Exercise 2.5.2 Find the equation of the line tangent to the graph of f (x) = (x 2 3 − 2) at x = −2 . Hint Use the preceding example as a guide. Answer y = −48x − 88 Combining the Chain Rule with Other Rules Now that we can combine the chain rule and the power rule, we examine how to combine the chain rule with the other rules we have learned. In particular, we can use it with the formulas for the derivatives of trigonometric functions or with the product rule. Example 2.5.4 : Using the Chain Rule on a General Cosine Function Find the derivative of h(x) = cos(g(x)). Solution Think of h(x) = cos(g(x)) as f (g(x)) where Then we do the following calculation. h'(x) = f '(g(x))g'(x) . Since f (x) = cos x f '(x) = − sin x . we have f '(g(x)) = − sin(g(x)) . Apply the chain rule. Substitutef′(g(x))=−\sin(g(x)). = − sin(g(x))g'(x) Thus, the derivative of h(x) = cos(g(x)) is given by h'(x) = − sin(g(x))g'(x). In the following example we apply the rule that we have just derived. Example 2.5.5 : Using the Chain Rule on a Cosine Function Find the derivative of h(x) = cos(5x 2 ). Solution 2 Letg(x) = 5x . Then g'(x) = 10x. Using the result from the previous example, h'(x) = − sin(5x2) ⋅ 10x = −10x sin(5x2) Example 2.5.6 : Using the Chain Rule on Another Trigonometric Function Find the derivative of h(x) = sec(4x 5 + 2x). Solution Apply the chain rule to h(x) = sec(g(x)) to obtain h'(x) = sec(g(x)tan(g(x))g'(x). In this problem, g(x) = 4x 5 + 2x, 5 h'(x) = sec(4 x so we have g'(x) = 20x 4 5 + 2x)tan(4 x 4 + 2x)(20 x + 2. Therefore, we obtain 4 + 2) = (20 x 5 + 2)sec(4 x 5 + 2x)tan(4 x + 2x). Example 2.5.1 : Using the Chain Rule on Another Trigonometric Function Find the derivative of h(x) = cot(sin(x). 10/21/2020 2.5.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10636 Solution Apply the chain rule to h(x) = cot(g(x)) to obtain \(h`(x)= -csc^2(sin(x)) cos(x). Exercise 2.5.3 Find the derivative of h(x) = sin(7x + 2). Hint Apply the chain rule to h(x) = sing(x) first and then use g(x) = 7x + 2 . Answer h'(x) = 7 cos(7x + 2) At this point we provide a list of derivative formulas that may be obtained by applying the chain rule in conjunction with the formulas for derivatives of trigonometric functions. Their derivations are similar to those used in Example and Example. For convenience, formulas are also given in Leibniz’s notation, which some students find easier to remember. (We discuss the chain rule using Leibniz’s notation at the end of this section.) It is not absolutely necessary to memorize these as separate formulas as they are all applications of the chain rule to previously learned formulas. Using the Chain Rule with Trigonometric Functions For all values of x for which the derivative is defined, d d ′ d dx ′ (cos(g(x)) = − sin(g(x))g (x) 2 du cosu = − sin u dx dx d du ′ 2 (tan(g(x)) = sec (g(x))g (x) tanu = sec u dx dx d d 2 ′ (cot(g(x)) = −csc (g(x))g (x) dx 2 du dx d ′ du (sec(g(x)) = sec(g(x)tan(g(x))g (x) secu = secutanu dx dx d dx cotu = −csc u dx d dx d dx d du sin u = cosu (sin(g(x)) = cos(g(x))g (x) dx dx d ′ (csc(g(x)) = −csc(g(x))cot(g(x))g (x) du cscu = −cscucotu dx dx . dx Example 2.5.7 : Combining the Chain Rule with the Product Rule Find the derivative of h(x) = (2x + 1) 5 7 (3x − 2 ) . Solution First apply the product rule, then apply the chain rule to each term of the product. d h'(x) = 5 7 ((2x + 1 ) ) ⋅ (3x − 2 ) dx d 7 + 5 ((3x − 2 ) ) ⋅ (2x + 1 ) dx 4 = 5(2x + 1 ) 7 ⋅ 2 ⋅ (3x − 2 ) 4 7 = 10(2x + 1 ) (3x − 2 ) 4 6 4 6 6 + 7(3x − 2 ) 6 5 ⋅ 3 ⋅ (2x + 1 ) 5 + 21(3x − 2 ) (2x + 1 ) = (2x + 1 ) (3x − 2 ) (10(3x − 7) + 21(2x + 1)) = (2x + 1 ) (3x − 2 ) (72x − 49) Exercise 2.5.4 Find the derivative of h(x) = x 3 . (2x + 3) 10/21/2020 2.5.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10636 Hint Start out by applying the quotient rule. Remember to use the chain rule to differentiate the denominator. Answer 3 − 4x h'(x) = 4 (2x + 3) Composites of Three or More Functions We can now combine the chain rule with other rules for differentiating functions, but when we are differentiating the composition of three or more functions, we need to apply the chain rule more than once. If we look at this situation in general terms, we can generate a formula, but we do not need to remember it, as we can simply apply the chain rule multiple times. In general terms, first we let k(x) = h(f (g(x))). (2.5.9) Then, applying the chain rule once we obtain d d ′ k'(x) = (h(f (g(x))) = h (f (g(x))) ⋅ dx f ((g(x))). (2.5.10) dx Applying the chain rule again, we obtain k'(x) = h'(f (g(x))f '(g(x))g'(x)). (2.5.11) Example 2.5.8 : Rule: Chain Rule for a Composition of Three Functions Solution For all values of x for which the function is differentiable, if k(x) = h(f (g(x))), then k'(x) = h'(f (g(x)))f '(g(x))g'(x). In other words, we are applying the chain rule twice. Notice that the derivative of the composition of three functions has three parts. (Similarly, the derivative of the composition of four functions has four parts, and so on.) Also, remember, we can always work from the outside in, taking one derivative at a time. Example 2.5.9 : Differentiating a Composite of Three Functions Find the derivative of k(x) = cos 4 2 (7 x + 1). Solution First, rewrite k(x) as 2 k(x) = (cos(7 x 4 + 1)) . Then apply the chain rule several times. 3 2 k'(x) = 4 cos (7 x d + 1)( 2 (cos(7 x + 1)) dx 3 2 = 4(cos (7 x d 2 + 1)(− sin(7 x + 1))( 2 (7 x + 1)) dx 3 2 = 4 cos (7 x 2 + 1)(− sin(7 x 2 = −56x sin(7 x 10/21/2020 3 + 1))(14x) 2 + 1)cos (7 x + 1) 2.5.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10636 Exercise 2.5.1 − −−−−−−−−− − − −−−− − Find the derivative of F (x) = √x + √x + √− x Answer: F ‘(x) = F ‘(x) = 1 2 − −−−− − −1/2 − (x + √x + √x ) (1 + 1 1 (1 + 1 2 − −1/2 (x + √x ) (1 + 1 (1 + 2 √x 2 √x+√x 2 √x+√x+√x 1 2 −1/2 x )) )) Exercise 2.5.5 Find the derivative of h(x) = si n 6 3 (x ). Hint Rewrite h(x) = si n 6 3 3 6 (x ) = (sin(x )) and use Example as a guide. Answer 2 5 3 3 h'(x) = 18 x si n (x ) cos(x ) Example 2.5.10 : Using the Chain Rule in a Velocity Problem A particle moves along a coordinate axis. Its position at time t is given by of the particle at time t = π 6 s(t) = sin(2t) + cos(3t) . What is the velocity ? Solution To find v(t) , the velocity of the particle at time t , we must differentiate s(t) . Thus, v(t) = s'(t) = 2 cos(2t) − 3 sin(3t). (2.5.12) Proof of Chain Rule At this point, we present a very informal proof of the chain rule. For simplicity’s sake we ignore certain issues: For example, we assume that g(x) ≠ g(a) for x ≠ a in some open interval containing a . We begin by applying the limit definition of the derivative to the function h(x) to obtain h'(a): f (g(x)) − f (g(a)) h'(a) = limx→a x −a . Rewriting, we obtain f (g(x)) − f (g(a)) g(x) − g(a) h'(a) = limx→a ⋅ x −a g(x) − g(a) . Although it is clear that g(x) − g(a) limx→a = g'(a) x −a , it is not obvious that f (g(x)) − f (g(a)) limx→a = f '(g(a)) . g(x) − g(a) To see that this is true, first recall that since g is differentiable at a, g is also continuous at a. Thus, limx→a g(x) = g(a) . Next, make the substitution y = g(x) and b = g(a) and use change of variables in the limit to obtain 10/21/2020 2.5.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10636 f (g(x)) − f (g(a)) limx→a f (y) − f (b) = limy→b = f '(b) = f '(g(a)). g(x) − g(a) y −b Finally, f (g(x)) − f (g(a)) h'(a) = limx→a g(x) − g(a) ⋅ = f '(g(a))g'(a) x −a g(x) − g(a) . □ Example 2.5.11 : Using the Chain Rule with Functional Values Let h(x) = f (g(x)). If g(1) = 4, g'(1) = 3 , and f '(4) = 7, find h'(1). Solution Use the chain rule, then substitute. h'(1) = f '(g(1))g'(1) = f '(4) ⋅ 3 =7⋅3 = 21 Apply the chain rule. Substitute g(1) = 4 and g'(1) = 3. Substitute f ′ (4) = 7. Simplify Exercise 2.5.6 Given h(x) = f (g(x)) . If g(2) = −3, g'(2) = 4, and f '(−3) = 7 , find h'(2). Hint Follow Example. Answer 28 The Chain Rule Using Leibniz’s Notation As with other derivatives that we have seen, we can express the chain rule using Leibniz’s notation. This notation for the chain rule is used heavily in physics applications. For h(x) = f (g(x)), let u = g(x) and y = h(x) = g(u). Thus, dy h'(x) = (2.5.13) dx dy f '(g(x)) = f '(u) = (2.5.14) du and du g'(x) = . (2.5.15) dx Consequently, dy dy = h'(x) = f '(g(x))g'(x) = dx du ⋅ du . dx Rule: Chain Rule Using Leibniz’s Notation If y is a function of u, and u is a function of x, then dy dy = dx 10/21/2020 du ⋅ du dx . 2.5.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10636 Example 2.5.12 : Taking a Derivative Using Leibniz’s Notation I Find the derivative of 5 x y =( ) . (2.5.16) 3x + 2 Solution x First, let u = . Thus, y = u . Next, find 5 3x + 2 du dx and dy du . Using the quotient rule, du 2 = 2 dx (3x + 2) and dy 4 = 5u du . Finally, we put it all together. dy dy = dx du Apply the chain rule. ⋅ du dx 4 = 5u 2 ⋅ 2 Substitutedydu=5u4anddudx=2(3x+2)2. (3x + 2) x 4 = 5( ) 3x + 2 2 ⋅ 2 Substituteu=x3x+2. (3x + 2) 4 10x = 6 (3x + 2) Simplify. It is important to remember that, when using the Leibniz form of the chain rule, the final answer must be expressed entirely in terms of the original variable given in the problem. Example 2.5.13 : Taking a Derivative Using Leibniz’s Notation II Find the derivative of 2 y = tan(4 x − 3x + 1). (2.5.17) Solution First, let u = 4x 2 − 3x + 1. Then y = tanu . Next, find du dx du and du : dy and = 8x − 3 dx dy 2 = sec u. du Finally, we put it all together. dy dy = dx du Apply the chain rule. ⋅ du dx 2 = sec u ⋅ (8x − 3) 2 2 = sec (4 x Use du = 8x − 3 dx − 3x + 1) ⋅ (8x − 3) and dy 2 = sec u du Substitute u = 4x 2 . − 3x + 1 . Exercise 2.5.7 Use Leibniz’s notation to find the derivative of y = cos(x ) . Make sure that the final answer is expressed entirely in terms of the variable x. 3 Hint 10/21/2020 2.5.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10636 Let u = x . 3 Answer dy 2 = −3 x 3 sin(x ). (2.5.18) dx Key Concepts The chain rule allows us to differentiate compositions of two or more functions. It states that for h(x) = f (g(x)), h'(x) = f '(g(x))g'(x). In Leibniz’s notation this rule takes the form dy dy = dx du ⋅ du dx . We can use the chain rule with other rules that we have learned, and we can derive formulas for some of them. The chain rule combines with the power rule to form a new rule: If h(x) = (g(x)) ,then h'(x) = n(g(x)) n n−1 g'(x) . When applied to the composition of three functions, the chain rule can be expressed as follows: If then h'(x) = f '(g(k(x))g'(k(x))k'(x). h(x) = f (g(k(x))), Key Equations The chain rule h'(x) = f '(g(x))g'(x) The power rule for functions n−1 h'(x) = n(g(x)) g'(x) Glossary chain rule the chain rule defines the derivative of a composite function as the derivative of the outer function evaluated at the inner function times the derivative of the inner function Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 2.5.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10636 2.5E: Exercises Exercise 2.5E. 1 For the following exercises, given dy dx = dy du y = f (u) and u = g(x) , find dy/dx by using Leibniz’s notation for the chain rule: . du dx 1) y = 3u − 6, u = 2x 2 2) y = 6u 3 , u = 7x − 4 3) y = sinu, u = 5x − 1 −x 4) y = cosu, u = 8 5) y = tanu, u = 9x + 2 −−−− − 6) y = √4u + 3 , u = x 2 − 6x Answers to even numbered questions 2. 18u 2 2 ⋅ 7 = 18(7x − 4 ) −1 4. −sinu ⋅ 6. 8x−24 −x − sin( 8= 8 ⋅7 )⋅ −1 8 4x−12 = 2 √4u+3 √4 x2 −24x+3 Exercise 2.5E. 2 For each of the following exercises, a. decompose each function in the form y = f (u) and u = g(x), and b. find dy as a function of x. dx 1) y = (3x − 2) 6 2) y = (3x 2 3) y = si n 5 4) y = ( x 3 + 1) (x) + 7 7 7 ) x 5) y = tan(secx) 6) y = csc(πx + 1) 7) y = cot 2 x 8) y = −6si n −3 x Answers to even numbered questions 2a. u = 3x +1 b. 18x(3x + 1) 2 2 2 4a. f (u) = u ,u = b. 7( ⋅( 7 x 7 + 7 x 6 ) 1 7 − x + 7 7 2 x 7 x ) 6a. f (u) = cscu, u = πx + 1 b. −πcsc(πx + 1) ⋅ cot(πx + 1) 10/21/2020 2.5E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10652 8a. f (u) = −6u −3 b. 18si n −4 , u = sinx x ⋅ cosx Exercise 2.5E. 3 dy For the following exercises, find 1) y = (3x 2 dx for each function. 4 + 3x − 1 ) 2) y = (5 − 2x) −2 3) y = cos 3 (πx) 4) y = (2x 3 5) y = 2 −x 3 + 6x + 1 ) 1 sin2 (x) 6) y = (tanx + sinx) −3 7) y = x 2 4 cos x 8) y = sin(cos7x) − − − − − − − − − 9) y = √6 + secπx 2 10) y = cot 3 (4x + 1) Answers to even numbered questions 2. 4 3 (5−2x) 4. y = (2x 3 2 −x 3 + 6x + 1 ) 6. −3(tanx + sinx ) −4 2 ⋅ (sec x + cosx) 8. −7cos(cos7x) ⋅ sin7x 10. −12cot 2 2 (4x + 1) ⋅ csc (4x + 1) Exercise 2.5E. 4 Let y = [f (x)] and suppose that f '(1) = 4 and 3 dy dx = 10 for x = 1 . Find f (1). Answer Under Construction Exercise 2.5E. 5 Let y = (f (x) + 5x 2 4 ) and suppose that f (−1) = −4 and dy dx =3 when x = −1 . Find f '(−1) Answer 10 3 4 Exercise 2.5E. 6 Let y = (f (u) + 3x) and u = x 2 3 − 2x . If f (4) = 6 and dy dx = 18 when x = 2 , find f '(4). Answer Under Construction 10/21/2020 2.5E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10652 Exercise 2.5E. 7 Find the equation of the tangent line to y = −sin( line together. x 2 ) at the origin. Use a calculator to graph the function and the tangent Answer y = −1 2 x Exercise 2.5E. 8 Find the equation of the tangent line to y = (3x + tangent line together. 1 x 2 ) at the point (1, 16). Use a calculator to graph the function and the Answer Under Construction Exercise 2.5E. 9 Find the x -coordinates at which the tangent line to y = (x − 6 x 8 ) is horizontal. Answer – x = ±√6 Exercise 2.5E. 10 Find an equation of the line that is normal to and the normal line together ( g(θ) = sin2 πθ) at the point ( 1 4 , 1 2 . Use a calculator to graph the function ) Answer Under Construction Exercise 2.5E. 11 For the following exercises, use the information in the following table to find h'(a) at the given value for a . ′ ′ x f (x) f (x) g(x) g (x) 0 2 5 0 2 1 1 −2 3 0 2 4 4 1 −1 3 3 −3 2 3 1) h(x) = f (g(x)); a = 0 2) h(x) = g(f (x)); a = 0 3) h(x) = (x 4 4) h(x) = ( −2 + g(x)) f (x) g(x) ;a =1 2 ) ;a =3 5) h(x) = f (x + f (x)); a = 1 10/21/2020 2.5E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10652 6) h(x) = (1 + g(x)) 3 7) h(x) = g(2 + f (x 2 ;a =2 )); a = 1 8) h(x) = f (g(sinx)); a = 0 Answer to odd numbered questions 1. 10 3. − 1 8 5. −4 7. −12 Exercise 2.5E. 12 The position function of a freight train is given by s(t) = 100(t + 1) −2 , with s in meters and t in seconds. At time t = 6 s, find the train’s a. velocity and b. acceleration. c. Using a. and b. is the train speeding up or slowing down? Answer a. − b. 200 343 600 2401 m/s m/s^2 c. The train is slowing down since velocity and acceleration have opposite signs Exercise 2.5E. 13 A mass hanging from a vertical spring is in simple harmonic motion as given by the following position function, where t is measured in seconds and s is in inches: s(t) = −3cos(πt + π 4 ). a. Determine the position of the spring at t = 1.5 s. b. Find the velocity of the spring at t = 1.5 s. Answer Under Construction Exercise 2.5E. 14 The total cost to 3 C = 0.0001 x produce 2 − 0.02 x x + 3x + 300. boxes of Thin Mint Girl Scout cookies is C dollars, In t weeks production is estimated to be x = 1600 + 100t boxes. where a. Find the marginal cost C '(x). b. Use Leibniz’s notation for the chain rule, changing. dC dt = dC dx ⋅ dx dt , to find the rate with respect to time t that the cost is c. Use b. to determine how fast costs are increasing when t = 2 weeks. Include units with the answer. Answer a. C '(x) = 0.0003x 2 10/21/2020 − 0.04x + 3 2.5E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10652 b. dC dt = 100 ⋅ (0.0003x 2 − 0.04x + 3) c. Approximately $90,300 per week Exercise 2.5E. 15 The formula for the area of a circle is A = πr , where r is the radius of the circle. Suppose a circle is expanding, meaning that both the area A and the radius r (in inches) are expanding. 2 a. Suppose r = 2 − where t is time in seconds. Use the chain rule 100 2 (t+7) dA dt = dA dr ⋅ dr dt to find the rate at which the area is expanding. b. Use a. to find the rate at which the area is expanding at t = 4 s. Answer Under Construction Exercise 2.5E. 16 The formula for the volume of a sphere is snowball is melting in the sun. a. Suppose r = 1 2 (t+1) − 1 12 S = 4 3 3 πr , where r (in feet) is the radius of the sphere. Suppose a spherical where t is time in minutes. Use the chain rule dS dt = dS dr ⋅ dr dt to find the rate at which the snowball is melting. b. Use a. to find the rate at which the volume is changing at t = 1 min. Answer a. dS dt 2 =− 8πr 3 (t+1) b. The volume is decreasing at a rate of − π 36 3 ft /min Exercise 2.5E. 17 The daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function T (x) = 94 − 10cos[ (x − 2)] , where x is hours after midnight. Find the rate at which the temperature is changing at 4 p.m. π 12 Answer Under Construction Exercise 2.5E. 18 The depth (in feet) of water at a dock changes with the rise and fall of tides. The depth is modeled by the function D(t) = 5sin( t − ) + 8 , where t is the number of hours after midnight. Find the rate at which the depth is changing at 6 a.m. π 7π 6 6 Answer 2.3 ft/hr Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 2.5E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10652 2.6: Implicit Differentiation dy In the last chapter we learned to find the derivative, , or y , when y is given explicitly as a function of x. That is, if we know y = f (x) for some function f , we can find y . For example, given y = 3 x − 7 , we can easily find y = 6x . (Here we explicitly state how x and y are related. Knowing x, we can directly find y .) ′ dx ′ 2 ′ Sometimes the relationship between y and x is not explicit; rather, it is implicit. For instance, we might know that x − y = 4 . This equality defines a relationship between x and y ; if we know x, we could figure out y . Can we still find y ? In this case, sure; we solve for y to get y = x − 4 (hence we now know y explicitly) and then differentiate to get y = 2x . 2 ′ 2 ′ Sometimes the implicit relationship between x and y is complicated. Suppose we are given sin(y) + y = 6 − x . A graph of this implicit function is given in Figure 2.19. In this case there is absolutely no way to solve for y in terms of elementary functions. The surprising thing is, however, that we can still find y via a process known as implicit differentiation. 3 3 ′ Figure 2.19:A graph of the implicit function sin(y) + y 3 2 = 6 −x . Implicit differentiation is a technique based on the Chain Rule that is used to find a derivative when the relationship between the variables is given implicitly rather than explicitly (solved for one variable in terms of the other). We begin by reviewing the Chain Rule. Let f and g be functions of x. Then d ′ ′ (f (g(x))) = f (g(x)) ⋅ g (x). (2.6.1) dx Suppose now that y = g(x) . We can rewrite the above as d ′ ′ (f (y))) = f (y)) ⋅ y , d or dx dy ′ (f (y))) = f (y) ⋅ dx . (2.1) dx These equations look strange; the key concept to learn here is that we can find y even if we don't exactly know how y and x relate. ′ We demonstrate this process in the following example. Example 2.6.1 :: Using Implicit Differentiation Find y given that sin(y) + y ′ 3 3 = 6 −x . Solution: We start by taking the derivative of both sides (thus maintaining the equality.) We have : d 3 d ( sin(y) + y ) = dx 3 (6 − x ). (2.6.2) dx The right hand side is easy; it returns −3x . 2 The left hand side requires more consideration. We take the derivative term--by--term. Using the technique derived from Equation 2.1 above, we can see that d ′ ( sin y) = cos y ⋅ y . (2.6.3) dx 10/21/2020 2.6.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10290 We apply the same process to the y term. 3 d d 3 3 (y ) = dx 2 ((y ) ) = 3(y ) ′ ⋅y . (2.6.4) dx Putting this together with the right hand side, we have ′ 2 cos(y)y + 3 y y ′ 2 = −3 x . (2.6.5) Now solve for y . ′ ′ 2 cos(y)y + 3 y y 2 ( cos y + 3 y )y ′ ′ 2 = −3 x . 2 = −3x 2 y ′ −3x = cos y + 3y This equation for next. y ′ probably seems unusual for it contains both x and y 2 terms. How is it to be used? We'll address that Exercise 2.6.1 Find y given that cos(y) + y ′ 3 3 = 6 −x . Answer Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page. Implicit functions are generally harder to deal with than explicit functions. With an explicit function, given an x value, we have an explicit formula for computing the corresponding y value. With an implicit function, one often has to find x and y values at the same time that satisfy the equation. It is much easier to demonstrate that a given point satisfies the equation than to actually find such a point. – For instance, we can affirm easily that the point (√6, 0) lies on the graph of the implicit function sin y + y = 6 − x . – Plugging in 0 for y , we see the left hand side is 0. Setting x = √6 , we see the right hand side is also 0; the equation is satisfied. The following example finds the equation of the tangent line to this function at this point. 3 3 3 3 Example 68: Using Implicit Differentiation to find a tangent line Find the equation of the line tangent to the curve of the implicitly defined function – (√6, 0). sin y + y 3 3 = 6 −x at the point 3 Solution: In Example 67 we found that 2 y ′ −3x = cos y + 3y 2 . – (2.6.6) We find the slope of the tangent line at the point (√6, 0) by substituting we have the slope as 3 y ′ 3 – 2 −3(√6) = 3 – √6 for x and 0 for y . Thus at the point 3 – (√6, 0) , − 3 − −3 √36 2 cos 0 + 3 ⋅ 0 = ≈ −9.91. (2.6.7) 1 Therefore the equation of the tangent line to the implicitly defined function sin y + y − 3 − 3 – y = −3 √36(x − √6) + 0 ≈ −9.91x + 18. 3 3 = 6 −x – at the point (√6, 0) is 3 (2.6.8) The curve and this tangent line are shown in Figure 2.20. 10/21/2020 2.6.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10290 Figure 2.20:The function sin y + y 3 2 = 6 −x – and its tangent line at the point (√6, 0). 3 Exercise 2.6.2 Find the equation of the line tangent to the curve of the implicitly defined function – (√7, 0). cos y + y 3 3 = 6 −x at the point 3 Hint: First check whether the point is on the curve. Answer Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page. This suggests a general method for implicit differentiation. For the steps below assume y is a function of x. 1. Take the derivative of each term in the equation. Treat the x terms like normal. When taking the derivatives of y terms, the usual rules apply except that, because of the Chain Rule, we need to multiply each term by y . 2. Get all the y terms on one side of the equal sign and put the remaining terms on the other side. 3. Factor out y ; solve for y by dividing. ′ ′ ′ ′ dy Practical Note: When working by hand, it may be beneficial to use the symbol confused for y or y . dx instead of y ′ , as the latter can be easily 1 Example 2.6.2 :: Using Implicit Differentiation Given the implicitly defined function y 3 2 +x y 4 = 1 + 2x , find y . ′ Solution: We will take the implicit derivatives term by term. The derivative of y is 3y 3 2 y ′ . The second term, x y , is a little tricky. It requires the Product Rule as it is the product of two functions of x: x and y . Its derivative is x (4y y ) + 2x y . The first part of this expression requires a y because we are taking the derivative of a y term. The second part does not require it because we are taking the derivative of x . 2 2 4 3 ′ 2 4 4 ′ 2 The derivative of the right hand side is easily found to be 2. In all, we get: 2 ′ 2 3 ′ 3 y y + 4 x y y + 2x y 4 = 2. (2.6.9) Move terms around so that the left side consists only of the y terms and the right side consists of all the other terms: ′ 2 ′ 2 3 3y y + 4x y y ′ 4 = 2 − 2x y . (2.6.10) Factor out y from the left side and solve to get ′ y ′ 2 − 2xy = 3y 10/21/2020 2 4 2 + 4x y 3 . 2.6.3 CC-BY-NC-SA (2.6.11) https://math.libretexts.org/@go/page/10290 To confirm the validity of our work, let's find the equation of a tangent line to this function at a point. It is easy to confirm that the point (0, 1) lies on the graph of this function. At this point, y = 2/3 . So the equation of the tangent line is y = 2/3(x − 0) + 1 . The function and its tangent line are graphed in Figure 2.21. ′ Figure 2.21:A graph of the implicitly defined function y 3 2 +x y 4 along with its tangent line at the point (0, 1). = 1 + 2x Notice how our function looks much different than other functions we have seen. For one, it fails the vertical line test. Such functions are important in many areas of mathematics, so developing tools to deal with them is also important. Example 2.6.3 : Using Implicit Differentiation Given the implicitly defined function sin(x 2 2 y )+y 3 , find y . ′ = x +y Solution: Differentiating term by term, we find the most difficulty in the first term. It requires both the Chain and Product Rules. d 2 2 2 d 2 2 ( sin(x y )) = cos(x y ) ⋅ dx 2 (x y ) dx 2 2 2 ′ 2 = cos(x y ) ⋅ (x (2y y ) + 2x y ) 2 ′ 2 2 2 = 2(x y y + x y ) cos(x y ). We leave the derivatives of the other terms to the reader. After taking the derivatives of both sides, we have 2 ′ 2 2 2 2 2(x y y + x y ) cos(x y ) + 3 y y ′ ′ = 1 +y . (2.6.12) We now have to be careful to properly solve for y , particularly because of the product on the left. It is best to multiply out the product. Doing this, we get ′ 2 2 2 ′ 2 x y cos(x y )y + 2x y 2 2 2 2 cos(x y ) + 3 y y ′ ′ = 1 +y . (2.6.13) From here we can safely move around terms to get the following: 2 2 2 ′ 2 ′ 2 x y cos(x y )y + 3 y y − y ′ = 1 − 2x y 2 2 2 cos(x y ). (2.6.14) . (2.6.15) Then we can solve for y to get ′ y ′ 1 − 2x y = 2 2 2 2 2 cos(x y ) 2 2 x y cos(x y ) + 3 y 2 −1 A graph of this implicit function is given in Figure 2.22. It is easy to verify that the points (0, 0), (0, 1) and (0, −1) all lie on the graph. We can find the slopes of the tangent lines at each of these points using our formula for y . ′ 10/21/2020 2.6.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10290 Figure 2.22:A graph of the implicitly defined function sin(x 2 2 y )+y 3 = x +y . At (0, 0), the slope is −1. At (0, 1), the slope is 1/2. At (0, −1), the slope is also 1/2. The tangent lines have been added to the graph of the function in Figure 2.23. Figure 2.23:A graph of the implicitly defined function sin(x 2 2 y )+y 3 = x +y and certain tangent lines. Quite a few "famous'' curves have equations that are given implicitly. We can use implicit differentiation to find the slope at various points on those curves. We investigate two such curves in the next examples. Example 2.6.3 : Finding slopes of tangent lines to a circle Find the slope of the tangent line to the circle x 2 +y 2 =1 – at the point (1/2, √3/2). Solution: Taking derivatives, we get 2x + 2y y ′ =0 . Solving for y gives: ′ y ′ −x = . (2.6.16) y This is a clever formula. Recall that the slope of the line through the origin and the point (x, y) on the circle will be y/x. We have found that the slope of the tangent line to the circle at that point is the opposite reciprocal of y/x, namely, −x/y. Hence these two lines are always perpendicular. – At the point (1/2, √3/2), we have the tangent line's slope as y ′ = −1/2 −1 = – – ≈ −0.577. √3/2 √3 (2.6.17) – A graph of the circle and its tangent line at (1/2, √3/2) is given in Figure 2.24, along with a thin dashed line from the origin that is perpendicular to the tangent line. (It turns out that all normal lines to a circle pass through the center of the circle.) 10/21/2020 2.6.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10290 – Figure 2.24:The unit circle with its tangent line at (1/2, √3/2). This section has shown how to find the derivatives of implicitly defined functions, whose graphs include a wide variety of interesting and unusual shapes. Implicit differentiation can also be used to further our understanding of "regular'' differentiation. One hole in our current understanding of derivatives is this: what is the derivative of the square root function? That is, d dx d − (√x ) = 1/2 (x ) =? (2.6.18) dx We allude to a possible solution, as we can write the square root function as a power function with a rational (or, fractional) power. We are then tempted to apply the Power Rule and obtain d 1/2 (x 1 ) = −1/2 x dx 2 1 = −. 2 √x (2.6.19) The trouble with this is that the Power Rule was initially defined only for positive integer powers, n > 0 . While we did not justify this at the time, generally the Power Rule is proved using something called the Binomial Theorem, which deals only with positive integers. The Quotient Rule allowed us to extend the Power Rule to negative integer powers. Implicit Differentiation allows us to extend the Power Rule to rational powers, as shown below. Let y = x , where m and n are integers with no common factors (so m = 2 and n = 5 is fine, but not). We can rewrite this explicit function implicitly as y = x . Now apply implicit differentiation. m/n n m =2 and n =4 is m m/n y =x y d (y n n m =x d n⋅y n−1 m ) = dx (x ) dx ⋅y ′ m−1 =m⋅x m−1 y ′ m x = n y m/n (now substitute x n−1 for y) m−1 m x = (apply lots of algebra) n (xm/n )n−1 m = (m−n)/n x n m = m/n−1 x . n The above derivation is the key to the proof extending the Power Rule to rational powers. Using limits, we can extend this once more to include all powers, including irrational (even transcendental!) powers, giving the following theorem. Theorem 2.6.1 :: Power Rule for Differentiation Let f (x) = x , where n ≠ 0 is a real number. Then f is a differentiable function, and f n 10/21/2020 2.6.6 CC-BY-NC-SA ′ n−1 (x) = n ⋅ x . https://math.libretexts.org/@go/page/10290 This theorem allows us to say the derivative of x is πx π π−1 . We now apply this final version of the Power Rule in the next example, the second investigation of a "famous'' curve. Example 2.6.4 :: Using the Power Rule Find the slope of x 2/3 +y 2/3 =8 at the point (8, 8). Solution: This is a particularly interesting curve called an astroid. It is the shape traced out by a point on the edge of a circle that is rolling around inside of a larger circle, as shown in Figure 2.25. Figure 2.25:An astroid, traced out by a point on the smaller circle as it rolls inside the larger circle. To find the slope of the astroid at the point (8, 8), we take the derivative implicitly. 2 −1/3 x 3 2 + y −1/3 y ′ =0 3 2 y −1/3 y ′ 2 −1/3 =− 3 x 3 −1/3 y ′ x =− y y ′ y =− −1/3 1/3 3 1/3 − − y . x = −√ x Plugging in x = 8 and y = 8 , we get a slope of −1. The astroid, with its tangent line at (8, 8), is shown in Figure 2.26. Figure 2.26:An astroid with a tangent line. Implicit Differentiation and the Second Derivative dy We can use implicit differentiation to find higher order derivatives. In theory, this is simple: first find , then take its derivative with respect to x. In practice, it is not hard, but it often requires a bit of algebra. We demonstrate this in an example. dx Example 2.6.5 :: Finding the second derivative 10/21/2020 2.6.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10290 Given x 2 +y 2 =1 , find d 2 y 2 dx =y ′′ . Solution: We found that y ′ dy = dx = −x/y in Example\PageIndex{3} . To find y , we apply implicit differentiation to y . ′′ y ′′ d ′ ′ = (y ) dx d x = (− dx ) (Now use the Quotient Rule.) y ′ y(1) − x(y ) =− y 2 replace y with −x/y: ′ y − x( y ′′ =− y −x y ) (2.6.20) 2 2 y +( y ′′ =− y y ′′ y 2 2 2 +x ) y . 2 (2.6.21) 2 +x =− y Now use the fact that y x (2.6.22) 3 = 1. y ′′ 1 =− y 3 . This is a particularly simple expression, and it is usable. We can see that Section 3.4, we will see how this relates to the shape of the graph. (2.6.23) y ′′ >0 when y < 0 and y ′′ <0 when y >0 . In Contributors Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution Noncommercial (BY-NC) License. http://www.apexcalculus.com/ Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 2.6.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10290 2.6E: Exercises Exercise 2.6E . 1 For the following exercises, use implicit differentiation to find 1) x 2 −y 2) 6x 2 3) x 2 2 dy dx . =4 + 3y 2 = 12 y = y −7 4) 3x 3 2 + 9x y 3 = 5x 5) xy − cos(xy) = 1 − − − − − 6) y √x + 4 = xy + 8 7) −xy − 2 = x 7 8) ysin(xy) = y 9) (xy ) 2 10) x 3 2 +2 + 3x = y y + xy 3 2 = −8 Answer 1. 2. 3. 4. 5. dy = dx dy = dx dy = dx x y −2x y −2xy 2 x −1 dy = dx dy x y − 3y 2x −y = dx x y y− 6. 7. 8. 9. dy 2 √x+ 4 = dx √x+4 −x dy 1+7y =− dx 7x dy y = dx dy dx 2 −3 2y( x2 −1) 2 dy 10. cos(xy) 2y−sin(xy)−xycosxy −2xy = dx 2 −3 x y−y = 3 x3 +3xy 2 Solution 1. x 2 −y 2 2x − 2y( =4 dy dx ) =0 dy 2y( dx dy dx = dy dx = 2. 6x 2 10/21/2020 ) = 2x 2x 2y x y + 3y 2 = 12 2.6E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10721 dy 12x + 6y( dy 6y( dx dy = dx dy = dx 3. x 2 dx ) = −12x −12x 6y −2x y y = y −7 dy 2 2xy + x ( dx dy 2 x ( dx dy 2 dy dy ) = −0 dx dy )− (x dx ) =0 = −2xy dx − 1) = −2xy −2xy = dx 2 x −1 4. 3x 3 2 9x + 9x y + 9y 2 2 3 = 5x + 18xy( dy 18xy( 2 ) = 3x dx 2 dy = dx dy = dx 3 x −9 y dy dx 2 ) = 15 x − 9y 2 2 18xy y x − 3y 2x 5. xy − cos(xy) = 1 dy x( x( dx dy dx dy ) + y + sin(xy)(x( dy ) + x( dx dx ) + y) = 0 )sin(xy) + ysin(xy) = −y dy x dx (1 + sin(xy)) = −y − ysin(xy) dy x dx (1 + sin(xy)) = −y(1 + sin(xy)) dy x dx = −y dy −y = dx 6. x − − − − − y √x + 4 = xy + 8 ( y(x + 4 ) 1/2) = xy + 8 1 2 dy ( y(x + 4 ) − 1/2) + dy dx dy dx − − − − − √x + 4 − x ( (x + 4 ) 1/2) = x dx dy dx dy dx +y = 0 y =y− 2 √x+4 − − − − − (√x + 4 − x) = y − y 2 √x+4 y y− dy dx 2 √x+ 4 = √x+4 −x 7. −xy − 2 = x 7 dy −x dx −y −0 = dy −x dx = 1 dy dx = 10/21/2020 7 1 7 1 7 +y +y −x 2.6E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10721 dy 1+7y =− dx 7x 8. ysin(xy) = y 2 +2 dy ycos(xy)(x dy + y) + dx dy dx dy dx dx dy sin(xy) − 2y 2 −y = y = 2 = −y cos(xy) cos(xy) 2 cos(xy) 2y−sin(xy)−xycosxy 9. (xy ) 2 + 3x = y 2 dy 2xy(x dy + y) + 3 = 2y dx dy 2 2x y + 2x y dx dy 2 2x y + 2x y dx − 2y 2 +3 = 0 dx dy − 2y +3 = 0 dx 2 dy −2xy = 2 dx dy 2 (2 x y − 2y) = −2x y dx 2 −3 −3 2 2y( x −1) 10. x 3 y + xy 2 3 + 3x y dx 3 (x 2 + 3x y 2 dy dx 2 2 2 + 3x y ) = −3 x y − y 3 x +3xy =0 = −3 x y − y dx −3 x y−y = 3 dy 2 dy dx = −8 +y dx dy 3 3 dy 3x y + x dy =0 −2y+sin(xy)+xycos(xy) dy dx dx dx 2 dx x dy sin(xy) − 2y (xycos(xy) + sin(xy) − 2y) = −y cos(xy) dx dy dx dy xycos(xy) + dx dy 2 dy dx dy sin(xy) = 2y xycos(xy) + y cos(xy) + dy dx dx 3 3 3 2 Exercise 2.6E . 2 For the following exercises, find the equation of the tangent line to the graph of the given equation at the indicated point. Use a calculator or computer software to graph the function and the tangent line. 1) x y − xy 2) x y 4 2 3) x y 4) x y 2 2 3 = −2, (−1, −1) + 5xy = 14, (2, 1) 2 + sin(πy) − 2 x 3 + 5x − 7 = − = 10, (2, −3) y, (1, 2) 4 5) xy + sin(x) = 1, ( π 2 , 0) Answer 1.y = − 5 4 x− 2. y = − 1 3. y = 1 11 2 4 x +2 π+12 4.y = − 10/21/2020 2 9 x− x+ 3π+38 π+12 15 2 2.6E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10721 5. y = 0 \ Solution 1. x 4 y − xy 3 = −2, (−1, −1) Take the derivative: 3 4 4x y + x dy dx −y 3 − 3x y dy 2 dx =0 dy Input x and y values to solve for 3 4 4(−1 ) (−1) + (−1 ) dy 4+ +1 +3 dx dy dx dy dx 3 − (−1 ) dx (this will be the slope of the tangent line): 2 − 3(−1)(−1 ) dy dx =0 =0 dy dx (1 + 3) = −4 − 1 dy dx −5 = 4 Now use this slope and the x and y values provided to solve the slope point intercept equation: −5 y = 4 x +b −5 −1 = 4 5 −1 = +b 4 5 −1 − (−1) + b =b 4 −9 b = 4 Therefore the equation of the tangent line is: 5 y =− 2. x 2 y 2 9 x− 4 4 + 5xy = 14, (2, 1) Take the derivative: 2x y 2 2 + 2y x dy dx dy + 5y + 5x dx =0 Input x and y values to solve for 2 2(2)(1 ) dy 4 +8 dx 2 + 2(1)(2 ) + 5 + 10 dy dx dy dx (this will be the slope of the tangent line): dy dy dx + 5(1) + 5(2) dx =0 =0 dy 18 = −9 dx dy dx −1 = 2 Now use this slope and the x and y values provided to solve the slope point intercept equation: y = 1 = −1 2 −1 2 x +b (2) + b 1 = −1 + b 2=b Therefore the equation of the tangent line is: y = −1 3. x y 10/21/2020 2 2 +2 2 + sin(πy) − 2 x = 10, (2, −3) 2.6E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10721 Take the derivative: y 2 dy + 2xy dx dy + πcos(πy) dx − 4x = 0 Input x and y values to solve for + 2(2)(−3) dy 9 − 12 dx = dx dy + πcos(−3π) dx − 4(2) = 0 dy 0 −π −12 − π( dy (this will be the slope of the tangent line): dy 2 (−3 ) dx dy dx dy dx dx −8 = 0 ) = −1 1 12+π Now use this slope and the x and y values provided to solve the slope point intercept equation: y = 1 12+π −3 = ( x +b 1 12+π )2 + b −3(12+π)−2 =b 12+π −38−3π 12+π =b Therefore the equation of the tangent line is: y = 1 π+12 x− 3π+38 π+12 Exercise 2.6E . 3 The graph of a folium of Descartes with equation 2x 3 + 2y 3 − 9xy = 0 is given in the following graph. a. Find the equation of the tangent line at the point (2, 1). Graph the tangent line along with the folium. b. Find the equation of the normal line to the tangent line in a. at the point (2, 1). Answer Under Construction Exercise 2.6E . 4 For the equation x 2 + 2xy − 3 y 2 = 0, a. Find the equation of the normal to the tangent line at the point (1, 1). b. At what other point does the normal line in a. intersect the graph of the equation? Answer a. y = −x + 2 10/21/2020 2.6E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10721 b. (3, −1) Exercise 2.6E . 5 Find all points on the graph of y For the equation x 2 + xy + y 2 3 2 − 27y = x =7 − 90 at which the tangent line is vertical. , a. Find the x-intercept(s). b.Find the slope of the tangent line(s) at the x-intercept(s). c. What does the value(s) in b. indicate about the tangent line(s)? Answer a. (±7√, 0) b. −2 c. They are parallel since the slope is the same at both intercepts. Exercise 2.6E . 6 Find the equation of the tangent line to the graph of the equation si n −1x −1 + si n y = π 6 at the point (0, 1 2 . ) Answer Under Construction Exercise 2.6E . 7 For the automatic number to work, you need to add the "AutoNum" template (preferably at the end) to the page. Answer y = −x + 1 Exercise 2.6E . 8 Find y' and y for x ′′ 2 + 6xy − 2 y 2 =3 . Answer Under Construction Exercise 2.6E . 9 The number of cell phones produced when x dollars is spent on labor and manufacturer can be modeled by the equation 60x y = 3240. 3/4 a. Find dy dx y dollars is spent on capital invested by a 1/4 and evaluate at the point (81, 16). b. Interpret the result of a. Answer a. −0.5926 b. When $81 is spent on labor and $16 is spent on capital, the amount spent on capital is decreasing by $0.5926 per $1 spent on labor. 10/21/2020 2.6E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10721 Exercise 2.6E . 10 The number of cars produced when x dollars is spent on labor and y dollars is spent on capital invested by a manufacturer can be modeled by the equation 30x y = 360. 1/3 2/3 (Both xand y are measured in thousands of dollars.) a. Find dy dx and evaluate at the point (27, 8). b. Interpret the result of a. Answer Under Construction Exercise 2.6E . 11 The volume of a right circular cone of radius x and height y is given by V is 85πcm . Find 3 dy dx = 1 3 2 πx y . Suppose that the volume of the cone when x = 4 and y = 16. Answer −8 Exercise 2.6E . 12 For the following exercises, consider a closed rectangular box with a square base with side x and height y . 1) Find an equation for the surface area of the rectangular box, S(x, y). 2) If the surface area of the rectangular box is 78 square feet, find dy dx when x = 3 feet and y = 5 feet. Answers to even numbered questions 2. −2.67 Exercise 2.6E . 13 For the following exercises, use implicit differentiation to determine y'. Does the answer agree with the formulas we have previously determined? 1) x = siny 2) x = cosy 3) x = tany Answers to even numbered questions 2. y' = − 10/21/2020 1 √1−x2 2.6E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10721 2.7: Derivatives of Inverse Trigonometric Functions In this section, we explore the relationship between the derivative of a function and the derivative of its inverse. For functions whose derivatives we already know, we can use this relationship to find derivatives of inverses without having to use the limit definition of the derivative. In particular, we will apply the formula for derivatives of inverse functions to trigonometric functions. This formula may also be used to extend the power rule to rational exponents. The Derivative of an Inverse Function We begin by considering a function and its inverse. If f (x) is both invertible and differentiable, it seems reasonable that the inverse of f (x) is also differentiable. Figure shows the relationship between a function f (x) and its inverse f (x). Look at the point (a, f (a)) on the graph of f (x) having a tangent line with a slope of −1 −1 −1 p (f − 1)'(a) = . (2.7.1) q This point corresponds to a point (f −1 (a), a) on the graph of f (x) having a tangent line with a slope of f '(f −1 q (a)) = . (2.7.2) p Thus, if f −1 (x) is differentiable at a , then it must be the case that (f −1 1 )'(a) = f '(f −1 . (a)) Figure 2.7.1 :The tangent lines of a function and its inverse are related; so, too, are the derivatives of these functions. We may also derive the formula for the derivative of the inverse by first recalling that x = f (f both sides of this equation (using the chain rule on the right), we obtain 1 = f '(f Solving for f −1 −1 (x))(f −1 −1 . Then by differentiating (x)) . )'(x)) , we obtain )'(x) (f −1 1 )'(x) = f '(f −1 . (x)) We summarize this result in the following theorem. Inverse Function Theorem Let f (x) be a function that is both invertible and differentiable. Let y = f f '(f (x)) ≠ 0 , −1 (x) be the inverse of f (x). For all x satisfying −1 dy d = dx 10/21/2020 (f dx −1 (x)) = (f −1 1 )'(x) = f '(f 2.7.1 CC-BY-NC-SA −1 . (2.7.3) (x)) https://math.libretexts.org/@go/page/10691 Alternatively, if y = g(x) is the inverse of f (x), then 1 ′ g (x) = . (2.7.4) f '(g(x)) Example 2.7.1 : Applying the Inverse Function Theorem x +2 Use the inverse function theorem to find the derivative of g(x) = . Compare the resulting derivative to that obtained x by differentiating the function directly. Solution x +2 The inverse of g(x) = x is f (x) = 2 x −1 . Since 1 g'(x) = , (2.7.5) f '(g(x)) begin by finding f '(x). Thus, −2 f '(x) = 2 −2 and f '(g(x)) = (x − 1) 2 −2 = 2 (g(x) − 1) x =− x +2 ( 2 2 − 1) . x Finally, 1 2 g'(x) = =− f '(g(x)) x2 . We can verify that this is the correct derivative by applying the quotient rule to g(x) to obtain 2 g'(x) = − 2 . x Exercise 2.7.1 Use the inverse function theorem to find the derivative of g(x) 1 g(x) = x +2 . Compare the result obtained by differentiating directly. Hint Use the preceding example as a guide. Answer 1 g'(x) = − 2 (x + 2) Example 2.7.2 : Applying the Inverse Function Theorem − Use the inverse function theorem to find the derivative of g(x) = √x . 3 Solution The function g(x) = √− x is the inverse of the function f (x) = x . Since g'(x) = 3 3 1 , begin by finding f '(x). Thus, f '(g(x)) 3 f '(x) = 3x (2.7.6) and 2/3 3 − 2 f '(g(x)) = 3(√x ) = 3 x 10/21/2020 2.7.2 CC-BY-NC-SA (2.7.7) https://math.libretexts.org/@go/page/10691 Finally, 1 g'(x) = 1 2/3 = −2/3 x . (2.7.8) 3 3x Exercise 2.7.2 Find the derivative of g(x) = √− x by applying the inverse function theorem. 5 Hint g(x) is the inverse of f (x) = x . 5 Answer 1 g(x) = −4/5 x 5 Theorem 2.7.1 Let f be a function. Then f is one to one on an interval I if f is increasing on I or f is decreasing on I . Proof Add proof here and it will automatically be hidden if you have a "AutoNum" template active on the page. Example 2.7.3 : 3 x Show that f (x) = 2 has an inverse and find (f −1 has an inverse and find (f −1 ′ ) (2). 1 +x Exercise 2.7.3 3 2x Show that f (x) = 2 ′ ) (2). 1 +x Answer Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page. Example 2.7.4 : Show that f (x) = x 1 +x has an inverse and find (f −1 ′ ) (x). From the previous example, we see that we can use the inverse function theorem to extend the power rule to exponents of the form 1 n , where n is a positive integer. This extension will ultimately allow us to differentiate q x , where q is any rational number. Extending the Power Rule to Rational Exponents The power rule may be extended to rational exponents. That is, if n is a positive integer, then d 1/n (x 1 ) = dx (1/n)−1 x . (2.7.9) n Also, if n is a positive integer and m is an arbitrary integer, then d m/n (x dx 10/21/2020 m ) = (m/n)−1 x n . 2.7.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10691 Proof The function 1/n g(x) = x is the inverse of the function n f (x) = x 1 . Since g'(x) = , begin by finding . f '(x) f '(g(x)) Thus, n−1 f '(x) = nx and f '(g(x)) = n(x 1/n n−1 ) (n−1)/n = nx . Finally, 1 1 g'(x) = x m/n we must rewrite it as (x 1/n d m/n (x d ) = dx m ) (1/n)−1 x n . and apply the chain rule. Thus, m ) 1/n ((x 1 = n nx To differentiate x (1−n)/n = (n−1)/n 1/n ) = m(x 1 m−1 ) ⋅ dx (1/n)−1 x n m = (m/n)−1 x . (2.7.10) n □ Example 2.7.3 : Applying the Power Rule to a Rational Power Find the equation of the line tangent to the graph of y = x at x = 8 . 2/3 Solution First find dy dx and evaluate it at x = 8 . Since dy 2 = dx the slope of the tangent line to the graph at x = 8 is −1/3 x 3 1 3 and dy 1 ∣x=8 = dx 3 . Substituting x = 8 into the original function, we obtain y = 4 . Thus, the tangent line passes through the point Substituting into the point-slope formula for a line, we obtain the tangent line 1 y = 4 x+ 3 3 . (8, 4) . Exercise 2.7.3 − − − − − Find the derivative of s(t) = √2t + 1 . Hint Use the chain rule. Answer −1/2 s'(t) = (2t + 1) Derivatives of Inverse Trigonometric Functions We now turn our attention to finding derivatives of inverse trigonometric functions. These derivatives will prove invaluable in the study of integration later in this text. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are actually algebraic functions. Previously, derivatives of algebraic functions have proven to be algebraic functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Here, for the first time, we see that the derivative of a function need not be of the same type as the original function. Example 2.7.4 : Derivative of the Inverse Sine Function Use the inverse function theorem to find the derivative of g(x) = si n −1 x . Solution 10/21/2020 2.7.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10691 Since for x in the interval [− π π , 2 ], f (x) = sinx is the inverse of g(x) = si n −1 , begin by finding f '(x). Since x 2 f '(x) = cosx and f '(g(x)) = cos(si n −1 − −−− − 2 x) = √1 − x , we see that d g'(x) = −1 (si n 1 1 x) = dx = f '(g(x)) − −−− − √1 − x2 . Analysis To see that π − −1 cos(si n π ≤θ ≤ 2 2 − −−− − x) = √1 − x2 , consider the following argument. Set . We begin by considering the case where π 0 <θ < 2 . Since θ −1 si n x =θ . In this case, sinθ = x where is an acute angle, we may construct a right triangle having acute angle θ , a hypotenuse of length 1 and the side opposite angle θ having length x. From the − −−− − Pythagorean theorem, the side adjacent to angle θ has length √1 − x . This triangle is shown in Figure. Using the triangle, − −−− − we see that cos(si n x) = cosθ = √1 − x . 2 −1 2 Figure 2.7.2 : Using a right triangle having acute angle θ , a hypotenuse of length 1, and the side opposite angle θ having − −−− − length x, we can see that cos(si n x) = cosθ = √1 − x . −1 In the case where − π <θ <0 2 , we make the observation that 0 < −θ < −1 cos(si n Now if θ = π 2 or θ = − π ,x =1 2 π 2 and hence − −−− − x) = cosθ = cos(−θ) = √1 − x2 . − −−− − or x = −1 , and since in either case cosθ = 0 and √1 − x 2 −1 cos(si n Consequently, in all cases, 2 − −−− − −1 2 cos(si n x) = √1 − x − −−− − 2 x) = cosθ = √1 − x =0 , we have . . Example 2.7.5 : Applying the Chain Rule to the Inverse Sine Function Apply the chain rule to the formula derived in Example to find the derivative of h(x) = si n find the derivative of h(x) = si n (2x ). −1 −1 (g(x)) and use this result to 3 Solution Applying the chain rule to h(x) = si n −1 (g(x)) , we have h'(x) = Now let g(x) = 2x 3 , 1 − − − − − − − − − g'(x) 2 √1 − (g(x)) . so g'(x) = 6x. Substituting into the previous result, we obtain h'(x) = 1 6x − −−−− − ⋅ 6x = − −−−− − √1 − 4x6 √1 − 4x6 Exercise 2.7.4 Use the inverse function theorem to find the derivative of g(x) = tan −1 x . Hint The inverse of g(x) is f (x) = tanx . Use Example as a guide. Answer 10/21/2020 2.7.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10691 1 g'(x) = 2 1 +x The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. These formulas are provided in the following theorem. Derivatives of Inverse Trigonometric Functions d −1 si n 1 x = − − − − − − − 2 √1 − (x) x = − − − − − − − 2 √1 − (x) dx d −1 cos −1 dx d −1 tan 1 x = dx d −1 cot −1 x = dx d 2 1 + (x) −1 sec 1 x = − − − − − − − |x| √(x )2 − 1 x = − − − − − − − 2 |x| √(x ) − 1 dx d 1 + (x)2 −1 csc −1 dx Example 2.7.6 : Applying Differentiation Formulas to an Inverse Tangent Function Find the derivative of f (x) = tan −1 2 (x ). Solution 2 Letg(x) = x , so g'(x) = 2x. Substituting into Equation, we obtain 1 f '(x) = 2 ⋅ (2x). 2 1 + (x ) Simplifying, we have 2x f '(x) = 4 . 1 +x Example 2.7.7 : Applying Differentiation Formulas to an Inverse Sine Function Find the derivative of h(x) = x 2 −1 si n x. Solution By applying the product rule, we have −1 h'(x) = 2xsi n 1 x+ − −−− − √1 − x2 ⋅ x2 Exercise 2.7.5 Find the derivative of h(x) = cos −1 (3x − 1). Hint Use Equation. with g(x) = 3x − 1 Answer 10/21/2020 2.7.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10691 −3 h'(x) = − −−−−− − √6x − 9x2 Example 2.7.8 : Applying the Inverse Tangent Function The position of a particle at time t is given by s(t) = tan −1 1 ( ) t for t ≥ 1 2 . Find the velocity of the particle at time t = 1 . Solution Begin by differentiating s(t) in order to find v(t) .Thus, 1 −1 v(t) = s'(t) = ⋅ 1 1 +( 2 . t 2 ) t Simplifying, we have 1 v(t) = − 2 t Thus, v(1) = − . +1 1 . 2 Exercise 2.7.6 Find the equation of the line tangent to the graph of f (x) = si n −1 x at x = 0. Hint f '(0) is the slope of the tangent line. Answer y =x Key Concepts The inverse function theorem allows us to compute derivatives of inverse functions without using the limit definition of the derivative. We can use the inverse function theorem to develop differentiation formulas for the inverse trigonometric functions. Key Equations Inverse function theorem 1 (f − 1)'(x) = f '(f −1 whenever f '(f −1 (x)) ≠ 0 and f (x) is differentiable. (x)) Power rule with rational exponents d m/n (x m ) = dx (m/n)−1 x . n Derivative of inverse sine function d −1 si n 1 x = dx − − − − − − − 2 √1 − (x) Derivative of inverse cosine function d −1 cos dx −1 x = − − − − − − − 2 √1 − (x) Derivative of inverse tangent function 10/21/2020 2.7.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10691 d −1 tan 1 x = dx 2 1 + (x) Derivative of inverse cotangent function d −1 cot −1 x = dx 1 + (x)2 Derivative of inverse secant function d −1 sec 1 x = dx − − − − − − − 2 |x| √(x ) − 1 Derivative of inverse cosecant function d −1 csc dx −1 x = − − − − − − − 2 |x| √(x ) − 1 Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 2.7.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10691 2.7E: Exercises Exercise 2.7E . 1 For the following exercises, use the graph of y = f (x) to a. sketch the graph of y = f b. use part a. to estimate (f −1 −1 (x) , and . )'(1) 1) 2) Answer a. b. (f −1 )'(1) 2 3) 10/21/2020 2.7E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10858 4) Answer a. b. (f −1 – )'(1) − 1/ √3 Exercise 2.7E . 2 For the following exercises, use the functions y = f (x) to find a. df dx at x = a and b. \(x=f^{−1}(y). c. Then use part b. to find df −1 dy at y = f (a). 1) f (x) = 6x − 1, x = −2 2) f (x) = 2x 3 − 3, x = 1 Answer 10/21/2020 2.7E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10858 a. 6 b. x = f c. −1 (y) = ( y+3 2 1/3 ) 1 6 3) f (x) = 9 − x 2 , 0 ≤ x ≤ 3, x = 2 4) f (x) = sinx, x = 0 Answer a. 1 b. x = f −1 −1 (y) = si n y c. 1 Exercise 2.7E . 3 For each of the following functions, find (f 1) f (x) = x + 3x + 2, x ≥ −1, a = 2 2) f (x) = x + 2x + 3, a = 0 2 3 −1 )'(a) . Answer 1 5 3) f (x) = x + √− x, a = 2 4) f (x) = x − 2 x , x < 0, a = 1 Answer 1 3 5) f (x) = x + sinx, a = 0 6) f (x) = tanx + 3x 2 ,a =0 Answer 1 Exercise 2.7E . 4 For each of the given functions y = f (x), a. find the slope of the tangent line to its inverse function f b. find the equation of the tangent line to the graph of f 1) f (x) = 4 2 1+x −1 −1 at the indicated point P , and at the indicated point. , P (2, 1) − − − − − 2) f (x) = √x − 4 , P (2, 8) Answer a. 4 b. y = 4x 3) f (x) = (x 3 10/21/2020 4 + 1 ) , P (16, 1) 2.7E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10858 4) f (x) = −x 3 − x + 2, P (−8, 2) Answer a. − 1 96 b. y = − 1 x+ 13 5) f (x) = x 5 3 + 3x 18 13 − 4x − 8, P (−8, 1) Exercise 2.7E . 5 For the following exercises, find 1) y = si n −1 dy dx for the given function. 2 (x ) Answer 2x √1−x4 2) y = cos − (√x ) 3) y = sec ( −1 −1 1 x ) Answer −1 √1−x2 − − − − − − −1 x 4) y = √csc 5) y = (1 + tan −1 3 x) Answer −1 3(1+tan 2 x) 2 1+x 6) y = cos −1 7) y = −1 (2x) ⋅ si n (2x) 1 −1 tan (x) Answer −1 2 −1 (1+x )(tan 2 x) 8) y = sec (−x) 9) y = cot − −−− − √4 − x2 −1 −1 Answer x (5−x2 ) √4−x2 10) y = x ⋅ csc −1 x Exercise 2.7E . 6 For the following exercises, use the given values to find (f 1) f (π) = 0, f ′ −1 )'(a) . (π) = −1, a = 0 Answer 10/21/2020 2.7E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10858 −1 2) f (6) = 2, f '(6) = 3) f ( 1 3 ′ 1 ) = −8, f ( 3 1 3 ,a =2 ) = 2, a = −8 Answer 1 2 – 4) f (√3) = 1 2 – ′ , f (√3) = 5) f (1) = −3, f ′ 2 3 ,a = 1 2 (1) = 10, a = −3 Answer 1 10 6) f (1) = 0, f ′ (1) = −2, a = 0 Exercise 2.7E . 7 The position of a moving hockey puck after t seconds is s(t) = tan −1 t where s is in meters. a. Find the velocity of the hockey puck at any time t . b. Find the acceleration of the puck at any time t . c. Evaluate a. and b. for t = 2, 4 ,and 6 seconds. d. What conclusion can be drawn from the results in c.? Answer a. v(t) = b. a(t) = 1 2 1+t −2t 2 2 (1+t ) c. (a)0.2, 0.06, 0.03; (b) − 0.16, −0.028, −0.0088 d. The hockey puck is decelerating/slowing down at 2, 4, and 6 seconds. Exercise 2.7E . 8 A building that is 225 feet tall casts a shadow of various lengths x as the day goes by. An angle of elevation θ is formed by lines from the top and bottom of the building to the tip of the shadow, as seen in the following figure. Find the rate of change of the angle of elevation when x = 272 feet. dθ dx Answer Under Construction 10/21/2020 2.7E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10858 Exercise 2.7E . 9 A pole stands 75 feet tall. An angle θ is formed when wires of various lengths of x feet are attached from the ground to the top of the pole, as shown in the following figure. Find the rate of change of the angle when a wire of length 90 feet is attached. dθ dx Answer radians per foot −0.0168 Exercise 2.7E . 10 A television camera at ground level is 2000 feet away from the launching pad of a space rocket that is set to take off vertically, as seen in the following figure. The angle of elevation of the camera can be found by θ = tan ( ) , where x is the height of the rocket. Find the rate of change of the angle of elevation after launch when the camera and the rocket are 5000 feet apart. −1 x 2000 Answer Under Construction Exercise 2.7E . 11 A local movie theater with a 30-foot-high screen that is 10 feet above a person’s eye level when seated has a viewing angle θ (in radians) given by θ = cot − cot , −1 x 40 −1 x 10 where x is the distance in feet away from the movie screen that the person is sitting, as shown in the following figure. 10/21/2020 2.7E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10858 a. Find dθ dx . b. Evaluate dθ dx for x = 5, 10, 15, and 20. c. Interpret the results in b.. d. Evaluate dθ dx for x = 25, 30, 35, and 40 e. Interpret the results in d. At what distance x should the person stand to maximize his or her viewing angle? Answer a. b. dθ = dx 18 325 , 10 100+x2 9 340 , − 42 4745 40 1600+x2 ,0 c. As a person moves farther away from the screen, the viewing angle is increasing, which implies that as he or she moves farther away, his or her screen vision is widening. d. − 54 12905 ,− 3 500 ,− 198 29945 ,− 9 1360 e. As the person moves beyond 20 feet from the screen, the viewing angle is decreasing. The optimal distance the person should stand for maximizing the viewing angle is 20 feet. 10/21/2020 2.7E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10858 2.8: Derivatives of Exponential and Logarithmic functions In this section, we explore derivatives of exponential and logarithmic functions. Exponential functions play an important role in modeling population growth and the decay of radioactive materials. Logarithmic functions can help rescale large quantities and are particularly helpful for rewriting complicated expressions. The graph of E(x) = e together with the line y = x + 1 are shown in Figure. This line is tangent to the graph of E(x) = e at x = 0 . x x Figure 2.8.2 : The tangent line to E(x) = e at x = 0 has slope 1. x Now that we have laid out our basic assumptions, we begin our investigation by exploring the derivative of B(x) = b Recall that we have assumed that B'(0) exists. By applying the limit definition to the derivative we conclude that x 0+h b B'(0) = li mh→0 0 −b h h = li mh→0 b −1 h ,b >0 . . Turning to B'(x), we obtain the following. x+h B'(x) = limh→0 b x = limh→0 x −b h h x b b −b h x h b ( b −1) = limh→0 h h x = b li mh→0 x = b B'(0) b −1 h Apply the limit definition of the derivative. Note thatbx+h=bxbh. Factor outbx. Apply a property of limits. Use B'(0) = li m 0+h b h→0 0 −b h h = li mh→0 We see that on the basis of the assumption that but its derivative is b −1 h x B(x) = b . is differentiable at 0, B(x) is not only differentiable everywhere, x B'(x) = b B'(0). For x Thus, we have will be derived later.) E(x) = e , E'(0) = 1. x B(x) = b , b > 0, E'(x) = e x . (The value of B'(0) for an arbitrary function of the form Derivative of the Natural Exponential Function Let E(x) = e be the natural exponential function. Then x x E'(x) = e . In general, d dx (e g(x) ) =e g(x) g'(x) . Example 2.8.1 : Derivative of an Exponential Function Find the derivative of f (x) = e tan(2x) . Solution: 10/21/2020 2.8.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10291 Using the derivative formula and the chain rule, f '(x) = e tan(2x) d dx (tan(2x)) = e tan(2x) 2 sec (2x) ⋅ 2 . Example 2.8.2 : Combining Differentiation Rules Find the derivative of y = e 2 x . x Solution Use the derivative of the natural exponential function, the quotient rule, and the chain rule. (e 2 x ⋅2)x⋅x−1⋅ e y' = 2 x 2 x e = 2 x 2 (2 x −1) x2 Apply the quotient rule. Simplify. Exercise 2.8.1 Find the derivative of h(x) = xe . 2x Hint Don’t forget to use the product rule. Answer h'(x) = e 2x + 2x e 2x Example 2.8.3 : Applying the Natural Exponential Function A colony of mosquitoes has an initial population of 1000. After t days, the population is given by A(t) = 1000e that the ratio of the rate of change of the population, A'(t), to the population, A(t) is constant. 0.3t . Show Solution First find A'(t). By using the chain rule, we have A'(t) = 300e to the population is given by A'(t) = 300e 0.3t . Thus, the ratio of the rate of change of the population 0.3t 1000e 0.3t = 0.3. The ratio of the rate of change of the population to the population is the constant 0.3. Exercise 2.8.2 If A(t) = 1000e describes the mosquito population after t days, as in the preceding example, what is the rate of change of A(t) after 4 days? 0.3t Hint Find A'(4). Answer 996 Derivative of the Logarithmic Function Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function. Definition: The Derivative of the Natural Logarithmic Function If x > 0 and y = lnx ,then 10/21/2020 2.8.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10291 dy 1 = dx x . More generally, let g(x) be a differentiable function. For all values of h(x) = ln(g(x)) is given by 1 h'(x) = g(x) x for which g'(x) > 0 , the derivative of g'(x). Proof If x > 0 and y = lnx , then e y = x. Differentiating both sides of this equation results in the equation e Solving for dy dx y dy dx = 1. yields dy dx 1 = e y . Finally, we substitute x = e to obtain y dy dx = 1 x . We may also derive this result by applying the inverse function theorem, as follows. Since y = g(x) = lnx is the inverse of f (x) = e , by applying the inverse function theorem we have x dy dx = 1 f '(g(x)) = 1 elnx = 1 x . Using this result and applying the chain rule to h(x) = ln(g(x)) yields h'(x) = 1 g(x) g'(x) . □ The graph of y = lnx and its derivative dy dx = 1 x are shown in Figure. Figure 2.8.3 : The function y = lnx is increasing on (0, +∞). Its derivative y ′ = 1 x is greater than zero on (0, +∞) Example 2.8.4 :Taking a Derivative of a Natural Logarithm Find the derivative of f (x) = ln(x 3 + 3x − 4) . Solution Use Equation directly. f '(x) = 1 3 x +3x−4 2 = 3 x +3 3 x +3x−4 2 ⋅ (3 x + 3) Use g(x) = x 3 + 3x − 4 in h'(x) = 1 g(x) g'(x) . Rewrite. Example 2.8.5 :Using Properties of Logarithms in a Derivative Find the derivative of f (x) = ln( 2 x sinx 2x+1 ) . Solution 10/21/2020 2.8.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10291 At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler. 2 x sinx f (x) = ln( ) = 2lnx + ln(sinx) − ln(2x + 1) 2x+1 f '(x) = 2 + cotx − x Apply properties of logarithms. Apply sum rule and h'(x) = 2 2x+1 1 g(x) g'(x) . Exercise 2.8.3 Differentiate: f (x) = ln(3x + 2) . 5 Hint Use a property of logarithms to simplify before taking the derivative. Answer f '(x) = 15 3x+2 Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of y = log y =b for b > 0, b ≠ 1 . bx and x Derivatives of General Exponential and Logarithmic Functions Let b > 0, b ≠ 1, and let g(x) be a differentiable function. i. If, y = log bx , then dy dx More generally, if h(x) = log b (g(x)) = x , . , then for all values of x for which g(x) > 0 , h'(x) = ii. If y = b 1 xlnb g'(x) g(x)lnb . then dy dx More generally, if h(x) = b g(x) , x = b lnb . then g(x) h'(x) = b g ′′ (x)lnb Proof If y = log x, then b = x. It follows that ln(b ) = lnx . Thus ylnb = lnx. Solving for y , we have y = and keeping in mind that lnb is a constant, we see that y y b dy dx = 1 xlnb lnx lnb . Differentiating . The derivative in Equation now follows from the chain rule. If x y =b 1 dy y dx . then = lnb lny = xlnb. . Solving for dy dx Using implicit differentiation, again keeping in mind that lnb is constant, it follows that and substituting y = b , we see that x dy dx x = ylnb = b lnb . The more general derivative (Equation) follows from the chain rule. □ Example 2.8.6 :Applying Derivative Formulas Find the derivative of h(x) = 10/21/2020 x 3 x 3 +2 . 2.8.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10291 Solution Use the quotient rule and Note. x h'(x) = x x x 3 ln3( 3 +2)−3 ln3( 3 ) 2 x Apply the quotient rule. ( 3 +2 ) x = 2⋅ 3 ln3 2 (3x+2) Simplify. Example 2.8.7 : Finding the Slope of a Tangent Line Find the slope of the line tangent to the graph of y = log 2 (3x + 1) at x = 1 . Solution dy To find the slope, we must evaluate dx at x = 1 . Using Equation, we see that dy dx = 3 ln2(3x+1) . By evaluating the derivative at x = 1 , we see that the tangent line has slope dy dx ∣x=1 = 3 4ln2 = . 3 ln16 Exercise 2.8.4 Find the slope for the line tangent to y = 3 at x = 2. x Hint Evaluate the derivative at x = 2. Answer 9ln(3) Logarithmic Differentiation At this point, we can take derivatives of functions of the form y = (g(x)) for certain values of n , as well as functions of the form y = b , where b > 0 and b ≠ 1 . Unfortunately, we still do not know the derivatives of functions such as y = x or y = x . These functions require a technique called logarithmic differentiation, which allows us to differentiate any function of the form h(x) = g(x) . It can also be used to convert a very complex differentiation problem into a simpler one, such as n g(x) x π f (x) finding the derivative of y = x√2x+1 x 3 e sin x . We outline this technique in the following problem-solving strategy. Problem-Solving Strategy: Using Logarithmic Differentiation 1. To differentiate y = h(x) using logarithmic differentiation, take the natural logarithm of both sides of the equation to obtain lny = ln(h(x)). 2. Use properties of logarithms to expand ln(h(x)) as much as possible. 3. Differentiate both sides of the equation. On the left we will have 4. Multiply both sides of the equation by y to solve for 5. Replace y by h(x). dy dx 1 dy y dx . . Example 2.8.8 : Using Logarithmic Differentiation Find the derivative of y = (2x 4 tanx + 1) . Solution Use logarithmic differentiation to find this derivative. 4 lny = ln(2 x 10/21/2020 tanx + 1) Step 1. Take the natural logarithm of both sides. 2.8.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10291 4 lny = tanxln(2 x 1 dy y dx 2 dy dx 4 = sec xln(2 x dy dx Step 2. Expand using properties of logarithms. + 1) 3 + 1) + 8x 3 8x 2 = y ⋅ (sec xln(2x4 + 1) + 4 = (2 x Step 3. Differentiate both sides. Use theproduct rule on the right. ⋅ tanx 4 2 x +1 tanx + 1) 4 2 x +1 2 4 (sec xln(2 x Step 4. Multiply byyon both sides. ⋅ tanx) 3 + 1) + 8x 2 x4 +1 ⋅ tanx) Step 5. Substitute y = (2x 4 tanx + 1) . Example 2.8.9 : Extending the Power Rule x√2x+1 Find the derivative of y = ex sin3 x . Solution This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter. x √2x+1 lny = ln lny = lnx + 1 dy y dx dy dx dy dx 1 2 Step 2. Expand using properties of logarithms. ln(2x + 1) − xlne − 3lnsinx = = y( 1 x 1 x + + x √2x+1 = Step 1. Take the natural logarithm of both sides. ex si n3 x ex si n3 x ( 1 x 1 −1 −3 2x+1 1 Step 4. Multiply by y on both sides. − 1 − 3cotx) 2x+1 + Step 3. Differentiate both sides. cosx sinx 1 2x+1 Step 5. Substitute y − 1 − 3cotx) x √2x+1 = ex si n3 x . Exercise 2.8.5 Use logarithmic differentiation to find the derivative of y = x . x Hint Follow the problem solving strategy. Answer Solution: dy dx x = x (1 + lnx) Exercise 2.8.6 Find the derivative of y = (tanx) . π Hint Use the result from Example. Answer π−1 y' = π(tanx ) 2 sec x Key Concepts On the basis of the assumption that the exponential function y = b , b > 0 is continuous everywhere and differentiable at 0, this function is differentiable everywhere and there is a formula for its derivative. We can use a formula to find the derivative of y = lnx , and the relationship log x = allows us to extend our differentiation formulas to include logarithms with arbitrary bases. Logarithmic differentiation allows us to differentiate functions of the form y = g(x) or very complex functions by taking the natural logarithm of both sides and exploiting the properties of logarithms before differentiating. x lnx b lnb f (x) Key Equations Derivative of the natural exponential function 10/21/2020 2.8.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10291 d dx (e g(x) ) =e g(x) g'(x) Derivative of the natural logarithmic function d dx (lng(x)) = 1 g(x) g'(x) Derivative of the general exponential function d dx g(x) (b g(x) ) =b g'(x)lnb Derivative of the general logarithmic function d dx g'(x) (logb g(x)) = g(x)lnb Glossary logarithmic differentiation is a technique that allows us to differentiate a function by first taking the natural logarithm of both sides of an equation, applying properties of logarithms to simplify the equation, and differentiating implicitly Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 2.8.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10291 2.8E: Exercises Exercise 2.8E . 1 For the following exercises, find f '(x) for each function. 1) f (x) = x 2 e x Answer 2x e x 2 +x e 2) f (x) = e x −x x 3) f (x) = e 3 x lnx Answer 3 e x 2 2 lnx(3 x lnx + x ) − − − − − − − 2x + 2x 4) f (x) = √e 5) f (x) = x −x x −x e −e e +e Answer 4 (e x +e −x 2 ) x 6) f (x) = 10 ln10 7) f (x) = 2 4x 2 + 4x Answer 4x+2 2 ⋅ ln2 + 8x 8) f (x) = 3 sin3x 9) f (x) = x π x ⋅π Answer π−1 πx x ⋅π π 10) f (x) = ln(4x 3 11) x +x ⋅ π lnπ + x) −−−− − f (x) = ln√5x − 7 Answer 5 2(5x−7) 12) f (x) = x 2 ln9x 13) f (x) = log(secx) Answer tanx ln10 14) f (x) = log 4 7 (6 x 10/21/2020 5 + 3) 2.8E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10857 15) f (x) = 2 x 2 x −4 ⋅ log3 7 Answer x 2 2 x −4 ⋅ ln2 ⋅ log3 7 x +2 ⋅ 2xln7 ln3 Exercise 2.8E . 2 For the following exercises, use logarithmic differentiation to find dy dx . 1) y = x √x 2) y = (sin2x) 4x Answer 4x (sin2x ) [4 ⋅ ln(sin2x) + 8x ⋅ cot2x] 3) y = (lnx) lnx 4) y = x lo g2 x Answer: lo g2 x x ⋅ 5) y = (x 2 2lnx xln2 lnx − 1) 6) y = x cotx Answer: cotx 2 x 7) y = ⋅ [−csc x ⋅ lnx + cotx x ] x+11 3 √x2 −4 8) y = x −1/2 2 (x 2/3 + 3) 4 (3x − 4 ) Answer −1/2 x 2/3 (x2 + 3 ) 4 (3x − 4 ) ⋅[ −1 2x + 4x 3( x2 +3) + 12 3x−4 ] Exercise 2.8E . 3 Find an equation of the tangent line to the graph of f (x) = 4xe x = −1. 2 ( x −1) at the point where Graph both the function and the tangent line. Answer Under Construction Exercise 2.8E . 4 Find the equation of the line that is normal to the graph of f (x) = x ⋅ 5 at the point where x = 1 . Graph both the function and the normal line. x Answer y = 10/21/2020 −1 5+5ln5 x + (5 + 1 5+5ln5 ) 2.8E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10857 Exercise 2.8E . 5 Find the equation of the tangent line to the graph of curve and the tangent line. (Hint: Use implicit differentiation to find dy dx 3 x − xlny + y 3 = 2x + 5 at the point where x =2 . Graph both the .) Answer Under Construction Exercise 2.8E . 6 Consider the function y = x 1/x for x > 0. a. Determine the points on the graph where the tangent line is horizontal. b. Determine the points on the graph where y' > 0 and those where y' < 0 . Solution: a. x = e 2.718 b. (e, ∞), (0, e) Answer a. x = e 2.718 b. (e, ∞), (0, e) Exercise 2.8E . 7 The formula I (t) = sint et is the formula for a decaying alternating current. a. Complete the following table with the appropriate values. t sint e t 0 (i) π/2 (ii) π (iii) 3π/2 (vi) 2π (v) 10/21/2020 2.8E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10857 2π (vi) 3π (vii) 7π/2 (viii) 4π (ix) b. Using only the values in the table, determine where the tangent line to the graph of (I(t)\) is horizontal. Answer Under Construction Exercise 2.8E . 8 The population of Toledo, Ohio, in 2000 was approximately 500,000. Assume the population is increasing at a rate of 5% per year. a. Write the exponential function that relates the total population as a function of t . b. Use a. to determine the rate at which the population is increasing in t years. c. Use b. to determine the rate at which the population is increasing in 10 years Answer a. P individuals t = 500, 000(1.05) b. P '(t) = 24395 ⋅ (1.05) individuals per year t c. 39, 737individuals per year Exercise 2.8E . 9 An isotope of the element erbium has a half-life of approximately 12 hours. Initially there are 9 grams of the isotope present. a. Write the exponential function that relates the amount of substance remaining as a function of t , measured in hours. b. Use a. to determine the rate at which the substance is decaying in t hours. c. Use b. to determine the rate of decay at t = 4 hours. Answer Under Construction Exercise 2.8E . 10 The number of cases of influenza in New York City from the beginning of 1960 to the beginning of 1961 is modeled by the function 2 N (t) = 5.3 e 0.093 t −0.87t , (0 ≤ t ≤ 4) , where N (t) gives the number of cases (in thousands) and t is measured in years, with t = 0 corresponding to the beginning of 1960. a. Show work that evaluates York City. 10/21/2020 N (0) and N (4). Briefly describe what these values indicate about the disease in New 2.8E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10857 b. Show work that evaluates United States. N '(0) and N '(3). Briefly describe what these values indicate about the disease in the Answer a. At the beginning of 1960 there were 5.3 thousand cases of the disease in New York City. At the beginning of 1963 there were approximately 723 cases of the disease in the United States. b. At the beginning of 1960 the number of cases of the disease was decreasing at rate of −4.611 thousand per year; at the beginning of 1963, the number of cases of the disease was decreasing at a rate of −0.2808thousand per year. Exercise 2.8E . 11 The relative rate of change of a differentiable function y = f (x) is given by Gompertz growth function, given by P (x) = ae −b⋅e −cx 100⋅f '(x) f (x) One model for population growth is a where a, b, and c are constants. a. Find the relative rate of change formula for the generic Gompertz function. b. Use a. to find the relative rate of change of a population in x = 20 months when a = 204, b = 0.0198, and c = 0.15. c. Briefly interpret what the result of b. means. Answer Under Construction Exercise 2.8E . 12 For the following exercises, use the population of New York City from 1790 to 1860, given in the following table. Year since 1790 Population 0 33,131 10 60,515 20 96,373 30 123,706 40 202,300 50 312,710 60 515,547 70 813,669 New York City Population Over TimeSource: http://en.wikipedia.org/wiki/Largest..._United_States _by_population_by_decade 1) Using a computer program or a calculator, fit a growth curve to the data of the form p = ab . t Answer 10/21/2020 2.8E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10857 t p = 35741(1.045) 2) Using the exponential best fit for the data, write a table containing the derivatives evaluated at each year. 3) Using the exponential best fit for the data, write a table containing the second derivatives evaluated at each year. Answer Year since 1790 P" 0 69.25 10 107.5 20 167.0 30 259.4 40 402.8 50 625.5 60 971.4 70 1508.5 4) Using the tables of first and second derivatives and the best fit, answer the following questions: a. Will the model be accurate in predicting the future population of New York City? Why or why not? b. Estimate the population in 2010. Was the prediction correct from a.? 10/21/2020 2.8E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10857 2.9: L'Hôpital's Rule In this section, we examine a powerful tool for evaluating limits. This tool, known as L’Hôpital’s rule, uses derivatives to calculate limits. With this rule, we will be able to evaluate many limits we have not yet been able to determine. Instead of relying on numerical evidence to conjecture that a limit exists, we will be able to show definitively that a limit exists and to determine its exact value. Applying L’Hôpital’s Rule L’Hôpital’s rule can be used to evaluate limits involving the quotient of two functions. Consider f (x) lim x→a If lim x→a f (x) = L1 and lim x→a g(x) = L2 ≠ 0, . then f (x) lim x→a However, what happens if lim x→a f (x) = 0 (2.9.1) g(x) and lim x→a = g(x) L1 . (2.9.2) L2 g(x) = 0 ? We call this one of the indeterminate forms, of type This is considered an indeterminate form because we cannot determine the exact behavior of f (x) 0 0 . as x → a without further g(x) analysis. We have seen examples of this earlier in the text. For example, consider 2 x −4 lim x→2 (2.9.3) x −2 and sin x lim . (2.9.4) x x→0 For the first of these examples, we can evaluate the limit by factoring the numerator and writing 2 x x→2 For lim sin x x→0 (x + 2)(x − 2) −4 lim = lim x −2 = lim(x + 2) = 2 + 2 = 4. x −2 x→2 (2.9.5) x→2 we were able to show, using a geometric argument, that x sin x lim x→0 = 1. (2.9.6) x Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could not calculate previously. The idea behind L’Hôpital’s rule can be explained using local linear approximations. Consider two differentiable functions and g such that lim f (x) = 0 = lim g(x) and such that g'(a) ≠ 0 For x near a ,we can write x→a f x→a f (x) ≈ f (a) + f '(a)(x − a) (2.9.7) g(x) ≈ g(a) + g'(a)(x − a). (2.9.8) and Therefore, f (x) f (a) + f '(a)(x − a) ≈ g(x) 10/21/2020 . (2.9.9) g(a) + g'(a)(x − a) 2.9.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 Figure 2.9.1 : If lim x→a Since f (x) = limx→a g(x) , then the ratio f (x)/g(x) is approximately equal to the ratio of their linear approximations near a. is differentiable at a , then f is continuous at a , and therefore f (a) = lim f (x) = 0 . g(a) = lim g(x) = 0 . If we also assume that f ' and g' are continuous at x = a , then f '(a) = lim g'(a) = lim g'(x) . Using these ideas, we conclude that f x→a x→a x→a Similarly, f '(x) and x→a f (x) limx→a g(x) f '(x)(x − a) = limx→a g'(x)(x − a) f '(x) = limx→a g'(x) . Note that the assumption that f ' and g' are continuous at a and g'(a) ≠ 0 can be loosened. We state L’Hôpital’s rule formally for the indeterminate form are using the notation 0 0 0 0 . Also note that the notation 0 does not mean we are actually dividing zero by zero. Rather, we 0 to represent a quotient of limits, each of which is zero. L’Hôpital’s Rule (0/0 Case) Suppose f and g are differentiable functions over an open interval containing a , except possibly at a . If and lim g(x) = 0, then limx→a f (x) = 0 x→a f (x) assuming the limit on the right exists or is ∞ f '(x) lim = lim x→a x→a g(x) , (2.9.10) g'(x) or −∞ . This result also holds if we are considering one-sided limits, or if a = ∞and − ∞. Proof: We provide a proof of this theorem in the special case when f , g, f ', and g' are all continuous over an open interval containing a. In that case, since lim f (x) = 0 = lim g(x) and f and g are continuous at a , it follows that f (a) = 0 = g(a) . Therefore, x→a f (x) limx→a f (x) − f (a) = limx→a g(x) x→a since f (a) = 0 = g(a) g(x) − g(a) f (x) − f (a) = limx→a x −a algebra g(x) − g(a) x −a 10/21/2020 2.9.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 f (x) − f (a) limx→a x −a = limit of a quotient g(x) − g(a) limx→a f '(a) x −a definition of the derivative = g'(a) limx→a f '(x) continuity of f ' and g' = limx→a g'(x) f '(x) = limx→a . limit of a quotient g'(x) Note that L’Hôpital’s rule states we can calculate the limit of a quotient f' derivatives g' f by considering the limit of the quotient of the g . It is important to realize that we are not calculating the derivative of the quotient f g .□ Example 2.9.1 : Applying L’Hôpital’s Rule (0/0 Case) Evaluate each of the following limits by applying L’Hôpital’s rule. 1− a. limx→0 b. lim cosx x sin(πx) x→1 c. ln x e 1/x −1 limx→∞ 1/x d. lim sin x − x x→0 x2 Solution a.. Since the numerator 1 − cos x → 0 and the denominator x → 0 , we can apply L’Hôpital’s rule to evaluate this limit. We have d (1 − cos x) 1 − cos x limx→0 = limx→0 x dx d limx→0 (sin x) sin x = limx→0 = 1 0 = = 0. 1 limx→0 (1) (x) dx b. As x → 1, the numerator obtain sin(πx) → 0 sin(πx) limx→1 ln x and the denominator ln(x) → 0. Therefore, we can apply L’Hôpital’s rule. We π cos(πx) = limx→1 1/x = limx→1 (πx) cos(πx) = (π ⋅ 1)(−1) = −π. c. As x → ∞ , the numerator e 1/x −1 → 0 and the denominator 1 ( obtain e lim x→∞ 1/x e −1 1 1/x ( x ) −1 x→∞ . Therefore, we can apply L’Hôpital’s rule. We −1 ( x2 = lim ) → 0 x 2 = lim e 1/x =e 0 e 1/x −1 = 1 ⋅ lim x→∞ ) x→∞ . (2.9.11) ln x x d. As x → 0, both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain 10/21/2020 2.9.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 sin x − x lim cos x − 1 = lim 2 x x→0 . (2.9.12) 2x x→0 Since the numerator and denominator of this new quotient both approach zero as In doing so, we see that cos x − 1 , we apply L’Hôpital’s rule again. − sin x lim = lim 2x x→0 x → 0 = 0. (2.9.13) 2 x→0 Therefore, we conclude that sin x − x lim 2 = 0. (2.9.14) . (2.9.15) x x→0 Exercise 2.9.1 Evaluate x lim x→0 tan x Hint d 2 tan x = sec x dx Answer 1 We can also use L’Hôpital’s rule to evaluate limits of quotients f (x) g(x) in which f (x) → ±∞ and g(x) → ±∞ . Limits of this form are classified as indeterminate forms of type ∞/∞. Again, note that we are not actually dividing ∞ by ∞. Since ∞ is not a real number, that is impossible; rather, ∞/∞. is used to represent a quotient of limits, each of which is ∞ or −∞ . L’Hôpital’s Rule (∞/∞ Case) Suppose f and g limx→a f (x) = ∞ are differentiable functions over an open interval containing (or −∞ ) and lim g(x) = ∞ (or −∞ ). Then, a , except possibly at a . Suppose x→a f (x) f '(x) lim = lim x→a x→a g(x) (2.9.16) g'(x) assuming the limit on the right exists or is ∞ or −∞ . This result also holds if the limit is infinite, if a = ∞ or −∞ , or the limit is one-sided. Example 2.9.2 : Applying L’Hôpital’s Rule (∞/∞) Case Evaluate each of the following limits by applying L’Hôpital’s rule. a. 3x + 5 limx→∞ 2x + 1 ln x b. lim + x→0 cot x Solution a. Since limx→∞ are first-degree polynomials with positive leading coefficients, (2x + 1) = ∞ . Therefore, we apply L’Hôpital’s rule and obtain 3x + 5 and 2x + 1 3 + 5/x 3x + 5 lim x→∞ 10/21/2020 = lim 2x + 1 x→∞ 3 = lim 2x + 1 x→∞ 2.9.4 CC-BY-NC-SA and 3 = 2 limx→∞ (3x + 5) = ∞ . (2.9.17) 2 https://math.libretexts.org/@go/page/10293 Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the chapter we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of x in the denominator. In doing so, we saw that 3x + 5 3 + 5/x lim x→∞ = lim 2x + 1 3 = 2x + 1 x→∞ . (2.9.18) 2 L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit. b. Here, lim + x→0 ln x = −∞ and lim + x→0 cot x = ∞ . Therefore, we can apply L’Hôpital’s rule and obtain 1/x ln x lim + x→0 = lim cot x 1 = lim −csc2 x + x→0 −xcsc2 x + x→0 . (2.9.19) Now as x → 0 , csc x → ∞ . Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of cscx to write + 2 2 1 sin lim + x→0 Now lim + x→0 2 si n x = 0 x = lim 2 −xcsc x + x→0 . (2.9.20) −x and \(\lim_{x→0^+}x=0\, so we apply L’Hôpital’s rule again. We find 2 si n x limx→0+ −x 2 sin x cos x = limx→0+ 0 = −1 = 0. −1 We conclude that ln x lim + x→0 = 0. (2.9.21) cot x Exercise 2.9.2 Evaluate ln x lim . 5x x→∞ Hint d 1 ln x = dx x Answer 0 As mentioned, L’Hôpital’s rule is an extremely useful tool for evaluating limits. It is important to remember, however, that to apply L’Hôpital’s rule to a quotient f(x)g(x), it is essential that the limit of f (x) g(x) be of the form 0 or . Consider the ∞/∞ 0 following example. Example 2.9.3 : When L’Hôpital’s Rule Does Not Apply Consider 2 x +5 lim x→1 . (2.9.22) 3x + 4 Show that the limit cannot be evaluated by applying L’Hôpital’s rule. Solution 10/21/2020 2.9.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply L’Hôpital’s rule. If we try to do so, we get d 2 (x + 5) = 2x (2.9.23) (3x + 4) = 3. (2.9.24) dx and d dx At which point we would conclude erroneously that 2 x +5 2x lim = lim 3x + 4 x→1 However, since lim 2 x→1 (x + 5) = 6 and lim x→1 (3x + 4) = 7, 2 x . (2.9.25) 3 we actually have +5 6 lim x→1 2 = 3 x→1 = . 3x + 4 (2.9.26) 7 We can conclude that d 2 x lim x→1 + 5) dx ≠ lim 3x + 4 2 (x +5 (2.9.27) d x→1 (3x + 4). dx Exercise 2.9.3 Explain why we cannot apply L’Hôpital’s rule to evaluate lim cos x + x→0 x . Evaluate lim cos x + x→0 x by other means. Hint Determine the limits of the numerator and denominator separately. Answer limx→0+ cos x = 1. Therefore, we cannot apply L’Hôpital’s rule. The limit of the quotient is ∞ Other Indeterminate Forms L’Hôpital’s rule is very useful for evaluating limits involving the indeterminate forms 0 0 and ∞/∞. However, we can also use L’Hôpital’s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits. The expressions 0 ⋅ ∞, ∞ − ∞, 1 , ∞ , and 0 are all considered indeterminate forms. These expressions are not real numbers. Rather, they represent forms that arise when trying to evaluate certain limits. Next we realize why these are indeterminate forms and then understand how to use L’Hôpital’s rule in these cases. The key idea is that we must rewrite the indeterminate forms in such a ∞ 0 0 way that we arrive at the indeterminate form 0 0 or ∞/∞. Indeterminate Form of Type 0⋅∞ Suppose we want to evaluate lim (f (x) ⋅ g(x)) , where f (x) → 0 and g(x) → ∞ (or −∞ ) as x → a . Since one term in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen to the product. We use the notation 0 ⋅ ∞ to denote the form that arises in this situation. The expression 0 ⋅ ∞ is considered indeterminate because we cannot determine without further analysis the exact behavior of the product f (x)g(x) as x → ∞ . For example, let n be a positive integer and consider x→a 1 f (x) = n (x 10/21/2020 + 1) and g(x) = 3x . 2 2.9.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 2 As x → ∞, f (x) → 0 and g(x) → ∞ . However, the limit as , then lim f (x)g(x) = 3 . If n = 1 , then lim consider another limit involving the indeterminate form L’Hôpital’s rule. n =2 x→∞ x → ∞ of 3x f (x)g(x) = 0⋅∞ varies, depending on n . If . If n = 3 , then lim f (x)g(x) = 0 . Here we and show how to rewrite the function as a quotient to use f (x)g(x) = ∞ x→∞ (xn + 1) x→∞ Example 2.9.4 : Indeterminate Form of Type 0 ⋅ ∞ Evaluate lim + x→0 x ln x. Solution First, rewrite the function x ln x as a quotient to apply L’Hôpital’s rule. If we write ln x x ln x = (2.9.28) 1/x we see that ln x → −∞ as x → 0 + and 1 → ∞ x as x → 0 . Therefore, we can apply L’Hôpital’s rule and obtain + d (ln x) ln x limx→0+ 1/x = limx→0+ 2 1/x dx = limx→0+ d −1/x = limx→0+ (−x) = 0. (1/x) dx We conclude that lim x ln x = 0. (2.9.29) + x→0 Figure 2.9.2 : Finding the limit at x = 0 of the function f (x) = x ln x. Exercise 2.9.4 Evaluate lim x cot x. (2.9.30) x→0 Hint Write x cos x x cot x = (2.9.31) sin x Answer 1 10/21/2020 2.9.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 Indeterminate Form of Type ∞ − ∞ Another type of indeterminate form is ∞ − ∞. Consider the following example. Let n be a positive integer and let f (x) = 3x and g(x) = 3x + 5 . As x → ∞, f (x) → ∞ and g(x) → ∞ . We are interested in lim (f (x) − g(x)) . Depending on whether f (x) grows faster, g(x) grows faster, or they grow at the same rate, as we see next, anything can happen in this limit. Since f (x) → ∞ and g(x) → ∞ , we write ∞ − ∞ to denote the form of this limit. As with our other indeterminate forms, ∞ − ∞ has no meaning on its own and we must do more analysis to determine the value of the limit. For example, suppose the exponent n in the function f (x) = 3x is n = 3 , then n 2 x→∞ n 3 lim (f (x) − g(x)) = lim (3 x x→∞ 2 − 3x − 5) = ∞. (2.9.32) − 5) = −5. (2.9.33) − 5) = −∞. (2.9.34) x→∞ On the other hand, if n = 2, then 2 lim (f (x) − g(x)) = lim (3 x x→∞ 2 − 3x x→∞ However, if n = 1 , then 2 lim (f (x) − g(x)) = lim (3x − 3 x x→∞ x→∞ Therefore, the limit cannot be determined by considering only ∞ − ∞ . Next we see how to rewrite an expression involving the indeterminate form ∞ − ∞ as a fraction to apply L’Hôpital’s rule. Example 2.9.5 : Indeterminate Form of Type ∞ − ∞ Evaluate 1 lim ( 2 1 − ). x + x→0 (2.9.35) tan x Solution By combining the fractions, we can write the function as a quotient. Since the least common denominator is have 1 2 tan x, we 2 (tan x) − x 1 − = 2 tan x x 2 x x . tan x As x → 0 , the numerator tan x − x → 0 and the denominator x tan x → 0. Therefore, we can apply L’Hôpital’s rule. Taking the derivatives of the numerator and the denominator, we have + 2 2 2 2 (tan x) − x limx→0+ As x → 0 , (sec x) − 2x → 1 and from the right, we conclude that + 2 2 2 x (sec x) − 2x = limx→0+ tan x 2 x sec x + 2x tan x → 0 2 2 . x sec x + 2x tan x . Since the denominator is positive as x approaches zero 2 (sec x) − 2x lim + x→0 2 = ∞. 2 (2.9.36) x sec x + 2x tan x Therefore, 1 lim ( + x→0 2 x 1 − ) = ∞. (2.9.37) tan x Exercise 2.9.5 Evaluate lim 1 + x→0 ( 1 − x ) sin x . Hint Rewrite the difference of fractions as a single fraction. 10/21/2020 2.9.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 Answer 0 Example 2.9.6 : Evaluate 1 lim ( 1 − ). x + x→0 (2.9.38) tan x Indeterminate Form of Types 0 , ∞ , and 1 Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions 0 , ∞ , and 1 are all indeterminate forms. On their own, these expressions are meaningless because we cannot actually evaluate these expressions as we would evaluate an expression involving real numbers. Rather, these expressions represent forms that arise when finding limits. Now we examine how L’Hôpital’s rule can be used to evaluate limits involving these indeterminate forms. 0 0 ∞ 0 0 ∞ Since L’Hôpital’s rule applies to quotients, we use the natural logarithm function and its properties to reduce a problem evaluating a limit involving exponents to a related problem involving a limit of a quotient. For example, suppose we want to evaluate lim f (x ) and we arrive at the indeterminate form ∞ . (The indeterminate forms 0 and 1 can be handled similarly.) We proceed as follows. Let g(x) 0 0 ∞ x→a g(x) y = f (x ) . (2.9.39) Then, g(x) ln y = ln(f (x ) ) = g(x) ln(f (x)). (2.9.40) Therefore, lim[ln(y)] = lim[g(x) ln(f (x))]. x→a (2.9.41) x→a Since lim f (x) = ∞, we know that lim ln(f (x)) = ∞ . Therefore, lim g(x) ln(f (x)) is of the indeterminate form 0 ⋅ ∞ , and we can use the techniques discussed earlier to rewrite the expression g(x) ln(f (x)) in a form so that we can apply L’Hôpital’s rule. Suppose lim g(x) ln(f (x)) = L , where L may be ∞ or −∞. Then x→a x→a x→a x→a lim[ln(y)] = L. (2.9.42) x→a Since the natural logarithm function is continuous, we conclude that ln( lim y) = L, (2.9.43) x→a which gives us g(x) lim y = lim f (x ) x→a =e L . (2.9.44) x→a Example 2.9.7 : Indeterminate Form of Type ∞ 0 Evaluate 1/x lim x . (2.9.45) x→∞ Solution Let y = x 1/x .Then, 1/x ln(x 1 ) = 10/21/2020 ln x ln x = x . (2.9.46) x 2.9.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 ln x We need to evaluate lim x→∞ x . Applying L’Hôpital’s rule, we obtain ln x lim ln y = lim x→∞ Therefore, lim x→∞ ln y = 0. 1/x = lim x x→∞ = 0. (2.9.47) 1 x→∞ Since the natural logarithm function is continuous, we conclude that ln( lim y) = 0, (2.9.48) x→∞ which leads to ln x lim y = lim x→∞ =e 0 = 1. (2.9.49) x x→∞ Hence, 1/x lim x = 1. (2.9.50) x→∞ Exercise 2.9.7 Evaluate lim x 1/ ln(x) . (2.9.51) x→∞ Answer Let y = x 1/ ln(x) and apply the natural logarithm to both sides of the equation. Answer e Example 2.9.8 : Evaluate π ) (x− lim π 2 (tan(x)) . (2.9.52) − x→ 2 Example 2.9.9 : Indeterminate Form of Type 0 0 Evaluate lim x sin x . (2.9.53) + x→0 Solution Let y =x sin x . (2.9.54) Therefore, ln y = ln(x sin x ) = sin x ln x. (2.9.55) We now evaluate lim sin x ln x. Since lim sin x = 0 and lim ln x = −∞ , we have the indeterminate form 0 ⋅ ∞ . To apply L’Hôpital’s rule, we need to rewrite sin x ln x as a fraction. We could write + x→0 + + x→0 x→0 sin x sin x ln x = (2.9.56) 1/ ln x 10/21/2020 2.9.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 or ln x ln x sin x ln x = . = 1/ sin x cscx Let’s consider the first option. In this case, applying L’Hôpital’s rule, we would obtain sin x limx→0+ sin x ln x = limx→0+ cos x 1/ ln x = limx→0+ 2 = limx→0+ (−x(ln x ) 2 −1/(x(ln x ) ) cos x). Unfortunately, we not only have another expression involving the indeterminate form 0 ⋅ ∞, but the new limit is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By writing ln x ln x sin x ln x = = cscx, 1/ sin x and applying L’Hôpital’s rule, we obtain 1/x ln x limx→0+ sin x ln x = limx→0+ Using the fact that cscx = 1 sin x and cot x = 2 −si n x limx→0+ x cos x We conclude that lim + x→0 cscx cos x sin x = limx→0+ −cscx cot x ln y = 0. x xcscx cot x . , we can rewrite the expression on the right-hand side as sin x = limx→0+ [ −1 = limx→0+ sin x ⋅ (− tan x)] = (limx→0+ Therefore, ln(lim + x→0 y) = 0 ) ⋅ (limx→0+ (− tan x)) = 1 ⋅ 0 = 0. and we have sin x limx→0+ y = limx→0+ x x =e 0 = 1. Hence, sin x limx→0+ x = 1. Exercise 2.9.9 Evaluate lim + x→0 x x . Hint Let y = x and take the natural logarithm of both sides of the equation. x Answer 1 Growth Rates of Functions Suppose the functions f and g both approach infinity as x → ∞ . Although the values of both functions become arbitrarily large as the values of x become sufficiently large, sometimes one function is growing more quickly than the other. For example, f (x) = x and g(x) = x both approach infinity as x → ∞ . However, as Table 2.9.1 shows, the values of x are growing much faster than the values of x . 2 3 3 2 Table 2.9.1: Comparing the Growth Rates of x and x 2 10 100 1000 10,000 2 100 10,000 1,000,000 100,000,000 3 1000 1,000,000 1,000,000,000 1,000,000,000,000 x f(x) = x g(x) = x 3 In fact, 3 x lim x→∞ 2 x = lim x = ∞. (2.9.57) x→∞ or, equivalently 10/21/2020 2.9.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 2 x lim x→∞x 3 1 = lim x→∞ x = 0. (2.9.58) x As a result, we say x is growing more rapidly than x as x → ∞ . On the other hand, for f (x) = x and g(x) = 3 x + 4x + 1 , although the values of g(x) are always greater than the values of f (x) for x > 0 , each value of g(x) is roughly three times the corresponding value of f (x) as x → ∞ , as shown in Table 2.9.2. In fact, 3 2 2 2 2 x lim x→∞ 2 3x 1 = + 4x + 1 . (2.9.59) 3 Table 2.9.2: Comparing the Growth Rates of x and 3x 2 2 + 4x + 1 1 01 1 ,01 0 0 0 0 0 x f(x) = x g(x) = 3 x In this case, we say that x and 3x 2 2 2 1 01 0, 1 ,0 0 01 , 0 0 0, 0 ,0 0 0 0 0 2 3 03 0, 3 ,0 0 03 , 4 4 0,1 0 ,0 1 0 01 1 + 4x + 1 are growing at the same rate as x → ∞. + 4x + 1 More generally, suppose f and g are two functions that approach infinity as x → ∞ . We say g grows more rapidly than f as x → ∞ if g(x) lim x→∞ f (x) = ∞\); or, equivalently, \( lim f (x) On the other hand, if there exists a constant M x→∞ ≠0 = 0. (2.9.60) g(x) such that f (x) lim x→∞ = M, (2.9.61) g(x) we say f and g grow at the same rate as x → ∞ . Next we see how to use L’Hôpital’s rule to compare the growth rates of power, exponential, and logarithmic functions. Example 2.9.10 : Comparing the Growth Rates of ln(x), x , and e 2 x For each of the following pairs of functions, use L’Hôpital’s rule to evaluate 10/21/2020 2.9.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 f (x) lim ( x→∞ ). (2.9.62) g(x) a. f (x) = x and g(x) = e b. f (x) = ln(x) and g(x) = x 2 x 2 Solution 2 a. Since lim 2 x→∞ x =∞ and lim x→∞ e x x→∞ 2 x limx→∞ Since lim x→∞ 2x = ∞ and li m x→∞ e x x , we can use L’Hôpital’s rule to evaluate lim =∞ e x 2x = limx→∞ e x [ e x . We obtain ] . , we can apply L’Hôpital’s rule again. Since 2x limx→∞ e x 2 = limx→∞ e x =0 , we conclude that 2 x limx→∞ e x = 0. Therefore, e grows more rapidly than x as x → ∞ (See Figure 2.9.3 and Table 2.9.3) x 2 Figure 2.9.3 : An exponential function grows at a faster rate than a power function. Table 2.9.3: Growth rates of a power function and an exponential function. 5 10 15 20 2 25 100 225 400 x 148 22,026 3,269,017 485,165,195 x x e b. Since lim x→∞ ln x = ∞ and lim x→∞ 2 x =∞ , we can use L’Hôpital’s rule to evaluate lim x→∞ ln x 2 . We obtain x 1/x ln x lim x→∞ 2 x = lim 1 = lim x→∞ 2x x→∞ 2 = 0. (2.9.63) 2x Thus, x grows more rapidly than ln x as x → ∞ (see Figure 2.9.4 and Table 2.9.4). 2 10/21/2020 2.9.13 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 Figure 2.9.4 : A power function grows at a faster rate than a logarithmic function. Table 2.9.4: Growth rates of a power function and a logarithmic function x 10 100 1000 10,000 ln(x) 2.303 4.605 6.908 9.10 100 10,000 1,000,000 100,000,000 2 x Example 2.9.11 : Comparing the Growth Rates of x 100 Compare the growth rates of x 100 and 2 x and 2 . x Hint: Apply L’Hôpital’s rule to x 100 x /2 Solution The function 2 grows faster than x x 100 . Using the same ideas as in Example a. it is not difficult to show that e grows more rapidly than Figure 2.9.5 and Table 2.9.5 , we compare e with x and x as x → ∞ . x x 3 p x for any p >0 . In 4 Figure 2.9.5 : The exponential function e grows faster than x for any p > 0 . (a) A comparison of e with x . (b) A comparison of e with x . x p x x 3 4 Table 2.9.5 : An exponential function grows at a faster rate than any power function 5 10 15 20 3 125 1000 3375 8000 4 625 10,000 50,625 160,000 x 148 22,026 3,326,017 485,165,195 x x x e 10/21/2020 2.9.14 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 Similarly, it is not difficult to show that compare ln x with [ ]x 3 p x grows more rapidly than ln x for any p >0 . In Figure 2.9.6 and Table, we and √− x. Figure 2.9.6 : The function y = ln(x) grows more slowly than x for any p > 0 as x → ∞ . p Table 2.9.6 : A logarithmic function grows at a slower rate than any root function x 10 100 1000 10,000 ln(x) 2.303 4.605 6.908 9.210 ]x 2.154 4.642 10 21.544 − √x 3.162 10 31.623 100 [ 3 Theorem 2.9.1 a 1. lim x x→∞ e x = 0.a > 0 . 2. $\lim_{x \to -\ infty} |x|â e^x=0, a>0$. 3. $\lim_{x \to \infty} \frac{\ln(x)}{xâ}=0, a>0$. 4. $\lim_{x \to 0^+} {\ln(x)}{xâ}=0, a>0$. Proof Add proof here and it will automatically be hidden if you have a "AutoNum" template active on the page. Example 2.9.12 : Find the following limits: 2 1. lim = 0. 2. $\lim_{x \to 0^+} \frac{\ln(x)}{x}=-\infty $. 3. $\lim_{x \to \infty} \frac{x-3e^{-x}}{x+2e^{-x}}= \lim_{x \to \infty} \frac{1-3/(xe^{x})} {1+2/((xe^{x}})=1$. x x→∞ e x Key Concepts L’Hôpital’s rule can be used to evaluate the limit of a quotient when the indeterminate form 0 0 or ∞/∞ arises. L’Hôpital’s rule can also be applied to other indeterminate forms if they can be rewritten in terms of a limit involving a quotient that has the indeterminate form 0 or ∞/∞. 0 The exponential function e grows faster than any power function x , p > 0 . The logarithmic function ln x grows more slowly than any power function x , p > 0 . x p p Glossary indeterminate forms 10/21/2020 2.9.15 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 when evaluating a limit, the forms 0 0 ,∞/∞, 0 ⋅ ∞, ∞ − ∞, 0 0 0 ,∞ , and 1 ∞ are considered indeterminate because further analysis is required to determine whether the limit exists and, if so, what its value is L’Hôpital’s rule if f and g are differentiable functions over an interval a , except possibly at a , and limx→a f (x) and lim x→a g(x) are infinite, then lim f (x) f '(x) = limx→a x→a g(x) limx→a f (x) = 0 = limx→a g(x) or , assuming the limit on the right exists or g'(x) is ∞ or −∞ Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 2.9.16 CC-BY-NC-SA https://math.libretexts.org/@go/page/10293 2.9E: Exercises Exercise 2.9E . 1 For the following exercises, evaluate the limit. 1) Evaluate the limit lim e x→∞ 2) Evaluate the limit lim x x e x→∞ . x k x . Answer ∞ 3) Evaluate the limit lim lnx x→∞ 4) Evaluate the limit lim x→a k x . x−a 2 2 x −a . Answer 1 2a 5) Evaluate the limit lim x→a 6) Evaluate the limit lim x→a x−a 3 3 x −a x−a n n x −a . . Answer 1 nan− 1 Exercise 2.9E . 2 For the following exercises, determine whether you can apply L’Hôpital’s rule directly. Explain why or why not. Then, indicate if there is some way you can alter the limit so you can apply L’Hôpital’s rule. 1) lim + x→0 2) lim 2 x lnx 1/x x→∞ x Answer Cannot apply directly; use logarithms 3) lim x→0 4) lim x→0 2/x x 2 x 1/x Answer Cannot apply directly; rewrite as lim 3 x→0 x 5) \(lim_{x→∞}\frac{e^x}{x}\ 10/21/2020 2.9E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10859 Exercise 2.9E . 3 For the following exercises, evaluate the limits with either L’Hôpital’s rule or previously learned methods. 2 1) lim x −9 x→3 x−3 Answer 6 2 2) lim x −9 3) lim (1+x ) x→3 x+3 −2 x→0 −1 x Answer −2 4) lim cosx x→π/2 π −x 2 5) lim x−π x→π sinx Answer −1 6) lim x−1 x→1 sinx n (1+x ) −1 7) lim x→0 x Answer n n 8) lim (1+x ) −1−nx 9) lim sinx−tanx x→0 2 x x→0 3 x Answer − 1 2 10) lim √1+x−√1−x 11) lim e −x−1 x→0 x x x→0 2 x Answer 1 2 12) lim tanx 13) lim x→1 x→0 x→1 √x lnx Answer 10/21/2020 2.9E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10859 1 14) lim 1/x x→0 (x + 1) 3 √x−√x 15) lim x→1 x−1 Answer 1 6 16) lim + x→0 2x x 17) lim x→∞ xsin( 1 x ) Answer 1 18) lim sinx−x x→0 19) lim 2 x + x→0 4 xln(x ) Answer 0 20) lim (x − e ) 21) lim x e x x→∞ 2 x→∞ −x Answer 0 x x 22) lim 3 −2 23) lim 1+1/x x→0 x→0 x 1−1/x Answer −1 24) lim x→π/4 (1 − tanx)cotx 25) lim 1/ x→∞ x e x Answer ∞ 26) lim 1/cosx x→0 x 27) lim 1/x x→0 x Answer 1 10/21/2020 2.9E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10859 28) lim x→0 1 (1 − 29) lim x→∞ (1 x x ) 1 − x x ) Answer 1 e Exercise 2.9E . 4 For the following exercises, use a calculator to graph the function and estimate the value of the limit, then use L’Hôpital’s rule to find the limit directly. x 1) lim e −1 x→0 2) lim x→0 x 1 xsin( x ) Answer 0 3) lim x−1 x→1 4) lim 1−cos(πx) e x→1 (x−1) −1 x−1 Answer 1 2 5) lim (x−1) 6) lim 1+cosx x→1 lnx x→π sinx Answer 0 7) lim x→0 (cscx 8) lim + x→0 − 1 x ) x tan(x ) Answer tan(1) 9) lim + x→0 10) lim x→0 lnx sinx x e −e −x x Answer 2 10/21/2020 2.9E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10859 2E: Exercises Chapter Review Exercises Exercise 1 Let f (x + y) = f (x) + f (y) + x 2 y + xy 2 f (x) and lim x→0 =1 x . Then find f^'(x) . Answer 2 1 +x Exercise 2 Use the limit definition of the derivative to find f (x), where f (x) = 3x . (You may use differentiation rules to check your answer). Find the tangent line to f (x) = 3x at x = 1. ′ 2 2 Answer Under Construction Exercise 3 Using the definition of the derivative (Newton's quotient) find f (x) for ′ 1 f (x) = −−−− − √ 2x + 5 (2E.1) Answer Under Construction Exercise 4 Use the limit denition of the derivative to find f ′ (x), where f (x) = 1 x+1 (You may using differentiation rules to check your answer). Then find the equation of the tangent line to f (x) at x = 2 Answer Under Construction Exercise 5 Find the derivatives of the following functions: − − − − − − − − − − 1) f (x) = √(x 2 2) g(x) = cos x x + sin x 3) h(t) = t 2 + csc t + + sec 2 4) y(x) = 3x 2 2x 2 5) f (x) = 10/21/2020 2 + x + 1) + x (x 2 t +5 +x −3 + 1) cot x 3 − cos x csc x 2E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10308 − −−−−−−− − 6) h(t) = √tan(2t − 1) x 7) y(x) = x2 − 1 2 ln(x 8) f (x) = + 1) 2 x 9) f (x) = ( 3 x −5 ) 2x + 1 10) h(t) = sin 2 − −−−−−−− − (√tan(2t − 1) ) sin(t) 11) h(t) = t − − 12) y = csc(cot(√4x)) 13) y(x) = (x + 5) 7 14) g(x) = 2 (x − 1) − tan(2x) + √x 2 cos(x) + x 15) g(x) = − cos(x) − √x 2 tan(x) + 3x 16) y = 3x + 4 2 x $ +5 17) y = sec(x) tan(x) − − − − − − − − − 18) y = √5 + √6x . 19) y = x 3 cos(x). 20) f (x) − tan(2x) 2x+1 Answer Under Construction Exercise 6 Find equations of the tangent and normal to 2 y = (2E.2) − 3 − 4 √x at the point (1, −2). Answer Under Construction Exercise 7 An object is released from rest (its initial velocity is zero) from the Empire State Building at a height of street level. The height of the object can 1250 ft above be modeled by the position functions s = f (t) = 1250 − 16t . 2 a. Verify the object is still falling at t = 5s : b. Find the object's instantaneous velocity at the time t = 5s 10/21/2020 2E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10308 Answer Under Construction Exercise 8 True or False? Justify the answer with a proof or a counterexample. 1) Every function has a derivative. 2) A continuous function has a continuous derivative. 3) A continuous function has a derivative. 4) If a function is differentiable, it is continuous. Answers to odd numbered questions 1. False. 3. False. Exercise 9 Use the limit definition of the derivative to exactly evaluate the derivative: − − − − − 1) f (x) = √x + 4 2) f (x) = 3 x Answers to odd numbered questions 1. 1 2 √x+4 Exercise 10 Find the derivatives of the following functions: 1) f (x) = 3x 3 4 − 2 x 2) f (x) = (4 − x 2 3) f (x) = e 3 ) sinx 4) f (x) = ln(x + 2) 5) f (x) = x 2 6) cosx + xtan(x) − −−−− − 2 f (x) = √3 x + 2 7) f (x) = 8) x 2 x 4 −1 si n (x) y = (y + 2) + xysin(x) Answers to odd numbered questions 1. 9x 2 3. e + sinx 2 1 4 ( 3 cosx 5. xsec 7. 8 x 2 (x) + 2xcos(x) + tan(x) − x sin(x) x √1−x2 −1 + si n (x)) Exercise 11 Find the following derivatives of various orders: 10/21/2020 2E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10308 1) First derivative of y = xln(x)cosx 2) Third derivative of y = (3x + 2) 2 3) Second derivative of y = 4 x 2 + x sin(x) Answers to odd numbered questions 1. cosx ⋅ (lnx + 1) − xln(x)sinx 3. 4 x 2 (ln4 ) 2 + 2sinx + 4xcosx − x sinx Exercise 12 Find the equation of the tangent line to the following equations at the specified point: 1) y = cos −1 2) y = x + e (x) + x x − 1 x at x = 0 at x = 1 Answer to even numbered questions 2. T = (2 + e)x − 2 Exercise 13 Draw the derivative for the following graphs. 21) Answer Exercise 14 10/21/2020 2E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10308 The following questions concern the water level in Ocean City, New Jersey, in January, which can be approximated by w(t) = 1.9 + 2.9cos( t), where t is measured in hours after midnight, and the height is measured in feet. π 6 1) Find and graph the derivative. What is the physical meaning? 2) Find w'(3). What is the physical meaning of this value? Answers to even numbered questions 2. w'(3) = − 2.9π 6 . At 3 a.m. the tide is decreasing at a rate of 1.514 ft/hr. Exercise 15 The following questions consider the wind speeds of Hurricane Katrina, which affected New Orleans, Louisiana, in August 2005. The data are displayed in a table. Hours after Midnight, August 26 Wind Speed (mph) 1 45 5 75 11 100 29 115 49 145 58 175 73 155 81 125 85 95 107 35 Wind Speeds of Hurricane KatrinaSource: http://news.nationalgeographic.com/n..._timeline.html. 1) Using the table, estimate the derivative of the wind speed at hour 39. What is the physical meaning? 2) Estimate the derivative of the wind speed at hour 83. What is the physical meaning? Answers to even numbered questions 2. −7.5. The wind speed is decreasing at a rate of 7.5 mph/hr Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 2E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10308 CHAPTER OVERVIEW 3: APPLICATIONS OF DERIVATIVES 3.0 INTRODUCTION TO APPLICATIONS OF DERIVATIVE 3.1: RELATED RATES 3.1E: EXERCISES 3.2 LINEAR APPROXIMATIONS AND DIFFERENTIALS 3.2E: EXERCISES 3.3: EXTREMAS 3.3E: EXERCISES 3.4 THE MEAN VALUE THEOREM 3.4E: EXERCISES 3.5 DERIVATIVE TESTS 3.5E: EXERCISES 3.6: APPLIED OPTIMIZATION PROBLEMS 3.6E: EXERCISES 3.7: CURVE SKECTCHING 3.7E: EXERCISES 3.8: NEWTON'S METHOD 3.8E: EXERCISES 3.9: ANTI DERIVATIVES AND RECTILINEAR MOTION 3.9E: EXERCISES 3E: CHAPTER EXERCISES 1 10/21/2020 3.0 Introduction to applications of Derivative A rocket is being launched from the ground and cameras are recording the event. A video camera is located on the ground a certain distance from the launch pad. At what rate should the angle of inclination (the angle the camera makes with the ground) change to allow the camera to record the flight of the rocket as it heads upward? (See [link].) Figure 1 : As a rocket is being launched, at what rate should the angle of a video camera change to continue viewing the rocket? (credit: modification of work by Steve Jurvetson, Wikimedia Commons) A rocket launch involves two related quantities that change over time. Being able to solve this type of problem is just one application of derivatives introduced in this chapter. We also look at how derivatives are used to find maximum and minimum values of functions. As a result, we will be able to solve applied optimization problems, such as maximizing revenue and minimizing surface area. In addition, we examine how derivatives are used to evaluate complicated limits, to approximate roots of functions, and to provide accurate graphs of functions. Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10890 3.1: Related Rates We have seen that for quantities that are changing over time, the rates at which these quantities change are given by derivatives. If two related quantities are changing over time, the rates at which the quantities change are related. For example, if a balloon is being filled with air, both the radius of the balloon and the volume of the balloon are increasing. In this section, we consider several problems in which two or more related quantities are changing and we study how to determine the relationship between the rates of change of these quantities. Setting up Related-Rates Problems Example 3.1.1 : An object is moving in the clockwise direction around the unit circle 1 ( – √2 1 , – √2 2 x +y 2 = 1. As it passes through the point , its y−coordinate is decreasing at the rate of 4 units per second. At what rate is the x−coordinate changing at ) this point. Solution: Exercise 3.1.1 An object is moving in the clockwise direction around the unit circle 1 ( – √2 −1 , – √2 2 x +y 2 = 1. As it passes through the point , its y−coordinate is decreasing at the rate of 4 units per second. At what rate is the x−coordinate changing at ) this point. Answer Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page. In many real-world applications, related quantities are changing with respect to time. For example, if we consider the balloon example again, we can say that the rate of change in the volume, V , is related to the rate of change in the radius, r. In this case, we say that and are related rates because V is related to r. Here we study several examples of related quantities that are changing with respect to time and we look at how to calculate one rate of change given another rate of change. dV dr dt dt Example 3.1.2 : Inflating a Balloon A spherical balloon is being filled with air at the constant rate of when the radius is 3cm? 3 2c m /sec (Figure). How fast is the radius increasing Figure 3.1.1 :As the balloon is being filled with air, both the radius and the volume are increasing with respect to time. Solution The volume of a sphere of radius r centimeters is 10/21/2020 3.1.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10301 4 V = 3 3 3 π r cm . Since the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, beginning to fill the balloon with air, the volume of air in the balloon is V (t) = 4 3 3 t seconds after 3 π[r(t)] c m . Differentiating both sides of this equation with respect to time and applying the chain rule, we see that the rate of change in the volume is related to the rate of change in the radius by the equation V ′ 2 (t) = 4π[r(t)] r'(t). The balloon is being filled with air at the constant rate of 2 cm3/sec, so V 3 2 2 ′ 3 (t) = 2c m /sec. Therefore, ′ 2c m /sec = (4π[r(t)] c m ) ⋅ (r (t)cm/s), which implies 1 ′ r (t) = 2 cm/sec . 2π[r(t)] When the radius r = 3cm, ′ r (t) = 1 18π cm/sec. Exercise 3.1.2 What is the instantaneous rate of change of the radius when r = 6cm ? Solution \ dr dt = 1 2 2πr Answer 1 72π , or approximately 0.0044 cm/sec cm/sec Before looking at other examples, let’s outline the problem-solving strategy we will be using to solve related-rates problems. Problem-Solving Strategy: Solving a Related-Rates Problem 1. Assign symbols to all variables involved in the problem. Draw a figure if applicable. 2. State, in terms of the variables, the information that is given and the rate to be determined. 3. Find an equation relating the variables introduced in step 1. 4. Using the chain rule, differentiate both sides of the equation found in step 3 with respect to the independent variable. This new equation will relate the derivatives. 5. Substitute all known values into the equation from step 4, then solve for the unknown rate of change Note that when solving a related-rates problem, it is crucial not to substitute known values too soon. For example, if the value for a changing quantity is substituted into an equation before both sides of the equation are differentiated, then that quantity will behave as a constant and its derivative will not appear in the new equation found in step 4. We examine this potential error in the following example. Examples of the Process Let’s now implement the strategy just described to solve several related-rates problems. The first example involves a plane flying overhead. The relationship we are studying is between the speed of the plane and the rate at which the distance between the plane and a person on the ground is changing. Example 3.1.3 : An Airplane Flying at a Constant Elevation An airplane is flying overhead at a constant elevation of 4000f t. A man is viewing the plane from a position 3000f t from the base of a radio tower. The airplane is flying horizontally away from the man. If the plane is flying at the rate of 10/21/2020 3.1.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10301 , at what rate is the distance between the man and the plane increasing when the plane passes over the radio 600f t/sec tower? Solution: Step 1. Draw a picture, introducing variables to represent the different quantities involved. Figure 3.1.2 : An airplane is flying at a constant height of 4000 ft. The distance between the person and the airplane and the person and the place on the ground directly below the airplane are changing. We denote those quantities with the variables s and x, respectively. As shown, x denotes the distance between the man and the position on the ground directly below the airplane. The variable s denotes the distance between the man and the plane. Note that both x and s are functions of time. We do not introduce a variable for the height of the plane because it remains at a constant elevation of 4000f t. Since an object’s height above the ground is measured as the shortest distance between the object and the ground, the line segment of length 4000 ft is perpendicular to the line segment of length x feet, creating a right triangle. Step 2. Since x denotes the horizontal distance between the man and the point on the ground below the plane, dx/dt represents the speed of the plane. We are told the speed of the plane is 600 ft/sec. Therefore, = 600 ft/sec. Since we are asked to find the rate of change in the distance between the man and the plane when the plane is directly above the radio tower, we need to find ds/dt when x = 3000f t. dx dt Step 3. From the figure, we can use the Pythagorean theorem to write an equation relating x and s : 2 [x(t)] 2 + 4000 2 = [s(t)] . Step 4. Differentiating this equation with respect to time and using the fact that the derivative of a constant is zero, we arrive at the equation x dx dt =s ds dt . Step 5. Find the rate at which the distance between the man and the plane is increasing when the plane is directly over the radio tower. That is, find when x = 3000f t. Since the speed of the plane is 600f t/sec, we know that = 600f t/sec . We are not given an explicit value for s ; however, since we are trying to find when x = 3000f t, we can use the Pythagorean theorem to determine the distance s when x = 3000 and the height is 4000f t. Solving the equation ds dt dx ds dt dt 2 3000 2 + 4000 2 =s for s , we have s = 5000f t at the time of interest. Using these values, we conclude that ds/dt is a solution of the equation (3000)(600) = (5000) ⋅ ds dt . Therefore, 10/21/2020 3.1.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10301 ds dt = 3000⋅600 5000 = 360f t/sec. Note: When solving related-rates problems, it is important not to substitute values for the variables too soon. For example, in step 3, we related the variable quantities x(t) and s(t) by the equation 2 2 [x(t)] + 4000 2 = [s(t)] . Since the plane remains at a constant height, it is not necessary to introduce a variable for the height, and we are allowed to use the constant 4000 to denote that quantity. However, the other two quantities are changing. If we mistakenly substituted x(t) = 3000 into the equation before differentiating, our equation would have been 2 3000 2 + 4000 2 = [s(t)] . After differentiating, our equation would become 0 = s(t) As a result, we would incorrectly conclude that ds dt ds dt . = 0. Exercise 3.1.3 What is the speed of the plane if the distance between the person and the plane is increasing at the rate of 300f t/sec? Hint ds dt = 300f t/sec Answer 500f t/sec We now return to the problem involving the rocket launch from the beginning of the chapter. Example 3.1.4 : Chapter Opener - A Rocket Launch Figure 3.1.3 : (credit: modification of work by Steve Jurvetson, Wikimedia Commons) A rocket is launched so that it rises vertically. A camera is positioned 5000f t from the launch pad. When the rocket is 1000f t above the launch pad, its velocity is 600f t/sec. Find the necessary rate of change of the camera’s angle as a function of time so that it stays focused on the rocket. Solution: Step 1. Draw a picture introducing the variables. 10/21/2020 3.1.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10301 Figure 3.1.4 : A camera is positioned 5000 ft from the launch pad of the rocket. The height of the rocket and the angle of the camera are changing with respect to time. We denote those quantities with the variables h and θ ,respectively. Let h denote the height of the rocket above the launch pad and θ be the angle between the camera lens and the ground. Step 2. We are trying to find the rate of change in the angle of the camera with respect to time when the rocket is 1000 ft off the ground. That is, we need to find when h = 1000f t. At that time, we know the velocity of the rocket is dθ dt dh dt = 600f t/sec. Step 3. Now we need to find an equation relating the two quantities that are changing with respect to time: h and θ . How can we create such an equation? Using the fact that we have drawn a right triangle, it is natural to think about trigonometric functions. Recall that tanθ is the ratio of the length of the opposite side of the triangle to the length of the adjacent side. Thus, we have tanθ = h 5000 . This gives us the equation h = 5000tanθ. Step 4. Differentiating this equation with respect to time t , we obtain dh dt 2 = 5000sec θ dθ dt . Step 5. We want to find when h = 1000f t. At this time, we know that = 600f t/sec . We need to determine sec θ . Recall that secθ is the ratio of the length of the hypotenuse to the length of the adjacent side. We know the length of the adjacent side is 5000f t. To determine the length of the hypotenuse, we use the Pythagorean theorem, where the length of one leg is 5000f t, the length of the other leg is h = 1000f t, and the length of the hypotenuse is c feet as shown in the following figure. dθ dh dt dt 2 We see that 2 1000 2 + 5000 2 =c and we conclude that the hypotenuse is 10/21/2020 3.1.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10301 − − c = 1000 √26f t. Therefore, when h = 1000, we have 2 sec θ = ( 1000 √26 5000 2 ) = 26 25 . Recall from step 4 that the equation relating f racdθdt to our known values is dh dt When h = 1000f t, we know that we arrive at the equation dh dt = 600f t/sec 2 = 5000sec θ and sec 2 θ = 600 = 5000( Therefore, dθ dt = 3 26 26 25 26 25 ) dθ dt . . Substituting these values into the previous equation, dθ dt . rad/sec. Exercise 3.1.4 What rate of change is necessary for the elevation angle of the camera if the camera is placed on the ground at a distance of 4000f t from the launch pad and the velocity of the rocket is 500 ft/sec when the rocket is 2000f t off the ground? Hint Find dθ dt when h = 2000f t. At that time, dh dt = 500f t/sec. Answer 1 10 rad/sec In the next example, we consider water draining from a cone-shaped funnel. We compare the rate at which the level of water in the cone is decreasing with the rate at which the volume of water is decreasing. Example 3.1.5 : Water Draining from a Funnel Water is draining from the bottom of a cone-shaped funnel at the rate of 0.03f t /sec. The height of the funnel is 2 ft and the radius at the top of the funnel is 1f t. At what rate is the height of the water in the funnel changing when the height of the water is f t? 3 1 2 (The volume of the cone is 1/3π r 2 h where r is the radius and h is the height.) Solution: Step 1: Draw a picture introducing the variables. 10/21/2020 3.1.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10301 Figure 3.1.5 : Water is draining from a funnel of height 2 ft and radius 1 ft. The height of the water and the radius of water are changing over time. We denote these quantities with the variables h and r,respectively. Let h denote the height of the water in the funnel, r denote the radius of the water at its surface, and V denote the volume of the water. Step 2: We need to determine dh dt when h = 1 2 ft . We know that dV = −0.03f t/sec. dt Step 3: The volume of water in the cone is 1 V = 2 π r h. 3 From the figure, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the same. Therefore, = or r = . Using this fact, the equation for volume can be simplified to r 1 h h 2 2 V = 1 3 π( h 2 2 ) h = π 12 3 h . Step 4: Applying the chain rule while differentiating both sides of this equation with respect to time t , we obtain dV dt Step 5: We want to find when = −0.03f t /sec. Therefore, dh dt dV h = 1 2 ft = π 4 2 dh h dt . . Since water is leaving at the rate of 3 , we know that 0.03f t /sec 3 dt −0.03 = π 4 ( 1 2 2 dh ) dt , which implies −0.03 = π dh 16 dt . It follows that dh dt =− 0.48 π = −0.153f t/sec. Exercise 3.1.5 At what rate is the height of the water changing when the height of the water is 1 4 ? ft Hint We need to find dh dt when h = 1 4 . Answer −0.61f t/sec Example 3.1.6 : Oil spilled from a tanker spreads in a circle whose area increases at a constant rate of 6 km increasing when the radius is 3 km. 2 /h . How fast the circumference Exercise 3.1.6 Oil spilled from a tanker spreads in a circle whose circumference increases at a constant rate of 6 km/h. How fast the area increasing when the radius is 3 km. Answer Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page. Example 3.1.7 : 10/21/2020 3.1.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10301 The altitude(height) of a right triangle is increasing a ta rate of 2 cm/min while the area of a triangle is increasing at a rate of 4 cm /min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100 cm . 2 2 Exercise 3.1.7 The altitude(height) of a right triangle is increasing a ta rate of 2 cm/min while the area of a triangle is increasing at a rate of 4 cm /min. At what rate is the base of the triangle changing when the altitude is 5 cm and the area is 25 cm . 2 2 Answer Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page. Example 3.1.8 : When air expands the pressure P and the volume V of the air satisfy the relationship P V = k where k is a constant. At a certain instant, the pressure is 10 lb/in , and the volume is 10 i n . If the volume is decreasing at the rate of 5 i n /s at this instant, what is the rate of change of the pressure? 7/5 2 4 3 3 Key Concepts To solve a related rates problem, first draw a picture that illustrates the relationship between the two or more related quantities that are changing with respect to time. In terms of the quantities, state the information given and the rate to be found. Find an equation relating the quantities. Use differentiation, applying the chain rule as necessary, to find an equation that relates the rates. Be sure not to substitute a variable quantity for one of the variables until after finding an equation relating the rates. Glossary related rates are rates of change associated with two or more related quantities that are changing over time Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 3.1.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10301 3.1E: Exercises Exercises 3.1E. 1 1) Find dy 2) Find dx 3) Find dz dt dt dt at x = 1 and y = x 2 +3 at x = −2 and y = 2x 2 at (x, y) = (1, 3) and z if +1 2 dx dt if 2 = x = 4. dy dt +y = −1. 2 if dx dt = 4 and dy dt = 3 . Answers to odd numbered questions 1. 8 3. $\frac{13}{\sqrt{10}}$ Exercise 3.1E. 2 If two electrical resistors are connected in parallel, the total resistance (measured in ohms, denoted by the Greek capital letter omega, Ω is given by the equation $ \frac{1}{R}= \frac{1} {R_1} + \frac{1}{R_2}$ If R is increasing at a rate of 0.5 and R decreases at a rate of 1.1Ω/min , at what rate does the total resistance change when $R_1=20Ω$ and $ R_2 = 50 \frac{Ω}{min} $? Ω 1 min 2 Answer Under Construction Exercise 3.1E. 3 A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the bottom moving along the ground when the bottom of the ladder is 5 ft from the wall? 10/21/2020 3.1E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10891 Answer $ 2\sqrt{3} ft/sec $ Exercise 3.1E. 4 A 25-ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of 1 ft/sec, and the bottom of the ladder is initially $20ft$ away from the wall, how fast does the ladder move up the wall $5sec$ after we start pushing? Answer Under Construction Exercise 3.1E. 5 Two airplanes are flying in the air at the same height: airplane A is flying east at 250 mi/h and airplane B is flying north at 300mi/h. If they are both heading to the same airport, located 30 miles east of airplane A and 40 miles north of airplane B, at what rate is the distance between the airplanes changing? 10/21/2020 3.1E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10891 Answer The distance is decreasing at $390mi/h.$ Exercise 3.1E. 6 You and a friend are riding your bikes to a restaurant that you think is east; your friend thinks the restaurant is north. You both leave from the same point, with you riding at 16 mph east and your friend riding $12mph$ north. After you traveled $4mi,$ at what rate is the distance between you changing? Answer Under Construction Exercise 3.1E. 7 Two buses are driving along parallel freeways that are $5mi$ apart, one heading east and the other heading west. Assuming that each bus drives a constant $55mph$, nd the rate at which the distance between the buses is changing when they are $13mi$ part, heading toward each other. Answer The distance between them shrinks at a rate of $\frac{1320}{13}≈101.5mph.$ Exercise 3.1E. 8 A 6-ft-tall person walks away from a 10-ft lamppost at a constant rate of 3ft/sec. What is the rate that the tip of the shadow moves away from the pole when the person is 10ft away from the pole? 10/21/2020 3.1E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10891 Using the previous problem, what is the rate at which the tip of the shadow moves away from the person when the person is 10 ft from the pole? Answer $\frac{9}{2} ft/sec$ Exercise 3.1E. 9 A 5-ft-tall person walks toward a wall at a rate of 2 ft/sec. A spotlight is located on the ground 40 ft from the wall. How fast does the height of the person’s shadow on the wall change when the person is 10 ft from the wall? Answer Under Construction Exercise 3.1E. 10 Using the previous problem, what is the rate at which the shadow changes when the person is 10 ft from the wall, if the person is walking away from the wall at a rate of 2 ft/sec? Answer It grows at a rate $\frac{4}{9}$ ft/sec Exercise 3.1E. 11 A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/sec. You are running on the ground starting directly under the helicopter at a rate of 10 ft/sec. Find the rate of change of the distance between the helicopter and yourself after 5 sec. Answer The distance is increasing at $(\frac{135\sqrt{26})}{26}$ ft/sec Exercise 3.1E. 12 For the following exercises, draw and label diagrams to help solve the related-rates problems. The side of a cube increases at a rate of m/sec. Find the rate at which the volume of the cube increases when the side of the cube is 4 m. 1 2 10/21/2020 3.1E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10891 The volume of a cube decreases at a rate of 10 m/sec. Find the rate at which the side of the cube changes when the side of the cube is 2 m. Answer $−\frac{5}{6}$ m/sec Exercise 3.1E. 13 The radius of a circle increases at a rate of 2 m/sec. Find the rate at which the area of the circle increases when the radius is 5 m. Answer Under Construction Exercise 3.1E. 14 The radius of a sphere decreases at a rate of 3 m/sec. Find the rate at which the surface area decreases when the radius is 10 m. Answer $240π m^2/sec$ Exercise 3.1E. 15 The radius of a sphere increases at a rate of 1 m/sec. Find the rate at which the volume increases when the radius is 20 m. Answer Under Construction Exercise 3.1E. 16 The radius of a sphere is increasing at a rate of 9 cm/sec. Find the radius of the sphere when the volume and the radius of the sphere are increasing at the same numerical rate. Answer $\frac{1}{2\sqrt{π}}$ cm Exercise 3.1E. 17 A triangle has two constant sides of length 3 ft and 5 ft. The angle between these two sides is increasing at a rate of 0.1 rad/sec. Find the rate at which the area of the triangle is changing when the angle between the two sides is π/6. Answer The area is increasing at a rate $\frac{(3\sqrt{3})}{8}ft_2/sec.$ Exercise 3.1E. 18 10/21/2020 3.1E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10891 A triangle has a height that is increasing at a rate of 2 cm/sec and its area is increasing at a rate of 4 cm /sec . Find the rate at which the base of the triangle is changing when the height of the triangle is 4 cm and the area is 20 cm . 2 2 Answer Under Construction Exercise 3.1E. 19 For the following exercises, consider a right cone that is leaking water. The dimensions of the conical tank are a height of 16 ft and a radius of 5 ft. 1) How fast does the depth of the water change when the water is 10 ft high if the cone leaks water at a rate of 10 f t /min ? 3 2) Find the rate at which the surface area of the water changes when the water is 10 ft high if the cone leaks water at a rate of 10 f t /min . 3 3) If the water level is decreasing at a rate of 3 in./min when the depth of the water is 8 ft, determine the rate at which water is leaking out of the cone. Answers to odd numbered questions 1. The depth of the water decreases at $\frac{128}{125π}$ ft/min. 3. The volume is decreasing at a rate of $\frac{(25π)}{16}ft^3/min.$ Exercise 3.1E. 20 A vertical cylinder is leaking water at a rate of 1 f t /sec . If the cylinder has a height of 10 ft and a radius of 1 ft, at what rate is the height of the water changing when the height is 6 ft? 3 Answer Under Construction Exercise 3.1E. 21 A cylinder is leaking water but you are unable to determine at what rate. The cylinder has a height of 2 m and a radius of 2 m. Find the rate at which the water is leaking out of the cylinder if the rate at which the height is decreasing is 10 cm/min when the height is 1 m Answer Under Construction Exercise 3.1E. 22 A trough has ends shaped like isosceles triangles, with width 3 m and height 4 m, and the trough is 10 m long. Water is being pumped into the trough at a rate of 5m /min . At what rate does the height of the water change when the water is 1 m deep? 3 10/21/2020 3.1E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10891 Answer Under Construction Exercise 3.1E. 23 A tank is shaped like an upside-down square pyramid, with base of 4 m by 4 m and a height of 12 m (see the following figure). How fast does the height increase when the water is 2 m deep if water is being pumped in at a rate of m/sec? 2 3 Answer: $\frac{3}{2} m/sec$ Exercise 3.1E. 24 For the following exercises, consider a pool shaped like the bottom half of a sphere, that is being lled at a rate of 25 f t /min. The radius of the pool is 10 ft. 3 1) Find the rate at which the depth of the water is changing when the water has a depth of 5 ft. 2) Find the rate at which the depth of the water is changing when the water has a depth of 1 ft. 3) If the height is increasing at a rate of 1 in./sec when the depth of the water is 2 ft, nd the rate at which water is being pumped in. Answers to even numbered questions 10/21/2020 3.1E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10891 2. $\frac{25}{19π} ft/min$ Exercise 3.1E. 25 Gravel is being unloaded from a truck and falls into a pile shaped like a cone at a rate of 10 f t /min . The radius of the cone base is three times the height of the cone. Find the rate at which the height of the gravel changes when the pile has a height of 5 ft. 3 Answer $\frac{2}{45π} ft/min$ Exercise 3.1E. 26 Using a similar setup from the preceding problem, nd the rate at which the gravel is being unloaded if the pile is 5 ft high and the height is increasing at a rate of 4 in./min. Answer Under Construction Exercise 3.1E. 27 You are stationary on the ground and are watching a bird y horizontally at a rate of 10 m/sec. The bird is located 40 m above your head. How fast does the angle of elevation change when the horizontal distance between you and the bird is 9 m? Answer The angle decreases at $\frac{400}{1681}rad/sec.$ Exercise 3.1E. 28 You stand 40 ft from a bottle rocket on the ground and watch as it takes off vertically into the air at a rate of 20 ft/sec. Find the rate at which the angle of elevation changes when the rocket is 30 ft in the air. Answer Under Construction Exercise 3.1E. 29 1) A lighthouse, L, is on an island 4 mi away from the closest point, P, on the beach (see the following image). If the lighthouse light rotates clockwise at a constant rate of 10 revolutions/min, how fast does the beam of light move across the beach 2 mi away from the closest point on the beach? 2) Determine at what rate the beam of light moves across the beach 1 mi away from the closest point on the beach. 10/21/2020 3.1E.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10891 Answer to odd numbered questions 1. $100π/min$ Exercise 3.1E. 30 You are walking to a bus stop at a right-angle corner. You move north at a rate of 2 m/sec and are 20 m south of the intersection. The bus travels west at a rate of 10 m/sec away from the intersection – you have missed the bus! What is the rate at which the angle between you and the bus is changing when you are 20 m south of the intersection and the bus is 10 m west of the intersection? Answer The angle is changing at a rate of $\frac{21}{25}rad/sec$. Exercises 3.1E. 31 For the following exercises, refer to the gure of baseball diamond, which has sides of 90 ft. 10/21/2020 3.1E.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10891 1) A batter hits a ball toward third base at 75 ft/sec and runs toward first base at a rate of 24 ft/sec. At what rate does the distance between the ball and the batter change when 2 sec have passed? 2) A batter hits a ball toward second base at 80 ft/sec and runs toward first base at a rate of 30 ft/sec. At what rate does the distance between the ball and the batter change when the runner has covered one-third of the distance to first base? (Hint: Recall the law of cosines.) 3) A batter hits the ball and runs toward first base at a speed of 22 ft/sec. At what rate does the distance between the runner and second base change when the runner has run 30 ft? 4) Runners start at first and second base. When the baseball is hit, the runner at first base runs at a speed of 18 ft/sec toward second base and the runner at second base runs at a speed of 20 ft/sec toward third base. How fast is the distance between runners changing 1 sec after the ball is hit? Answers to even numbered questions 2. The distance is increasing at a rate of 62.50 ft/sec. 4. The distance is decreasing at a rate of 11.99 ft/sec Source Calculus Applets using GeoGebra by Marc Renault is licensed under a Creative Commons Attribution-NonCommercialShareAlike 3.0 Unported License. Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 3.1E.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10891 3.2 Linear approximations and Differentials We have just seen how derivatives allow us to compare related quantities that are changing over time. In this section, we examine another application of derivatives: the ability to approximate functions locally by linear functions. Linear functions are the easiest functions with which to work, so they provide a useful tool for approximating function values. In addition, the ideas presented in this section are generalized later in the text when we study how to approximate functions by higher-degree polynomials Introduction to Power Series and Functions. Linear Approximation of a Function at a Point Consider a function f that is differentiable at a point x = a . Recall that the tangent line to the graph of f at a is given by the equation ′ y = f (a) + f (a)(x − a). (1) For example, consider the function f (x) = at a = 2 . Since f is differentiable at x = 2 and f (2) = − . Therefore, the tangent line to the graph of f at a = 2 is given by the equation 1 x ′ ′ f (x) = − 1 2 x , we see that 1 4 1 y = 1 − 2 (x − 2). (2) 4 Figure(a) shows a graph of f (x) = along with the tangent line to f at x = 2 . Note that for x near 2, the graph of the tangent line is close to the graph of f . As a result, we can use the equation of the tangent line to approximate f (x) for x near 2. For example, if x = 2.1, the y value of the corresponding point on the tangent line is 1 x 1 y = 1 − 2 (2.1 − 2) = 0.475. (3) 4 The actual value of f (2.1) is given by 1 f (2.1) = ≈ 0.47619. (4) 2.1 Therefore, the tangent line gives us a fairly good approximation of f (2.1) (Figure(b)). However, note that for values of x far from 2, the equation of the tangent line does not give us a good approximation. For example, if x = 10, the y -value of the corresponding point on the tangent line is 1 y = 1 − 2 1 (10 − 2) = 4 − 2 = −1.5, (5) 2 whereas the value of the function at x = 10 is f (10) = 0.1. 10/21/2020 1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10892 Figure 1 : (a) The tangent line to f (x) = 1/x at x = 2 provides a good approximation to f for x near 2. (b) At x = 2.1, the value of y on the tangent line to f (x) = 1/x is 0.475. The actual value of f (2.1) is 1/2.1, which is approximately 0.47619. In general, for a differentiable function f , the equation of the tangent line to f at x = a can be used to approximate f (x) for x near a . Therefore, we can write ′ f (x) ≈ f (a) + f (a)(x − a) for x near a . We call the linear function ′ L(x) = f (a) + f (a)(x − a) (6) the linear approximation, or tangent line approximation, of f at x = a . This function L is also known as the linearization of f at x = a. To show how useful the linear approximation can be, we look at how to find the linear approximation for − f (x) = √x at x = 9. − Example 1 : Linear Approximation of √x − − − Find the linear approximation of f (x) = √− x at x = 9 and use the approximation to estimate √9.1. Solution: Since we are looking for the linear approximation at x = 9, using Equation we know the linear approximation is given by ′ L(x) = f (9) + f (9)(x − 9). We need to find f (9) and f ′ (9). – − f (x) = √x ⇒ f (9) = √9 = 3 1 ′ f (x) = 2 √x ′ ⇒ f (9) = 1 2 √9 = 1 6 Therefore, the linear approximation is given by Figure. L(x) = 3 + Using the linear approximation, we can estimate − − − √9.1 1 6 (x − 9) by writing − − − √9.1 = f (9.1) ≈ L(9.1) = 3 + 1 6 (9.1 − 9) ≈ 3.0167. Figure 2 :The local linear approximation to f (x) = √− x at x = 9 provides an approximation to f for x near 9. Analysis − − − Using a calculator, the value of √9.1 to four decimal places is 3.0166. The value given by the linear approximation, 3.0167, is very close to the value obtained with a calculator, so it appears that using this linear approximation is a good way to estimate √− x , at least for x near 9 . At the same time, it may seem odd to use a linear approximation when we can just − − − − − − push a few buttons on a calculator to evaluate √9.1. However, how does the calculator evaluate √9.1? The calculator uses an approximation! In fact, calculators and computers use approximations all the time to evaluate mathematical expressions; they just use higher-degree approximations. Exercise 1 Find the local linear approximation to f (x) = √− x at x = 8 . Use it to approximate 3 10/21/2020 2 CC-BY-NC-SA [ 3 ]8.1 to five decimal places. https://math.libretexts.org/@go/page/10892 Hint ′ L(x) = f (a) + f (a)(x − a) Answer $L(x)=2+\frac{1}{1}2(x−8);$ 2.00833 Example 2 : Linear Approximation of sinx Find the linear approximation of f (x) = sinx at x = and use it to approximate sin(62°). π 3 Solution First we note that since rad is equivalent to 60°, using the linear approximation at x = π/3 seems reasonable. The linear approximation is given by π 3 L(x) = f ( π 3 ′ )+f ( π 3 )(x − π 3 ). We see that f (x) = sinx ⇒ f ( ′ π 3 ′ f (x) = cosx ⇒ f ( ) = sin( π 3 π ) = 3 ) = cos( π 3 √3 2 ) = 1 2 Therefore, the linear approximation of f at x = π/3 is given by Figure. L(x) = To estimate sin(62°) using sin(62°) is given by sin(62°) = f ( L , we must first convert 62π 180 ) ≈ L( 62π 180 ) = √3 2 + 1 2 √3 2 62° ( 62π 180 + 1 2 (x − π 3 ) to radians. We have − π 3 ) = √3 2 + 1 2 ( 2π 180 62° = ) = √3 2 62π 180 + radians, so the estimate for π 180 ≈ 0.88348. Figure 3 : The linear approximation to f (x) = sinx at x = π/3 provides an approximation to sinx for x near π/3. Exercise 2 Find the linear approximation for f (x) = cosx at x = π 2 . Hint ′ L(x) = f (a) + f (a)(x − a) Answer L(x) = −x + π 2 Linear approximations may be used in estimating roots and powers. In the next example, we find the linear approximation for f (x) = (1 + x) at x = 0 , which can be used to estimate roots and powers for real numbers near 1. The same idea can be extended to a function of the form f (x) = (m + x) to estimate roots and powers near a different number m. n n 10/21/2020 3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10892 Example 3 : Approximating Roots and Powers Find the linear approximation of f (x) = (1 + x) at x = 0 . Use this approximation to estimate (1.01) n 3 . Solution The linear approximation at x = 0 is given by ′ L(x) = f (0) + f (0)(x − 0). Because n f (x) = (1 + x ) ′ ⇒ f (0) = 1 n−1 f (x) = n(1 + x ) ′ ⇒ f (0) = n, the linear approximation is given by Figure(a). L(x) = 1 + n(x − 0) = 1 + nx We can approximate (1.01) by evaluating L(0.01) when n = 3 . We conclude that 3 3 (1.01 ) = f (1.01) ≈ L(1.01) = 1 + 3(0.01) = 1.03. Figure 4 : (a) The linear approximation of f (x) at x = 0 is L(x). (b) The actual value of 1.01 is 1.030301. The linear approximation of f (x) at x = 0 estimates 1.01 to be 1.03. 3 3 Exercise 3 Find the linear approximation of f (x) = (1 + x) at x = 0 without using the result from the preceding example. 4 Hint ′ 3 f (x) = 4(1 + x ) Answer L(x) = 1 + 4x Differentials We have seen that linear approximations can be used to estimate function values. They can also be used to estimate the amount a function value changes as a result of a small change in the input. To discuss this more formally, we define a related concept: differentials. Differentials provide us with a way of estimating the amount a function changes as a result of a small change in input values. When we first looked at derivatives, we used the Leibniz notation dy/dx to represent the derivative of y with respect to x. Although we used the expressions dy and dx in this notation, they did not have meaning on their own. Here we see a meaning 10/21/2020 4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10892 to the expressions dy and dx. Suppose y = f (x) is a differentiable function. Let dx be an independent variable that can be assigned any nonzero real number, and define the dependent variable dy by ′ dy = f (x)dx. It is important to notice that dy is a function of both divide both sides of Equation by dx, which yields x and dx dy (7) . The expressions dy and dx are called differentials. We can ′ = f (x). (8) dx This is the familiar expression we have used to denote a derivative. Equation is known as the differential form of Equation. Example 4 : Computing differentials For each of the following functions, find dy and evaluate when x = 3 and dx = 0.1. a. y = x + 2x b. y = cosx 2 Solution The key step is calculating the derivative. When we have that, we can obtain dy directly. a. Since f (x) = x 2 + 2x, we know f ′ (x) = 2x + 2 , and therefore dy = (2x + 2)dx. When x = 3 and dx = 0.1, dy = (2 ⋅ 3 + 2)(0.1) = 0.8. b. Since f (x) = cosx, f ′ (x) = −sin(x). This gives us dy = −sinxdx. When x = 3 and dx = 0.1, dy = −sin(3)(0.1) = −0.1sin(3). Exercise 4 For y = e , find dy . 2 x Hint ′ dy = f (x)dx Answer 2 dy = 2x e x dx We now connect differentials to linear approximations. Differentials can be used to estimate the change in the value of a function resulting from a small change in input values. Consider a function f that is differentiable at point a . Suppose the input x changes by a small amount. We are interested in how much the output y changes. If x changes from a to a + dx , then the change in x is dx (also denoted Δx), and the change in y is given by Δy = f (a + dx) − f (a). (9) Instead of calculating the exact change in y , however, it is often easier to approximate the change in approximation. For x near a, f (x) can be approximated by the linear approximation ′ L(x) = f (a) + f (a)(x − a). y by using a linear (10) Therefore, if dx is small, ′ f (a + dx) ≈ L(a + dx) = f (a) + f (a)(a + dx − a). 10/21/2020 5 CC-BY-NC-SA (11) https://math.libretexts.org/@go/page/10892 That is, ′ f (a + dx) − f (a) ≈ L(a + dx) − f (a) = f (a)dx. (12) In other words, the actual change in the function f if x increases from a to a + dx is approximately the difference between L(a + dx) and f (a) , where L(x) is the linear approximation of f at a . By definition of L(x), this difference is equal to f (a)dx . In summary, ′ ′ Δy = f (a + dx) − f (a) ≈ L(a + dx) − f (a) = f (a)dx = dy. Therefore, we can use the differential dy = f We can see this in the following graph. ′ (a)dx (13) to approximate the change in y if x increases from x = a to x = a + dx . Figure 5 : differential dy=f'(a)dx is used to approximate the actual change in y if x increases from a to a+dx. We now take a look at how to use differentials to approximate the change in the value of the function that results from a small change in the value of the input. Note the calculation with differentials is much simpler than calculating actual values of functions and the result is very close to what we would obtain with the more exact calculation. Example 5 : Approximating Change with Differentials Let y = x 2 + 2x. Compute Δy and dy at x = 3 if dx = 0.1. Solution The actual change in y if x changes from x = 3 to x = 3.1is given by 2 Δy = f (3.1) − f (3) = [(3.1 ) The approximate change in y is given by dy = f ′ (3)dx 2 + 2(3.1)] − [ 3 . Since f ′ + 2(3)] = 0.81. (x) = 2x + 2, we have ′ dy = f (3)dx = (2(3) + 2)(0.1) = 0.8. Exercise 5 For y = x 2 + 2x, find Δy and dy at x = 3 if dx = 0.2. Hint ′ dy = f (3)dx, Δy = f (3.2) − f (3) Answer dy = 1.6, Δy = 1.64 Calculating the Amount of Error Any type of measurement is prone to a certain amount of error. In many applications, certain quantities are calculated based on measurements. For example, the area of a circle is calculated by measuring the radius of the circle. An error in the 10/21/2020 6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10892 measurement of the radius leads to an error in the computed value of the area. Here we examine this type of error and study how differentials can be used to estimate the error. Consider a function f with an input that is a measured quantity. Suppose the exact value of the measured quantity is a , but the measured value is a + dx . We say the measurement error is dx (or Δx). As a result, an error occurs in the calculated quantity f (x). This type of error is known as a propagated error and is given by Δy = f (a + dx) − f (a). (14) Since all measurements are prone to some degree of error, we do not know the exact value of a measured quantity, so we cannot calculate the propagated error exactly. However, given an estimate of the accuracy of a measurement, we can use differentials to approximate the propagated error Δy. Specifically, if f is a differentiable function at a,the propagated error is ′ Δy ≈ dy = f (a)dx. (15) Unfortunately, we do not know the exact value a. However, we can use the measured value a + dx, and estimate ′ Δy ≈ dy ≈ f (a + dx)dx. (16) In the next example, we look at how differentials can be used to estimate the error in calculating the volume of a box if we assume the measurement of the side length is made with a certain amount of accuracy. Example 6 : Volume of a Cube Suppose the side length of a cube is measured to be 5 cm with an accuracy of 0.1 cm. a. Use differentials to estimate the error in the computed volume of the cube. b. Compute the volume of the cube if the side length is (i) 4.9 cm and (ii) 5.1 cm to compare the estimated error with the actual potential error. Solution a. The measurement of the side length is accurate to within ±0.1cm. Therefore, −0.1 ≤ dx ≤ 0.1. The volume of a cube is given by V 3 =x , which leads to 2 dV = 3 x dx. Using the measured side length of 5 cm, we can estimate that 2 2 −3(5 ) (0.1) ≤ dV ≤ 3(5 ) (0.1). Therefore, −7.5 ≤ dV ≤ 7.5. b. If the side length is actually 4.9 cm, then the volume of the cube is 3 V (4.9) = (4.9 ) 3 = 117.649c m . If the side length is actually 5.1 cm, then the volume of the cube is 3 V (5.1) = (5.1 ) 3 = 132.651c m . Therefore, the actual volume of the cube is between 117.649 and 132.651. Since the side length is measured to be 5 cm, the computed volume is V (5) = 5 = 125. Therefore, the error in the computed volume is 3 117.649 − 125 ≤ ΔV ≤ 132.651 − 125. That is, −7.351 ≤ ΔV ≤ 7.651. We see the estimated error dV is relatively close to the actual potential error in the computed volume. Exercise 6 10/21/2020 7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10892 Estimate the error in the computed volume of a cube if the side length is measured to be 6 cm with an accuracy of 0.2 cm. Hint 2 dV = 3 x dx Answer The volume measurement is accurate to within 21.6cm 3 . The measurement error dx (=Δx) and the propagated error Δy are absolute errors. We are typically interested in the size of an error relative to the size of the quantity being measured or calculated. Given an absolute error Δq for a particular quantity, we define the relative error as , where q is the actual value of the quantity. The percentage error is the relative error expressed as a percentage. For example, if we measure the height of a ladder to be 63 in. when the actual height is 62 in., the absolute error is 1 in. but the relative error is = 0.016 , or 1.6. By comparison, if we measure the width of a piece of cardboard to be 8.25 in. when the actual width is 8 in., our absolute error is in., whereas the relative error is = , or 3.1 Therefore, the percentage error in the measurement of the cardboard is larger, even though 0.25 in. is less than 1 in. Δq q 1 62 1 0.25 1 4 8 32 Example 7 : Relative and Percentage Error An astronaut using a camera measures the radius of Earth as 4000 mi with an error of ±80 mi. Let’s use differentials to estimate the relative and percentage error of using this radius measurement to calculate the volume of Earth, assuming the planet is a perfect sphere. Solution: If the measurement of the radius is accurate to within ±80, we have −80 ≤ dr ≤ 80. Since the volume of a sphere is given by V =( 4 3 )πr3, we have 2 dV = 4π r dr. Using the measured radius of 4000 mi, we can estimate 2 2 −4π(4000 ) (80) ≤ dV ≤ 4π(4000 ) (80). To estimate the relative error, consider . Since we do not know the exact value of the volume V , use the measured radius r = 4000mi to estimate V . We obtain V ≈ ( )π(4000 ) . Therefore the relative error satisfies dV V 4 3 3 2 2 −4π(4000 ) (80) 3 ≤ 4π(4000 ) /3 dV V 4π(4000 ) (80) ≤ 3 , 4π(4000 ) /3 which simplifies to −0.06 ≤ dV V ≤ 0.06. The relative error is 0.06 and the percentage error is 6 Exercise 7 Determine the percentage error if the radius of Earth is measured to be 3950 mi with an error of ±100 mi. Hint Use the fact that dV 2 = 4π r dr to find dV /V . Answer 7.6% Key Concepts 10/21/2020 8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10892 A differentiable function y = f (x) can be approximated at a by the linear function ′ L(x) = f (a) + f (a)(x − a). For a function y = f (x), if x changes from a to a + dx , then ′ dy = f (x)dx is an approximation for the change in y . The actual change in y is Δy = f (a + dx) − f (a). A measurement error dx can lead to an error in a calculated quantity f (x). The error in the calculated quantity is known as the propagated error. The propagated error can be estimated by ′ dy ≈ f (x)dx. To estimate the relative error of a particular quantity q, we estimate Δq q . Key Equations Linear approximation ′ L(x) = f (a) + f (a)(x − a) A differential ′ dy = f (x)dx Glossary differential the differential dx is an independent variable that can be assigned any nonzero real number; the differential to be dy = f (x)dx dy is defined ′ differential form given a differentiable function respect to x ′ y = f (x), linear approximation the linear function L(x) = f (a) + f ′ the equation (a)(x − a) ′ dy = f (x)dx is the differential form of the derivative of y with is the linear approximation of f at x = a percentage error the relative error expressed as a percentage propagated error the error that results in a calculated quantity f (x) resulting from a measurement error dx relative error given an absolute error Δq for a particular quantity, Δq q is the relative error. tangent line approximation (linearization) since the linear approximation of f at x = a is defined using the equation of the tangent line, the linear approximation of f at x = a is also known as the tangent line approximation to f at x = a Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10892 3.2E: Exercises Exercise 3.2E. 1 For the following exercises, find the linear approximation L(x) to y = f (x) near x = a for the function. 1) f (x) = x + x 4 2) f (x) = 1 x ,a =0 ,a =2 3) f (x) = tanx, a = π 4) f (x) = sinx, a = π 4 2 5) f (x) = xsinx, a = 2π 6) f (x) = si n 2 x, a = 0 Answers to even numbered questions 2. L(x) = 1 2 − 1 4 (x − 2) 4. L(x) = 1 6. L(x) = 0 Exercise 3.2E. 2 For the following exercises, compute the values given within 0.01 by deciding on the appropriate evaluating L(x) = f (a) + f '(a)(x − a). Check your answer using a calculator. f (x) and a , and 1) (2.001) 6 2) sin(0.02) 3) cos(0.03) 4) (15.99) 1/4 5) 1 0.98 6) sin(3.14) Answers to even numbered questions 2. 0.02 4. 1.9996875 6. 0.001593 Exercise 3.2E. 3 For the following exercises, determine the appropriate f (x) and a , and evaluate the numerical error in the linear approximations that follow. L(x) = f (a) + f '(a)(x − a). Calculate 1) (1.01) 3 2) cos(0.01) 3) (sin(0.01)) 2 4) (1.01) −3 5) (1 + 10/21/2020 1 10 10 ) 3.2E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10898 − − − − 6) √8.99 Answers to even numbered questions 2. 1; error, 0.00005 4. 0.97;error, 6. 3 − 1 error, ; 600 0.0006 −7 4.632 × 10 Exercise 3.2E. 4 For the following exercises, find the differential of the function. 1) y = 3x 4 2 +x − 2x + 1 2) y = xcosx − − − − − 3) y = √1 + x 4) y = 2 x +2 x−1 Answers to even numbered questions 2. dy = (cosx − xsinx)dx 4. dy = ( 2 x −2x−2 2 )dx (x−1) Exercise 3.2E. 5 For the following exercises, find the differential and evaluate for the given x and dx. 1) y = 3x − x + 6, x = 2, dx = 0.1 2) y = , x = 1, dx = 0.25 2 1 x+1 3) y = tanx, x = 0, dx = 2 4) y = 3 x +2 5) y = sin(2x) √x+1 , x=0, dx=0.1\) , x = π, dx = 0.25 x 6) y = x 3 π 10 + 2x + 1 x , x = 1, dx = 0.05 Answers to even numbered questions 2. dy = − 1 2 dx, − (x+1) 4. dy = 1 16 2 9 x +12x−2 3/2 dx, −0.1 2(x+1) 6. dy = (3x 2 +2 − 1 2 )dx, 0.2 x Exercise 3.2E. 6 For the following exercises, find the change in volume dV or in surface area dA. 1) dV if the sides of a cube change from 10 to 10.1. 2) dA if the sides of a cube change from x to x + dx . 3) dA if the radius of a sphere changes from r by dr. 4) dV if the radius of a sphere changes from r by dr. 10/21/2020 3.2E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10898 5) dV if a circular cylinder with r = 2 changes height from 3 cm to 3.05cm. 6) dV if a circular cylinder of height 3 changes from r = 2 to r = 1.9cm. Answers to even numbered questions 2. 12xdx 4. 4π r 2 dr 6. −1.2πcm 3 Exercise 3.2E. 7 For the following exercises, use differentials to estimate the maximum and relative error when computing the surface area or volume. 1) A spherical golf ball is measured to have a radius of possible change in volume? 5mm, with a possible measurement error of 0.1mm. What is the 2) A pool has a rectangular base of 10 ft by 20 ft and a depth of 6 ft. What is the change in volume if you only fill it up to 5.5 ft? Answer 3 −100f t 3) An ice cream cone has height 4 in. and radius 1 in. If the cone is 0.1 in. thick, what is the difference between the volume of the cone, including the shell, and the volume of the ice cream you can fit inside the shelf Exercise 3.2E. 8 For the following exercises, confirm the approximations by using the linear approximation at x = 0. − − − − − 1) √1 − x 2) 1 √1−x2 ≈1− 1 2 x ≈1 − − − − − − 2 2 +x ≈c 3) √c Answer Under Construction Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 3.2E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10898 3.3: Extremas Given a particular function, we are often interested in determining the largest and smallest values of the function. This information is important in creating accurate graphs. Finding the maximum and minimum values of a function also has practical significance because we can use this method to solve optimization problems, such as maximizing profit, minimizing the amount of material used in manufacturing an aluminum can, or finding the maximum height a rocket can reach. In this section, we look at how to use derivatives to find the largest and smallest values for a function. Absolute Extrema Consider the function f (x) = x + 1 over the interval (−∞, ∞). As x → ±∞, f (x) → ∞. Therefore, the function does not have a largest value. However, since x + 1 ≥ 1 for all real numbers x and x + 1 = 1 when x = 0 , the function has a smallest value, 1, when x = 0 . We say that 1 is the absolute minimum of f (x) = x + 1 and it occurs at x = 0 . We say that f (x) = x + 1 does not have an absolute maximum (Figure 3.3.1). 2 2 2 2 2 Figure 3.3.1 : The given function has an absolute minimum of 1 at x=0. The function does not have an absolute maximum. Definition: absolute extrema Let f be a function defined over an interval I and let c ∈ I . We say f has an absolute maximum on I at c if f (c) ≥ f (x) for all x ∈ I . We say f has an absolute minimum on I at c if f (c) ≤ f (x) for all x ∈ I . If f has an absolute maximum on I at c or an absolute minimum on I at c , we say f has an absolute extremum on I at c . Before proceeding, let’s note two important issues regarding this definition. First, the term absolute here does not refer to absolute value. An absolute extremum may be positive, negative, or zero. Second, if a function f has an absolute extremum over an interval I at c , the absolute extremum is f (c). The real number c is a point in the domain at which the absolute extremum occurs. For example, consider the function f (x) = 1/(x + 1) over the interval (−∞, ∞). Since 2 1 f (0) = 1 ≥ 2 x = f (x) (3.3.1) +1 for all real numbers x, we say f has an absolute maximum over occurs at x = 0 , as shown in Figure(b). (−∞, ∞) at x = 0 . The absolute maximum is f (0) = 1 . It A function may have both an absolute maximum and an absolute minimum, just one extremum, or neither. Figure 3.3.2 shows several functions and some of the different possibilities regarding absolute extrema. However, the following theorem, called the Extreme Value Theorem, guarantees that a continuous function f over a closed, bounded interval [a, b] has both an absolute maximum and an absolute minimum. 10/21/2020 3.3.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10302 Figure 3.3.2 : Graphs (a), (b), and (c) show several possibilities for absolute extrema for functions with a domain of Graphs (d), (e), and (f) show several possibilities for absolute extrema for functions with a domain that is a bounded interval. (−∞, ∞). Extreme Value Theorem If f is a continuous function over the closed, bounded interval [a, b], then there is a point in [a, b] at which absolute maximum over [a, b] and there is a point in [a, b] at which f has an absolute minimum over [a, b]. f has an The proof of the extreme value theorem is beyond the scope of this text. Typically, it is proved in a course on real analysis. There are a couple of key points to note about the statement of this theorem. For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. If the interval I is open or the function has even one point of discontinuity, the function may not have an absolute maximum or absolute minimum over I . For example, consider the functions shown in Figure(d), (e), and (f). All three of these functions are defined over bounded intervals. However, the function in graph (e) is the only one that has both an absolute maximum and an absolute minimum over its domain. The extreme value theorem cannot be applied to the functions in graphs (d) and (f) because neither of these functions is continuous over a closed, bounded interval. Although the function in graph (d) is defined over the closed interval [0, 4], the function is discontinuous at x = 2 . The function has an absolute maximum over [0, 4] but does not have an absolute minimum. The function in graph (f) is continuous over the half-open interval [0, 2), but is not defined at x = 2 , and therefore is not continuous over a closed, bounded interval. The function has an absolute minimum over [0, 2), but does not have an absolute 10/21/2020 3.3.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10302 maximum over [0, 2). These two graphs illustrate why a function over a bounded interval may fail to have an absolute maximum and/or absolute minimum. Before looking at how to find absolute extrema, let’s examine the related concept of local extrema. This idea is useful in determining where absolute extrema occur. Local Extrema and Critical Points Consider the function f shown in Figure 3.3.3. The graph can be described as two mountains with a valley in the middle. The absolute maximum value of the function occurs at the higher peak, at x = 2 . However, x = 0 is also a point of interest. Although f (0) is not the largest value of f , the value f (0) is larger than f (x) for all x near 0. We say f has a local maximum at x = 0 . Similarly, the function f does not have an absolute minimum, but it does have a local minimum at x = 1 because f (1) is less than f (x) for x near 1. Figure 3.3.3 : This function f has two local maxima and one local minimum. The local maximum at x = 2 is also the absolute maximum. Definition: local extrema A function f has a local maximum at c if there exists an open interval I containing c such that I is contained in the domain of f and f (c) ≥ f (x) for all x ∈ I . A function f has a local minimum at c if there exists an open interval I containing c such that I is contained in the domain of f and f (c) ≤ f (x) for all x ∈ I . A function f has a local extremum at c if f has a local maximum at c or f has a local minimum at c . Note that if f has an absolute extremum at c and f is defined over an interval containing c , then f (c) is also considered a local extremum. If an absolute extremum for a function f occurs at an endpoint, we do not consider that to be a local extremum, but instead refer to that as an endpoint extremum. Given the graph of a function f , it is sometimes easy to see where a local maximum or local minimum occurs. However, it is not always easy to see, since the interesting features on the graph of a function may not be visible because they occur at a very small scale. Also, we may not have a graph of the function. In these cases, how can we use a formula for a function to determine where these extrema occur? To answer this question, let’s look at Figure 3.3.3 again. The local extrema occur at x = 0, x = 1, and x = 2. Notice that at x = 0 and x = 1 , the derivative f (x) = 0 . At x = 2 , the derivative f (x) does not exist, since the function f has a corner there. In fact, if f has a local extremum at a point x = c , the derivative f (c) must satisfy one of the following conditions: either f (c) = 0 or f (c) is undefined. Such a value c is known as a critical point and it is important in finding extreme values for functions. ′ ′ ′ ′ ′ Definition: critical points Let c be an interior point in the domain of f . We say that c is a critical point of f if f 10/21/2020 3.3.3 CC-BY-NC-SA ′ (c) = 0 or f ′ (c) is undefined. https://math.libretexts.org/@go/page/10302 As mentioned earlier, if Fermat’s theorem. f has a local extremum at a point x =c , then c must be a critical point of f . This fact is known as Fermat’s Theorem If f has a local extremum at c and f is differentiable at c , then f ′ (c) = 0. PROOF: Suppose has a local extremum at c and f is differentiable at c. We need to show that f (c) = 0 . To do this, we will show that f (c) ≥ 0 and f (c) ≤ 0 , and therefore f (c) = 0 . Since f has a local extremum at c, f has a local maximum or local minimum at c. Suppose f has a local maximum at c. The case in which f has a local minimum at c can be handled similarly. There then exists an open interval I such that f(c) ≥ f(x) for all x ∈ I . Since f is differentiable at c, from the definition of the derivative, we know that f ′ ′ ′ ′ f(x) − f(c) ′ f (c) = lim x→c . (3.3.2) x −c Since this limit exists, both one-sided limits also exist and equal f ′ (c) . Therefore, f(x) − f(c) ′ f (c) = lim + x→c (3.3.3) x − c, and f(x) − f(c) ′ f (c) = lim − x→c Since near f(c) c is a local maximum, we see that , but x > c , we have Similarly, it can be shown that f f (x)−f (c) x−c ′ (c) ≥ 0. ≤ 0 . (3.3.4) x −c f(x) − f(c) ≤ 0 for x near c. Therefore, for . From Equation we conclude that Therefore, f ′ (c) = 0. ′ x f (c) ≤ 0 . □ From Fermat’s theorem, we conclude that if f has a local extremum at c , then either f words, local extrema can only occur at critical points. ′ (c) = 0 or f ′ (c) is undefined. In other Note this theorem does not claim that a function f must have a local extremum at a critical point. Rather, it states that critical points are candidates for local extrema. For example, consider the function f (x) = x . We have f (x) = 3x = 0 when x = 0 . Therefore, x = 0 is a critical point. However, f (x) = x is increasing over (−∞, ∞) , and thus f does not have a local extremum at x = 0 . In Figure, we see several different possibilities for critical points. In some of these cases, the functions have local extrema at critical points, whereas in other cases the functions do not. Note that these graphs do not show all possibilities for the behavior of a function at a critical point. 3 ′ 2 3 10/21/2020 3.3.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10302 Figure 3.3.4 : (a–e) A function f has a critical point at c if f (c) = 0 or f (c) is undefined. A function may or may not have a local extremum at a critical point. ′ ′ Later in this chapter we look at analytical methods for determining whether a function actually has a local extremum at a critical point. For now, let’s turn our attention to finding critical points. We will use graphical observations to determine whether a critical point is associated with a local extremum. Example 3.3.1 : Locating Critical Points For each of the following functions, find all critical points. Use a graphing utility to determine whether the function has a local extremum at each of the critical points. a. f (x) = x − x b. f (x) = (x − 1) c. f (x) = 1 3 3 5 2 2 2 + 4x 3 4x 2 1+x Solution a. The derivative f (x) = x − 5x + 4 is defined for all real numbers x. Therefore, we only need to find the values for x where f (x) = 0 . Since f (x) = x − 5x + 4 = (x − 4)(x − 1) , the critical points are x = 1 and x = 4. From the graph of f in Figure, we see that f has a local maximum at x = 1 and a local minimum at x = 4 . ′ ′ 2 ′ 2 Figure 3.3.5 : This function has a local maximum and a local minimum. b. Using the chain rule, we see the derivative is ′ 2 f (x) = 3(x 10/21/2020 2 2 − 1 ) (2x) = 6x(x 2 − 1) . 3.3.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10302 Therefore, f has critical points when x = 0 and when x − 1 = 0 . We conclude that the critical points are x = 0, ±1. From the graph of f in Figure, we see that f has a local (and absolute) minimum at x = 0 , but does not have a local extremum at x = 1 or x = −1 . 2 Figure 3.3.6 : This function has three critical points: x=0, x=1, and x=−1. The function has a local (and absolute) minimum at x=0, but does not have extrema at the other two critical points. c. By the chain rule, we see that the derivative is 2 ′ 2 (1+x 4)−4x(2x) f (x) = 2 2 (1+x ) = 4−4x 2 2 . (1+x ) The derivative is defined everywhere. Therefore, we only need to find values for x where f (x) = 0 . Solving f (x) = 0 , we see that 4 − 4x = 0, which implies x = ±1 . Therefore, the critical points are x = ±1 . From the graph of f in Figure, we see that f has an absolute maximum at x = 1 and an absolute minimum at x = −1. Hence, f has a local maximum at x = 1 and a local minimum at x = −1 . (Note that if f has an absolute extremum over an interval I at a point c that is not an endpoint of I , then f has a local extremum at c. ) ′ ′ 2 Figure 3.3.7 : This function has an absolute maximum and an absolute minimum. Exercise 3.3.1 Find all critical points for f (x) = x 3 − 1 2 2 x − 2x + 1. HInt Calculate f ′ (x). Answer x = −23, x = 1 Locating Absolute Extrema The extreme value theorem states that a continuous function over a closed, bounded interval has an absolute maximum and an absolute minimum. As shown in Figure, one or both of these absolute extrema could occur at an endpoint. If an absolute extremum does not occur at an endpoint, however, it must occur at an interior point, in which case the absolute extremum is a local extremum. Therefore, by Note, the point c at which the local extremum occurs must be a critical point. We summarize this result in the following theorem. 10/21/2020 3.3.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10302 Location of Absolute Extrema Let f be a continuous function over a closed, bounded interval I . The absolute maximum of minimum of f over I must occur at endpoints of I or at critical points of f in I . f over I and the absolute With this idea in mind, let’s examine a procedure for locating absolute extrema. Problem-Solving Strategy: Locating Absolute Extrema over a Closed Interval Consider a continuous function f defined over the closed interval [a, b]. 1. Evaluate f at the endpoints x = a and x = b. 2. Find all critical points of f that lie over the interval (a, b) and evaluate f at those critical points. 3. Compare all values found in (1) and (2). From Note, the absolute extrema must occur at endpoints or critical points. Therefore, the largest of these values is the absolute maximum of f . The smallest of these values is the absolute minimum of f . Now let’s look at how to use this strategy to find the absolute maximum and absolute minimum values for continuous functions. Example 3.3.2 : Locating Absolute Extrema For each of the following functions, find the absolute maximum and absolute minimum over the specified interval and state where those values occur. a. f (x) = −x + 3x − 2 over [1, 3]. b. f (x) = x − 3x over [0, 2]. 2 2 2/3 Solution a. Step 1. Evaluate f at the endpoints x = 1 and x = 3 . f (1) = 0 and f (3) = −2 Step 2. Since f (x) = −2x + 3, f is defined for all real numbers x. Therefore, there are no critical points where the derivative is undefined. It remains to check where f (x) = 0 . Since f (x) = −2x + 3 = 0 at x = and is in the interval [1, 3], f ( ) is a candidate for an absolute extremum of f over [1, 3]. We evaluate f ( ) and find ′ ′ ′ ′ 3 3 2 2 f( 3 2 ) = 1 4 3 3 2 2 . Step 3. We set up the following table to compare the values found in steps 1 and 2. x f (x) 0 0 3 1 2 4 3 −2 Conclusion Absolute maximum Absolute minimum From the table, we find that the absolute maximum of f over the interval [1, 3] is , and it occurs at x = minimum of f over the interval [1, 3] is −2, and it occurs at x = 3 as shown in Figure 3.3.8. 10/21/2020 3.3.7 CC-BY-NC-SA 1 3 4 2 . The absolute https://math.libretexts.org/@go/page/10302 Figure 3.3.8 : This function has both an absolute maximum and an absolute minimum. b. Step 1. Evaluate f at the endpoints x = 0 and x = 2 . f (0) = 0 and f (2) = 4 − 3 [ 3 ]4 ≈ −0.762 Step 2. The derivative of f is given by ′ 2 f (x) = 2x − 1/3 4/3 = x 2x −2 1/3 x for x ≠ 0 . The derivative is zero when 2x − 2 = 0 , which implies x = ±1 . The derivative is undefined at x = 0 . Therefore, the critical points of f are x = 0, 1, −1. The point x = 0 is an endpoint, so we already evaluated f (0) in step 1. The point x = −1 is not in the interval of interest, so we need only evaluate f (1). We find that 4/3 f (1) = −2. Step 3. We compare the values found in steps 1 and 2, in the following table. x f (x) Conclusion 0 0 Absolute maximum 1 −2 Absolute minimum 2 −0.762 We conclude that the absolute maximum of f over the interval [0, 2] is zero, and it occurs at x = 0 . The absolute minimum is −2, and it occurs at x = 1 as shown in Figure 3.3.9. Figure 3.3.9 : This function has an absolute maximum at an endpoint of the interval. 10/21/2020 3.3.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10302 Exercise 3.3.2 Find the absolute maximum and absolute minimum of (x) = x 2 − 4x + 3 over the interval [1, 4]. Hint Look for critical points. Evaluate f at all critical points and at the endpoints. Answer The absolute maximum is 3 and it occurs at x = 4 . The absolute minimum is −1 and it occurs at x = 2 . Example 3.3.3 : Find the absolute maximum and absolute minimum values for the functions over the specified domain: 3x − −−−− − , x ∈ [−1, 1] √4 x2 + 1 1. f (x) = . 2. f (x) = 2xe −x/2 3. f (x) = x − on [0, 6]. sin(2x) on [0, π] Solution: 1. Critical points: Since f (x) = 3x − −−−− − √4 x2 + 1 , by using the quotient rule we have, (3x)(8x) − −−−− − 2 3 √4 x + 1 − − −−−− − √4 x2 + 1 ′ f (x) = 2 4x 2 +1 2 3(4 x + 1) − 24 x = 3 2 (4 x 2 12 x + 1) 2 2 + 3 − 24 x = 3 2 (4 x 2 −12 x − 1) +3 = 3 2 (4 x − 1) 2 . 2 The derivative is zero when −12x 2 2 + 3 = −3(4 x Therefore, the critical points of f are x = 1 − 1) = 0, which implies x = ± . 2 1 1 ,− 2 . 2 We compare the values in the following table:x Conclusion f (x) x −3 −1 – √5 1 −3 2 – 2 √2 − 1 3 2 – 2 √2 = = −3 absolute minimum – √8 3 – √8 3 1 10/21/2020 absolute maximum – √5 3.3.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10302 We conclude that the absolute maximum of minimum is −3 – √8 f over the interval [-1, 1] is 3 – √5 , and it occurs at x =1 . The absolute 1 , and it occurs at x = − . 2 2. Critical points: At this point, we know how to locate absolute extrema for continuous functions over closed intervals. We have also defined local extrema and determined that if a function f has a local extremum at a point c , then c must be a critical point of f . However, c being a critical point is not a sufficient condition for f to have a local extremum at c . Later in this chapter, we show how to determine whether a function actually has a local extremum at a critical point. First, however, we need to introduce the Mean Value Theorem, which will help as we analyze the behavior of the graph of a function. Key Concepts A function may have both an absolute maximum and an absolute minimum, have just one absolute extremum, or have no absolute maximum or absolute minimum. If a function has a local extremum, the point at which it occurs must be a critical point. However, a function need not have a local extremum at a critical point. A continuous function over a closed, bounded interval has an absolute maximum and an absolute minimum. Each extremum occurs at a critical point or an endpoint. Glossary absolute extremum if f has an absolute maximum or absolute minimum at c, we say f has an absolute extremum at c absolute maximum if f (c) ≥ f (x) for all x in the domain of f , we say f has an absolute maximum at c absolute minimum if f (c) ≤ f (x) for all x in the domain of f , we say f has an absolute minimum at c critical point if f (c) = 0 or f ′ ′ (c) is undefined, we say that c is a critical point of f extreme value theorem if f is a continuous function over a finite, closed interval, then f has an absolute maximum and an absolute minimum Fermat’s theorem if f has a local extremum at c, then c is a critical point of f local extremum if f has a local maximum or local minimum at c, we say f has a local extremum at c local maximum if there exists an interval I such that f (c) ≥ f (x) for all x ∈ I , we say f has a local maximum at c local minimum if there exists an interval I such that f (c) ≤ f (x) for all x ∈ I , we say f has a local minimum at c Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 3.3.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10302 Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 3.3.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10302 3.3E: Exercises Exercise 3.3E . 1 In precalculus, you learned a formula for the position of the maximum or minimum of a quadratic equation y = ax + bx + c , which was m = − . Prove this formula using calculus. 2 b (2a) Answer Under Construction Exercise 3.3E . 2 If you are finding an absolute minimum over an interval [a, b], why do you need to check the endpoints? Draw a graph that supports your hypothesis. Answer Under Construction Exercise 3.3E . 3 If you are examining a function over an interval (a, b), for a and b finite, is it possible not to have an absolute maximum or absolute minimum? Answer Under Construction Exercise 3.3E . 4 When you are checking for critical points, explain why you also need to determine points where f (x) is undefined. Draw a graph to support your explanation. Answer Under Construction Exercise 3.3E . 5 Can you have a finite absolute maximum for y = ax arguments. 2 + bx + c over (−∞, ∞)? Explain why or why not using graphical Answer Under Construction Exercise 3.3E . 6 Can you have a finite absolute maximum for why or why not using graphical arguments. 3 y = ax 2 + bx + cx + d over (−∞, ∞) assuming a is non-zero? Explain Answer No 10/21/2020 3.3E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10901 Exercise 3.3E . 7 Let m be the number of local minima and M be the number of local maxima. Can you create a function where ? Draw a graph to support your explanation. M > m +2 Answer Under Construction Exercise 3.3E . 8 Is it possible to have more than one absolute maximum? Use a graphical argument to prove your hypothesis. Answer Since the absolute maximum is the function (output) value rather than the x value, the answer is no Exercise 3.3E . 9 Is it possible to have no absolute minimum or maximum for a function? If so, construct such a function. If not, explain why this is not possible. Answer Under Construction Exercise 3.3E . 10 Graph the function y = e absolute maximum? ax . For which values of a , on any infinite domain, will you have an absolute minimum and Answer When a = 0 Exercise 3.3E . 11 For the following exercises, determine where the local and absolute maxima and minima occur on the graph given. Assume domains are closed intervals unless otherwise specified. 1) 2) 10/21/2020 3.3E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10901 3) 4) Answers to even numbered questions 2. Absolute minimum at 3; Absolute maximum at −2.2; local minima at −2, 1; local maxima at −1, 2 4. Absolute minima at −2, 2; absolute maxima at −2.5, 2.5; local minimum at 0; local maxima at −1, 1 Exercise 3.3E . 12 For the following problems, draw graphs of properties: f (x), which is continuous, over the interval [−4, 4] with the following 1) Absolute maximum at x = 2 and absolute minima at x = ±3 2) Absolute minimum at x = 1 and absolute maximum at x = 2 3) Absolute maximum at x = 4, absolute minimum at x = −1, local maximum at x = −2, and a critical point that is not a maximum or minimum at x = 2 4) Absolute maxima at x = 2 and x = −3 , local minimum at x = 1 , and absolute minimum at x = 4 Answer Under Construction 10/21/2020 3.3E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10901 Exercise 3.3E . 13 For the following exercises, find the critical points in the domains of the following functions. 1) y = 4x 3 2) − 3x − 2 y = 4 √x − x 3) y = 1 x−1 4) y = ln(x − 2) 5) y = tan(x) − −−− − 6) y = √4 − x 2 7) y = x 3/2 8) y = 5/2 − 3x 2 x −1 2 x +2x−3 9) y = si n 2 (x) 10) y = x + 1 x Answers to even numbered questions 2. x = 1 4. None 6. x = 0 8. None 10. x = −1, 1 Exercise 3.3E . 14 For the following exercises, find the local and/or absolute maxima for the functions over the specified domain. 1) f (x) = x 2 + 2) y = x 2 + 3) y = (x − x 5) 1 2 (x−x ) over [−1, 4] over [1, 4] x 2 4) y = 3 2 over [−1, 1] 2 ) over [0, 1] − − − − − y = √9 − x over [1, 9] 6) y = x + sin(x) over [0, 2π] 7) y = x 1+x over [0, 100] 8) y = |x + 1| + |x − 1| over [−3, 2] − − 9) y = √− x − √x 3 over [0, 4] 10) y = sinx + cosx over [0, 2π] 11) y = 4sinθ − 3cosθ over [0, 2π] Answers to even numbered questions 2. Absolute maximum: x = 4, y = 4. Absolute minimum: x = 10/21/2020 1 2 33 2 ; absolute minimum: x = 1, y = 3 ,y =4 3.3E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10901 6. Absolute maximum: x = 2π, y = 2π; absolute minimum: x = 0, y = 0 8. Absolute maximum: x = −3; absolute minimum: −1 ≤ x ≤ 1, y = 2 10. Absolute maximum: x = π 4 – , y = √2 ; absolute minimum: x = 5π 4 – , y = −√2 Exercise 3.3E . 15 For the following exercises, find the local and absolute minima and maxima for the functions over (−∞, ∞). 1) y = x 2 + 4x + 5 2) y = x 3 − 12x 3) y = 3x 4 4) y = x 3 5) y = 6) y = 3 2 + 8x − 18 x 6 (1 − x ) 2 x +x+6 x−1 2 x −1 x−1 Answers to odd numbered questions 1. Absolute minimum: x = −2, y = 1 3. Absolute minimum: x = −3, y = −135; local maximum: x = 0, y = 0 ; local minimum: x = 1, y = −7 – – – – 5. Local maximum: x = 1 − 2√2, y = 3 − 4√2 ; local minimum: x = 1 + 2√2, y = 3 + 4√2 Exercise 3.3E . 16 For the following functions, use a calculator to graph the function and to estimate the absolute and local maxima and minima. Then, solve for them explicitly. − −−− − 1) y = 3x √1 − x 2 2) y = x + sin(x) 3) y = 12x 5 4) y = 5) y = 3 4 + 45 x 3 + 20 x 2 − 90 x − 120x + 3 2 x +6 x −x−30 x−2 √4−x2 √4+x2 Answers to odd numbered questions 1. Absolute maximum: x = √2 2 ,y = 3 2 ; absolute minimum: x = − √2 2 ,y =− 3 2 3. Local maximum: x = −2, y = 59 ; local minimum: x = 1, y = −130 5. Absolute maximum: x = 0, y = 1; absolute minimum: x = −2, 2, y = 0 Exercise 3.3E . 17 A company that produces cell phones has a cost function of C = x − 1200x + 36, 400, where C is cost in dollars and x is number of cell phones produced (in thousands). How many units of cell phone (in thousands) minimizes this cost function? 2 Answer Under Construction 10/21/2020 3.3E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10901 Exercise 3.3E . 18 A ball is thrown into the air and its position is given by h(t) = −4.9t ascending. How long after it is thrown does this happen? 2 + 60t + 5m. Find the height at which the ball stops Answer h = 9245 49 m, t = 300 49 s Exercise 3.3E . 19 For the following exercises, consider the production of gold during the California gold rush (1848–1888). The production of gold can be modeled by (25t) G(t) = 2 ( t +16) , where t is the number of years since the rush began (0 ≤ t ≤ 40) and G is ounces of gold produced (in millions). A summary of the data is shown in the following figure. 1) Find when the maximum (local and global) gold production occurred, and the amount of gold produced during that maximum. 2) Find when the minimum (local and global) gold production occurred. What was the amount of gold produced during this minimum? Answer to even numbered questions 2. The global minimum was in 1848, when no gold was produced. Exercise 3.3E . 20 Find the critical points, maxima, and minima for the following piecewise functions. 1) y = { x − 4x 2) y = { x +1 2 2 2 0 ≤ x ≤ 1// x 2 x ≤ 1// x −4 − 4x + 5 1 <x ≤2 x >1 Answers to even numbered questions 2. Absolute minima: x = 0, x = 2, y = 1 ; local maximum at x = 1, y = 2 Exercise 3.3E . 21 For the following exercises, find the critical points of the following generic functions. Are they maxima, minima, or neither? State the necessary conditions. 1) y = ax 2 10/21/2020 + bx + c, given that a > 0 3.3E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10901 2) y = (x − 1) , given that a > 1 a Answers to even numbered questions 2. No maxima/minima if a is odd, minimum at x = 1 if a is even 10/21/2020 3.3E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10901 3.4 The Mean Value Theorem The Mean Value Theorem is one of the most important theorems in calculus. We look at some of its implications at the end of this section. First, let’s start with a special case of the Mean Value Theorem, called Rolle’s theorem. Rolle’s Theorem Informally, Rolle’s theorem states that if the outputs of a differentiable function f are equal at the endpoints of an interval, then there must be an interior point c where f (c) = 0 . Figure illustrates this theorem. ′ Figure 1 : If a differentiable function f satisfies f(a)=f(b), then its derivative must be zero at some point(s) between a and b. Rolle’s Theorem Let f be a continuous function over the closed interval [a, b] and differentiable over the open interval . There then exists at least one c ∈ (a, b) such that f (c) = 0. (a, b) such that ′ f (a) = f (b) PROOF: Let k = f(a) = f(b). We consider three cases: f (x) = k for all x ∈ (a, b). There exists x ∈ (a, b) such that f (x) > k. There exists x ∈ (a, b) such that f (x) < k. Case 2: Since f is a continuous function over the closed, bounded interval [a, b], by the extreme value theorem, it has an absolute maximum. Also, since there is a point x ∈ (a, b) such that f(x) > k , the absolute maximum is greater than k . Therefore, the absolute maximum does not occur at either endpoint. As a result, the absolute maximum must occur at an interior point c ∈ (a, b) . Because f has a maximum at an interior point c , and f is differentiable at (c\), by Fermat’s theorem, f (c) = 0. ′ Case 3: The case when there exists a point x ∈ (a, b) such that f(x) < k is analogous to case 2, with maximum replaced by minimum. Case 1: If f(x) = 0 for all x ∈ (a, b) , then f ′ (x) = 0 for all x ∈ (a, b). □ An important point about Rolle’s theorem is that the differentiability of the function f is critical. If f is not differentiable, even at a single point, the result may not hold. For example, the function f(x)=|x|−1 is continuous over [−1, 1] and f (−1) = 0 = f (1) , but f (c) ≠ 0 for any c ∈ (−1, 1) as shown in the following figure. ′ 10/21/2020 1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10902 Figure 2 : Since f (x) = |x| − 1 is not differentiable at x = 0 , the conditions of Rolle’s theorem are not satisfied. In fact, the conclusion does not hold here; there is no c ∈ (−1, 1) such that f (c) = 0. ′ Let’s now consider functions that satisfy the conditions of Rolle’s theorem and calculate explicitly the points c where ′ f (c) = 0. Example 1 : Using Rolle’s Theorem For each of the following functions, verify that the function satisfies the criteria stated in Rolle’s theorem and find all values c in the given interval where f (c) = 0. ′ a. f (x) = x b. f (x) = x 2 3 + 2x − 4x over [−2, 0] over [−2, 2] Solution Since f is a polynomial, it is continuous and differentiable everywhere. In addition, f (−2) = 0 = f (0). Therefore, f satisfies the criteria of Rolle’s theorem. We conclude that there exists at least one value c ∈ (−2, 0) such that f (c) = 0 . Since f (x) = 2x + 2 = 2(x + 1), we see that f (c) = 2(c + 1) = 0 implies c = −1 as shown in the following graph. ′ ′ ′ Figure 3 : This function is continuous and differentiable over [−2,0], f'(c)=0 when c=−1. b. As in part a. f is a polynomial and therefore is continuous and differentiable everywhere. Also, f (−2) = 0 = f (2). That said, f satisfies the criteria of Rolle’s theorem. Differentiating, we find that f (x) = 3x − 4. Therefore, f (c) = 0 when x =± . Both points are in the interval [−2, 2], and, therefore, both points satisfy the conclusion of Rolle’s theorem as ′ 2 ′ 2 √3 shown in the following graph. 10/21/2020 2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10902 Figure 4 : For this polynomial over [−2, 2], f ′ (c) = 0 – at x = ±2/√3 . Exercise 1 Verify that the function f (x) = 2x − 8x + 6 defined over the interval Find all points c guaranteed by Rolle’s theorem. 2 [1, 3] satisfies the conditions of Rolle’s theorem. Hint Find all values c, where f ′ (c) = 0 . Answer c =2 The Mean Value Theorem and Its Meaning Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions f that are zero at the endpoints. The Mean Value Theorem generalizes Rolle’s theorem by considering functions that are not necessarily zero at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s theorem (Figure). The Mean Value Theorem states that if f is continuous over the closed interval [a, b] and differentiable over the open interval (a, b), then there exists a point c ∈ (a, b) such that the tangent line to the graph of f at c is parallel to the secant line connecting (a, f (a)) and (b, f (b)). 10/21/2020 3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10902 Figure 5 : The Mean Value Theorem says that for a function that meets its conditions, at some point the tangent line has the same slope as the secant line between the ends. For this function, there are two values c and c such that the tangent line to f at c and c has the same slope as the secant line. 1 1 2 2 Mean Value Theorem Let f be continuous over the closed interval [a, b] and differentiable over the open interval (a, b). Then, there exists at least one point c ∈ (a, b) such that f (b) − f (a) ′ f (c) = (1) b −a PROOF: The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem. Consider the line connecting (a, f(a)) and (b, f(b)). Since the slope of that line is f(b) − f(a) (2) b−a and the line passes through the point (a, f(a)), the equation of that line can be written as f(b) − f(a) y = (x − a) + f(a). (3) b−a Let g(x) denote the vertical difference between the point (x, f(x)) and the point (x, y) on that line. Therefore, f(b) − f(a) g(x) = f(x) − [ (x − a) + f(a)]. (4) b−a Figure 6: The value g(x) is the vertical difference between the point (x,f(x)) and the point (x,y) on the secant line connecting (a,f(a)) and (b,f(b)). Since the graph of f intersects the secant line when x = a and x = b , we see that g(a) = 0 = g(b) . Since f is a differentiable function over (a, b) , g is also a differentiable function over (a, b) . Furthermore, since f is continuous over [a, b], g is also continuous over [a, b]. Therefore, g satisfies the criteria of Rolle’s theorem. Consequently, there exists a point c ∈ (a, b) such that g (c) = 0. Since ′ 10/21/2020 4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10902 ′ f(b) − f(a) ′ g (x) = f (x) − , (5) b−a we see that ′ f(b) − f(a) ′ g (c) = f (c) − . (6) b−a Since g ′ (c) = 0, we conclude that f(b) − f(a) ′ f (c) = . (7) b−a □ In the next example, we show how the Mean Value Theorem can be applied to the function f (x) = √− x over the interval [0, 9]. The method is the same for other functions, although sometimes with more interesting consequences. Example 2 : Verifying that the Mean Value Theorem Applies For f (x) = √− x over the interval [0, 9], show that f satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value c ∈ (0, 9) such that f '(c) is equal to the slope of the line connecting (0, f (0)) and (9, f (9)). Find these values c guaranteed by the Mean Value Theorem. Solution We know that f (x) = √− x is continuous over [0, 9] and differentiable over (0, 9). Therefore, f satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value c ∈ (0, 9) such that f '(c) is equal to the slope of the line connecting (0, f (0)) and (9, f (9)) (Figure). To determine which value(s) of c are guaranteed, first calculate the derivative of f . The derivative f '(x) = . The slope of the line connecting (0, f (0)) and (9, f (9)) is given by 1 (2 √x) – – √9 − √0 f (9) − f (0) = 3 = 9 −0 9 −0 1 = 9 . (8) 3 We want to find c such that f '(c) = . That is, we want to find c such that 1 3 1 1 = 2 √c Solving this equation for c , we obtain c = the endpoints. 10/21/2020 9 4 . (9) 3 . At this point, the slope of the tangent line equals the slope of the line joining 5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10902 Figure 7 : The slope of the tangent line at c=9/4 is the same as the slope of the line segment connecting (0,0) and (9,3). One application that helps illustrate the Mean Value Theorem involves velocity. For example, suppose we drive a car for 1 h down a straight road with an average velocity of 45 mph. Let s(t) and v(t) denote the position and velocity of the car, respectively, for 0 ≤ t ≤ 1 h. Assuming that the position function s(t) is differentiable, we can apply the Mean Value Theorem to conclude that, at some time c ∈ (0, 1), the speed of the car was exactly s(1) − s(0) v(c) = s'(c) = = 45mph. (10) 1 −0 Example 3 : Mean Value Theorem and Velocity If a rock is dropped from a height of 100 ft, its position t seconds after it is dropped until it hits the ground is given by the function s(t) = −16t + 100. 2 a. Determine how long it takes before the rock hits the ground. b. Find the average velocity v of the rock for when the rock is released and the rock hits the ground. c. Find the time t guaranteed by the Mean Value Theorem when the instantaneous velocity of the rock is v avg avg . Solution a. When the rock hits the ground, its position is s(t) = 0. Solving the equation −16t + 100 = 0 for t , we find that t = ± sec . Since we are only considering t ≥ 0 , the ball will hit the ground sec after it is dropped. 2 5 5 2 2 b. The average velocity is given by s(5/2)−s(0) vavg = 5/2−0 = 1−100 5/2 = −40f t/sec. c. The instantaneous velocity is given by the derivative of the position function. Therefore, we need to find a time t such that v(t) = s'(t) = vavg = −40f t/sec. Since s(t) is continuous over the interval [0, 5/2] and differentiable over the interval (0, 5/2), by the Mean Value Theorem, there is guaranteed to be a point c ∈ (0, 5/2) such that s(5/2)−s(0) s'(c) = 5/2−0 = −40. Taking the derivative of the position function s(t) , we find that s'(t) = −32t. Therefore, the equation reduces to s'(c) = −32c = −40. Solving this equation for c , we have c = . Therefore, sec after the rock is dropped, the instantaneous velocity equals the average velocity of the rock during its free fall: −40 ft/sec. 10/21/2020 6 5 5 4 4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10902 Figure 8 : At time t = 5/4 sec, the velocity of the rock is equal to its average velocity from the time it is dropped until it hits the ground. Exercise 2 Suppose a ball is dropped from a height of 200 ft. Its position at time instantaneous velocity of the ball equals its average velocity. t is s(t) = −16t 2 + 200. Find the time t when the Hint First, determine how long it takes for the ball to hit the ground. Then, find the average velocity of the ball from the time it is dropped until it hits the ground. Answer 5 2 √2 sec Corollaries of the Mean Value Theorem Let’s now look at three corollaries of the Mean Value Theorem. These results have important consequences, which we use in upcoming sections. At this point, we know the derivative of any constant function is zero. The Mean Value Theorem allows us to conclude that the converse is also true. In particular, if f '(x) = 0 for all x in some interval I , then f (x) is constant over that interval. This result may seem intuitively obvious, but it has important implications that are not obvious, and we discuss them shortly. Corollary 1: Functions with a Derivative of Zero Let f be differentiable over an interval I . If f '(x) = 0 for all x ∈ I , then f (x) = constant for all x ∈ I . PROOF: Since f is differentiable over I , f must be continuous over I . Suppose f(x) is not constant for all x in I . Then there exist a, b ∈ I , where a ≠ b and f(a) ≠ f(b). Choose the notation so that a < b. Therefore, f (b)−f (a) b−a Since that f ≠ 0. s a differentiable function, by the Mean Value Theorem, there exists f (b)−f (a) f'(c) = Therefore, there exists c ∈ I such that f'(x) = 0 for all x ∈ I . 10/21/2020 b−a f'(c) ≠ 0 7 c ∈ (a, b) such . , which contradicts the assumption that CC-BY-NC-SA https://math.libretexts.org/@go/page/10902 From Note, it follows that if two functions have the same derivative, they differ by, at most, a constant. Corollary 2: Constant Difference Theorem If f and g are differentiable over an interval C. I and f '(x) = g'(x) for all x ∈ I , then f (x) = g(x) + C for some constant PROOF: Let h(x) = f(x) − g(x). Then, h'(x) = f'(x) − g'(x) = 0 for all x ∈ I. By Corollary 1, there is a constant C such that h(x) = C for all x ∈ I . Therefore, f(x) = g(x) + C for all x ∈ I. □ The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall that a function f is increasing over I if f (x ) < f (x ) whenever x < x , whereas f is decreasing over I if f (x ) > f (x ) whenever x < x . Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing; if the derivative is negative, then the function is decreasing (Figure). We make use of this fact in the next section, where we show how to use the derivative of a function to locate local maximum and minimum values of the function, and how to determine the shape of the graph. 1 1 2 1 2 1 2 2 This fact is important because it means that for a given function f , if there exists a function F such that F '(x) = f (x); then, the only other functions that have a derivative equal to f are F (x) + C for some constant C . We discuss this result in more detail later in the chapter. Figure 9 : If a function has a positive derivative over some interval I , then the function increases over that interval I ; if the derivative is negative over some interval I , then the function decreases over that interval I . Corollary 3: Increasing and Decreasing Functions Let f be continuous over the closed interval [a, b] and differentiable over the open interval (a, b). i. If f '(x) > 0 for all x ∈ (a, b), then f is an increasing function over [a, b]. ii. If f '(x) < 0 for all x ∈ (a, b), then f is a decreasing function over [a, b]. PROOF: We will prove i.; the proof of ii. is similar. Suppose f is not an increasing function on I . Then there exist a and b in I such that a < b , but f(a) ≥ f(b) . Since f is a differentiable function over I , by the Mean Value Theorem there exists c ∈ (a, b) such that f(b) − f(a) f'(c) = . (11) b−a 10/21/2020 8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10902 Since f(a) ≥ f(b) , we know that conclude that f(b) − f(a) ≤ 0 . Also, a 0. We f(b) − f(a) f'(c) = ≤ 0. (12) b−a However, f'(x) > 0 for all increasing function over I . x ∈ I . This is a contradiction, and therefore f must be an □ Key Concepts If f is continuous over [a, b] and differentiable over (a, b) and f (a) = 0 = f (b) , then there exists a point c ∈ (a, b) such that f '(c) = 0. This is Rolle’s theorem. If f is continuous over [a, b] and differentiable over (a, b), then there exists a point c ∈ (a, b) such that f (b)−f (a) ′ f (c) = b−a . This is the Mean Value Theorem. If f (x) = 0 over an interval I , then f is constant over I . If two differentiable functions f and g satisfy f '(x) = g'(x) over I , then f (x) = g(x) + C for some constant C . If f '(x) > 0 over an interval I , then f is increasing over I . If f '(x) < 0 over I , then f is decreasing over I . ′ Glossary mean value theorem if f is continuous over [a, b] and differentiable over (a, b) , then there exists c ∈ (a, b) such that f (b)−f (a) f '(c) = b−a rolle’s theorem if f is continuous over [a, b] and differentiable over (a, b) , and if f (a) = f (b) , then there exists c ∈ (a, b) such that f '(c) = 0 Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10902 3.4E: Exercises Exercise 3.4E. 1 1)Why do you need continuity to apply the Mean Value Theorem? Construct a counterexample. 2) Why do you need differentiability to apply the Mean Value Theorem? Find a counterexample. 3) When are Rolle’s theorem and the Mean Value Theorem equivalent? 4) If you have a function with a discontinuity, is it still f'(c)(b − a) = f(b) − f(a)? Draw such an example or prove why not. possible to have Answers 2. One example is f(x) = |x| + 3, −2 ≤ x ≤ 2 , 4. Yes, but the Mean Value Theorem still does not apply Exercise 3.4E. 2 For the following exercises, determine over what intervals (if any) the Mean Value Theorem applies. Justify your answer. 1) y = sin(πx) 2) y = 1 3 x − −−− − 3) y = √4 − x 2 − −−− − 2 −4 4) y = √x 5) y = ln(3x − 5) Answers 2. (−∞, 0), (0, ∞) , 4. (−∞, −2), (2, ∞) Exercise 3.4E. 3 For the following exercises, graph the functions on a calculator and draw the secant line that connects the endpoints. Estimate the number of points c such that f'(c)(b − a) = f(b) − f(a). 1) y = 3x 3 + 2x + 1 2) y = tan( 3) y = x 2 4) y = x 6 π 4 x) over [− cos(πx) − 3 4 5 x over [−1, 1] 3 2 , 3 2 ] over [−2, 2] − 9 8 4 x + 15 16 3 x + 3 32 2 x + 3 16 x + 1 32 over [−1, 1] Answers 10/21/2020 3.4E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10903 1. 2 points 3. 5 points Exercise 3.4E. 4 For the following exercises, use the Mean Value Theorem and find all points 0 < c < 2 such that f(2) − f(0) = f'(c)(2 − 0) . 1) f(x) = x 3 2) f(x) = sin(πx) 3) f(x) = cos(2πx) 4) f(x) = 1 + x + x 2 5) f(x) = (x − 1) 10 6) f(x) = (x − 1) 9 Answers 1. $c=\frac{2\sqrt{3}}{3}$ 3. $c=\frac{1}{2},1,\frac{3}{2}$, 5. Solution: c = 1 Exercise 3.4E. 5 For the following exercises, show there is no c such that f(1) − f(−1) = f'(c)(2) . Explain why the Mean Value Theorem does not apply over the interval [−1, 1]. 1) f(x) =∣ x − 2) f(x) = 1 2 ∣ ∣ 1 2 x −− 3) f(x) = √|x| 4) f(x) = [x] Hint: This is called the oor function and it is de ned so that f(x) is the largest integer less than or equal to \(x\ Answers 1. Not differentiable 3. Not differentiable Exercise 3.4E. 6 10/21/2020 3.4E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10903 For the following exercises, determine whether the Mean Value Theorem applies for the functions over the given interval \ (([a,b]\. Justify your answer. 1) y = e over [0, 1] x 2) y = ln(2x + 3) over [− 3 2 , 0] 3) f(x) = tan(2πx) over [0, 2] − −−− − 4) y = √9 − x over [−3, 3] 2 5) y = 1 |x+1| 6) y = x 3 7) y = 8) y = over [0, 3] + 2x + 1 2 x +3x+2 x x sin(πx)+1 over [0, 6] over [−1, 1] over [0, 1] 9) y = ln(x + 1) over [0, e − 1] 10) y = xsin(πx) over [0, 2] 11) y = 5 + |x| over [−1, 1] Answer 1, 3, 5, 9. Yes, 7.The Mean Value Theorem does not apply; discontinuous at x = 0. 11. The Mean Value Theorem does not apply; not differentiable at x , = 0 . Exercise 3.4E. 7 For the following exercises, consider the roots of the equation. 1) Show that the equation y = x 3 2 + 3x + 16 has exactly one real root. What is it? 2) Find the conditions for exactly one root (double root) for the equation y = x 2 + bx + c Answer: $b=±2\sqrt{c}$ 3) Find the conditions for y = e x −b to have one root. Is it possible to have more than one root? Answer Under Construction Exercise 3.4E. 8 For the following exercises, use a calculator to graph the function over the interval [a, b] and graph the secant line from a to b . Use the calculator to estimate all values of c as guaranteed by the Mean 10/21/2020 3.4E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10903 Value Theorem. Then, find the exact value of c , if possible, or write the final equation and use a calculator to estimate to four digits. 1) y = tan(πx) over [− 2) y = 2 + 2x − 4∣ ∣ 4) y = x + 4 , 1 4 ] over [0, 3] 1 √x+1 3) y =∣ x 1 over [ 1 x − − − − − 5) y = √x + 1 + 1 2 x 1 2 over [−4, 0] , 4] over [3, 8] Answers to odd numbered questions 1. c = ± 1 π −1 cos ( √π ), c = ±0.1533 2 3. The Mean Value Theorem does not apply. 5. 1 2 √c+1 − 2 c3 = 521 2880 ; c = 3.133, 5.867 Exercise 3.4E. 9 At 10:17 a.m., you pass a police car at 55 mph that is stopped on the freeway. You pass a second police car at 55 mph at 10:53 a.m., which is located 39 mi from the rst police car. If the speed limit is 60 mph, can the police cite you for speeding? Answer Under Construction Exercise 3.4E. 10 Two cars drive from one spotlight to the next, leaving at the same time and arriving at the same time. Is there ever a time when they are going the same speed? Prove or disprove. Answer Yes. Exercise 3.4E. 11 Show that 2 y = sec x 2 2 y = sec x − tan x and 2 y = tan x have the same derivative. What can you say about ? Answer Under Construction Exercise 3.4E. 12 Show that $y=csc^2x$ and (y=csc^2x−cot^2x\)? 10/21/2020 2 y = cot x have the same derivative. What can you say about \ 3.4E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10903 Answer It is constant. Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 3.4E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10903 3.5 Derivative tests Earlier in this chapter we stated that if a function f has a local extremum at a point c , then c must be a critical point of f. However, a function is not guaranteed to have a local extremum at a critical point. For example, f(x) = x has a critical point at x = 0 since f (x) = 3x is zero at x = 0 , but f does not have a local extremum at x = 0 . Using the results from the previous section, we are now able to determine whether a critical point of a function actually corresponds to a local extreme value. In this section, we also see how the second derivative provides information about the shape of a graph by describing whether the graph of a function curves upward or curves downward. 3 ′ 2 The First Derivative Test Corollary 3 of the Mean Value Theorem showed that if the derivative of a function is positive over an interval I then the function is increasing over I . On the other hand, if the derivative of the function is negative over an interval I , then the function is decreasing over I as shown in the following figure.. Figure 1 : Both functions are increasing over the interval (a, b) . At each point x , the derivative f (x) > 0 . Both functions are decreasing over the interval (a, b) . At each point x , the derivative f (x) < 0. ′ ′ A continuous function f has a local maximum at point c if and only if f switches from increasing to decreasing at point c . Similarly, f has a local minimum at c if and only if f switches from decreasing to increasing at c . If f is a continuous function over an interval I containing c and differentiable over I , except possibly at c , the only way f can switch from increasing to decreasing (or vice versa) at point c is if f' changes sign as x increases through c . If f is differentiable at c , the only way that f' . can change sign as x increases through c is if f'(c) = 0 . Therefore, for a function f that is continuous over an interval I containing c and differentiable over I , except possibly at c , the only way f can switch from increasing to decreasing (or vice versa) is if f (c) = 0 or f'(c) is undefined. Consequently, to locate local extrema for a function f , we look for points c in the domain of f such that f (c) = 0 or f'(c) is undefined. Recall that such points are called critical points of f . ′ ′ Note that f need not have a local extrema at a critical point. The critical points are candidates for local extrema only. In Figure, we show that if a continuous function f has a local extremum, it must occur at a critical point, but a function may not have a local extremum at a critical point. We show that if f has a local extremum at a critical point, then the sign of f' switches as x increases through that point. Figure 2 : The function f has four critical points: a, b, c ,and d . The function f has local maxima at a and d , and a local minimum at b . The function f does not have a local extremum at c . The sign of f changes at all local extrema. ′ Using Figure, we summarize the main results regarding local extrema. If a continuous function f has a local extremum, it must occur at a critical point c . The function has a local extremum at the critical point c if and only if the derivative f' switches sign as x increases through c . Therefore, to test whether a function has a local extremum at a critical point c , we must determine the sign of f'(x) to the left and right of c . 10/21/2020 1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10905 This result is known as the first derivative test. FIRST DERIVATIVE TEST Suppose that f is a continuous function over an interval I containing a critical point c . If f is differentiable over I , except possibly at point c , then f(c) satisfies one of the following descriptions: i. If f' changes sign from positive when x < c to negative when x > c, then f(c) is a local maximum of f . ii. If f' changes sign from negative when x < c to positive when x > c, then f(c) is a local minimum of f . iii. If f has the same sign for x < c and x > c, then f(c) is neither a local maximum nor a local minimum of f ′ Now let’s look at how to use this strategy to locate all local extrema for particular functions. EXAMPLE 1 : USING THE FIRST DERIVATIVE TEST TO FIND LOCAL EXTREMA Use the first derivative test to find the location of all local extrema for f(x) = x 3 2 − 3x − 9x − 1. Use a graphing utility to confirm your results. Solution Step 1. The derivative is f'(x) = 3x that the critical points must satisfy 2 − 6x − 9. To find the critical points, we need to find where f'(x) = 0. Factoring the polynomial, we conclude 2 3(x − 2x − 3) = 3(x − 3)(x + 1) = 0. (1) Therefore, the critical points are x= 3, −1. Now divide the interval (−∞, ∞) into the smaller intervals (−∞, −1), (−1, 3) and (3, ∞). Step 2. Since f is a continuous function, to determine the sign of f'(x) over each subinterval, it suffices to choose a point over each of the intervals (−∞, −1), (−1, 3) and (3, ∞) and determine the sign of f' at each of these points. For example, let’s choose x = −2, x = 0, and x = 4 as test points. ′ Interval Test Point Sign of Conclusion f'(x) = 3(x − 3)(x + 1) at Test Point (−∞, −1) x = −2 (+)(−)(−)=+ f is increasing. (−1, 3) x = 0 (+)(−)(+)=+ f is increasing. (3, ∞) x = 4 (+)(+)(+)=+ f is increasing. Step 3. Since f' switches sign from positive to negative as x increases through 1, f has a local maximum at x = −1 . Since f' switches sign from negative to positive as x increases through 3, f has a local minimum at x = 3 . These analytical results agree with the following graph. Figure 3 : The function f has a maximum at x = −1 and a minimum at x = 3 10/21/2020 2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10905 EXERCISE 1 Use the first derivative test to locate all local extrema for f(x) = −x 3 + 3 2 2 x + 18x. Hint Find all critical points of f and determine the signs of f'(x) over particular intervals determined by the critical points. Answer f has a local minimum at −2 and a local maximum at 3. EXAMPLE 2 : USING THE FIRST DERIVATIVE TEST Use the first derivative test to find the location of all local extrema for f(x) = 5x 1/3 5/3 −x . Step 1. The derivative is 5 f'(x) = −2/3 x 5 − 3 2/3 x 3 2/3 5 = 5(1 − x 5 − 5x = 3 3x 4/3 4/3 5x − 2/3 = ) . 2/3 (2) 2/3 3x 3x The derivative f'(x) = 0 when 1 − x = 0. Therefore, f'(x) = 0 at x = ±1 . The derivative f'(x) is undefined at x = 0. Therefore, we have three critical points: x = 0, x = 1, and x = −1 . Consequently, divide the interval (−∞, ∞) into the smaller intervals (−∞, −1), (−1, 0), (0, 1), and (1, ∞) . 4/3 Step 2: Since f' is continuous over each subinterval, it suffices to choose a test point x in each of the intervals from step 1 and determine the sign of f' at each of these points. The points x = −2, x = − , x = , andx = 2 are test points for these intervals. Interval 1 1 2 2 Test Point Sign of f '(x) = 4/3 5(1−x 2/3 3x ) at Conclusion Test Point (−∞, −1) (+)(−) x = −2 + (−1, 0) (0, 1) (1, ∞) x = − x = 1 (+)(+) 2 + (+)(+) 1 2 + (+)(−) x = 2 + = − f is decreasing. = + f is increasing. = + f is increasing. = − f is decreasing. Step 3: Since f is decreasing over the interval (−∞, −1) and increasing over the interval (−1, 0), f has a local minimum at x = −1 . Since f is increasing over the interval (−1, 0) and the interval (0, 1) , f does not have a local extremum at x = 0 . Since f is increasing over the interval (0, 1) and decreasing over the interval (1, ∞), f has a local maximum at x = 1 . The analytical results agree with the following graph. 10/21/2020 3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10905 Figure 4 : The function f has a local minimum at x = −1 and a local maximum at x = 1 EXERCISE 2 Use the first derivative test to find all local extrema for (x) = 3 x−1 . Hint The only critical point of f is x = 1. Answer f has no local extrema because f' does not change sign at x = 1 . Example 3 : Consider the function 2 −x f (x) = e 2 . (3) 1. Find the interval(s) on which $f(x)$ is increasing. 2. Find the interval(s) on which $f(x)$ is decreasing. 3. Find the local (relative) maximum(s) (if any). 4. Find the local (relative) minimum(s) (if any). Solution: Exercise 1 Consider the function f (x) = x e −x . (4) 1. Find the interval(s) on which $f(x)$ is increasing. 2. Find the interval(s) on which $f(x)$ is decreasing. 3. Find the local (relative) maximum(s) (if any). 4. Find the local (relative) minimum(s) (if any). Answer Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page. Concavity and Points of Inflection We now know how to determine where a function is increasing or decreasing. However, there is another issue to consider regarding the shape of the graph of a function. If the graph curves, does it curve upward or curve downward? This notion is called the concavity of the function. Figure(a) shows a function f with a graph that curves upward. As x increases, the slope of the tangent line increases. Thus, since the derivative increases as x increases, f' is an increasing function. We say this function f is concave up. Figure(b) shows a function f that curves downward. As x increases, the slope of the tangent line decreases. Since the derivative decreases as x increases, f' is a decreasing function. We say this function f is concave down. DEFINITION: CONCAVITY TEST Let f be a function that is differentiable over an open interval I . If f' is increasing over I , we say f is concave up over I . If f' is decreasing over I , we say f is concave down over I . 10/21/2020 4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10905 Figure 5 : (a), (c) Since f is increasing over the interval (a, b) , we say f is concave up over (a, b). (b), (d) Since f is decreasing over the interval (a, b) , we say f is concave down over (a, b). ′ ′ In general, without having the graph of a function f, how can we determine its concavity? By definition, a function f is concave up if f' is increasing. From Corollary 3, we know that if f' is a differentiable function, then f' is increasing if its derivative f (x) > 0 . Therefore, a function f that is twice differentiable is concave up when f (x) > 0 . Similarly, a function f is concave down if f' is decreasing. We know that a differentiable function f' is decreasing if its derivative f (x) < 0 . Therefore, a twice-differentiable function f is concave down when f (x) < 0 . Applying this logic is known as the concavity test. ′′ ′′ ′′ ′′ TEST FOR CONCAVITY Let f be a function that is twice differentiable over an interval I . i. If f ii. If f ′′ ′′ (x) > 0 (x) < 0 for all x ∈ I , then f is concave up over I for all x ∈ I, then f is concave down over I . We conclude that we can determine the concavity of a function f by looking at the second derivative of f . In addition, we observe that a function f can switch concavity (Figure). However, a continuous function can switch concavity only at a point x if f (x) = 0 or f (x) is undefined. Consequently, to determine the intervals where a function f is concave up and concave down, we look for those values of x where f (x) = 0 or f (x) is undefined. When we have determined these points, we divide the domain of f into smaller intervals and determine the sign of f ' over each of these smaller intervals. If f changes sign as we pass through a point x , then f changes concavity. It is important to remember that a function f may not change concavity at a point x even if f (x) = 0 or f (x) is undefined. If, however, f does change concavity at a point a and f is continuous at a , we say the point (a, f(a)) is an inflection point of f . ′′ ′′ ′′ ′′ ′ ′′ ′′ ′′ DEFINITION: INFLECTION POINT If f is continuous at a and f changes concavity at a , the point (a, f(a)) is an inflection point of f . Figure 6 : Since f ′′ (x) > 0 for x < a , the function f is concave up over the interval (−∞, a) . Since f (x) < 0 for x > a , the function f is concave down over the interval (a, ∞) . The point (a, f(a)) is an inflection point of f . ′′ EXAMPLE 4 : TESTING FOR CONCAVITY 10/21/2020 5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10905 For the function f(x) = x − 6x + 9x + 30, determine all intervals where inflection points for f . Use a graphing utility to confirm your results. 3 2 f is concave up and all intervals where f is concave down. List all Solution To determine concavity, we need to find the second derivative f (x). The first derivative is f (x) = 3x − 12x + 9, so the second derivative is f (x) = 6x − 12. If the function changes concavity, it occurs either when f (x) = 0 or f (x) is undefined. Since f is defined for all real numbers x , we need only find where f (x) = 0 . Solving the equation 6x − 12 = 0 , we see that x = 2 is the only place where f could change concavity. We now test points over the intervals (−∞, 2) and (2, ∞) to determine the concavity of f . The points x = 0 and x = 3 are test points for these intervals. ′′ ′ ′′ ′′ 2 ′′ ′′ ′′ Interval Test Point Sign of f (x) = 6x − 12 at Test Point (−∞, 2) x = 0 − (2, ∞) x = 3 + Conclusion ′′ We conclude that f is concave down over the interval (−∞, 2) and concave up over the interval point (2, f(2)) = (2, 32) is an inflection point. Figure confirms the analytical results. f is concave down f (2, ∞) . Since f is concave up changes concavity at x = 2 , the Figure 7 : The given function has a point of inflection at (2, 32) where the graph changes concavity. EXERCISE 4 For f(x) = −x 3 + 3 2 2 x + 18x , find all intervals where f is concave up and all intervals where f is concave down. Hint Find where f ′′ (x) = 0 Answer f is concave up over the interval (−∞, 1 2 ) and concave down over the interval ( 1 2 , ∞) Example 5 : Consider the function 2 g(x) = e −x /2 . (5) Find 1. the interval(s) on which g(x) is concave up and 2. the interval(s) on which g(x) is concave down and 10/21/2020 6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10905 3. inflection points ( if any). Exercise 5 Consider the function g(x) = x e −x . (6) Find 1. the interval(s) on which g(x) is concave up and 2. the interval(s) on which g(x) is concave down and 3. inflection points ( if any). Answer Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page. Example 1 : Consider the function g(x) = sec(x) tan(x), x ∈ (−π/2, π/2). (7) Find 1. the interval(s) on which g(x) is concave up and 2. the interval(s) on which g(x) is concave down and 3. inflection points ( if any). We now summarize, in Table, the information that the first and second derivatives of a function information in Figure. f provide about the graph of f , and illustrate this What Derivatives Tell Us about Graphs Sign of f Sign of f ′ Is f increasing or decreasing? ′′ Positive Positive Increasing Positive Negative Increasing Negative Positive Decreasing Negative Negative Decreasing Figure 8 :Consider a twice-differentiable I over an open interval I . If f (x) > 0 for all x ∈ I , the function is increasing over I . If f (x) < 0 for all , the function is decreasing over I . If f (x) > 0 for all x ∈ I , the function is concave up. If f (x) < 0 for all x ∈ I , the function is concave down on I . ′ ′ ′′ x ∈ I ′′ The Second Derivative Test The first derivative test provides an analytical tool for finding local extrema, but the second derivative can also be used to locate extreme values. Using the second derivative can sometimes be a simpler method than using the first derivative. We know that if a continuous function has a local extrema, it must occur at a critical point. However, a function need not have a local extrema at a critical point. Here we examine how the second derivative test can be used to determine whether a function has a local extremum at a critical point. Let f be a twice-differentiable function such that f'(a) = 0 and f is continuous over an open interval I containing a . Suppose f (a) < 0 . Since f is continuous over I, f (x) < 0 for all x ∈ I (Figure). Then, by Corollary 3, f' is a decreasing function over I . Since f'(a) = 0 , we conclude that for all x ∈ I, f'(x) > 0 if x < a and f'(x) < 0 if x > a . Therefore, by the first derivative test, f has a local maximum at x = a . On the other hand, suppose there exists a point b such that f'(b) = 0 but f (b) > 0 . Since f is continuous over an open interval I containing b , then f (x) > 0 for all x ∈ I ′′ ′′ ′′ ′′ ′′ 10/21/2020 ′′ ′′ 7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10905 (Figure). Then, by Corollary 3, f' is an increasing function over I . Since f'(b) = 0 , we conclude that for all x ∈ I , f'(x) < 0 if x 0 if x > b . Therefore, by the first derivative test, f has a local minimum at x = b. Figure 9 : Consider a twice-differentiable function f such that f is continuous. Since f (a) = 0 and f (a) < 0 , there is an interval I containing a such that for all x in I , f is increasing if x < a and f is decreasing if x > a . As a result, f has a local maximum at x = a . Since f (b) = 0 and f (b) > 0 , there is an interval I containing b such that for all x in I , f is decreasing if x b . As a result, f has a local minimum at x = b . ′′ ′ ′′ ′ ′′ SECOND DERIVATIVE TEST Suppose f'(c) = 0, f i. If f ii. If f iii. If f ′′ ′′ ′′ ′′ is continuous over an interval containing c . , then f has a local minimum at c . , then f has a local maximum at c . (c) = 0, then the test is inconclusive. (c) > 0 (c) < 0 Note that for case iii. when f (c) = 0 , then f may have a local maximum, local minimum, or neither at c . For example, the functions f(x) = x , f(x) = x , and f(x) = −x all have critical points at x = 0 . In each case, the second derivative is zero at x = 0 . However, the function f(x) = x has a local minimum at x = 0 whereas the function f(x) = −x has a local maximum at x , and the function f(x) = x does not have a local extremum at x = 0 . ′′ 3 4 4 4 4 3 Let’s now look at how to use the second derivative test to determine whether f has a local maximum or local minimum at a critical point c where f'(c) = 0. EXAMPLE 4 : USING THE SECOND DERIVATIVE TEST Use the second derivative to find the location of all local extrema for f(x) = x 5 3 − 5x . Solution apply the second derivative test, we first need to find critical points f'(x) = 5x − 15x = 5x (x − 3) = 0 when $x=0,±\sqrt{3}\$. 4 2 2 c where f'(c) = 0 . The derivative is 4 f'(x) = 5x 2 − 15x . Therefore, 2 To determine whether f has a local extrema at any of these points, we need to evaluate the sign of f at these points. The second derivative is ′′ f ′′ 3 (x) = 20x 2 − 30x = 10x(2x − 3). In the following table, we evaluate the second derivative at each of the critical points and use the second derivative test to determine whether f has a local maximum or local minimum at any of these points. x f ′′ Conclusion (x) – −√3 – −30√3 Local maximum 0 0 Second derivative test is inconclusive – √3 – 30√3 Local minimum – – By the second derivative test, we conclude that f has a local maximum at x = −√3 and f has a local minimum at x = √3 . The second derivative test is inconclusive at x = 0 . To determine whether f has a local extrema at x = 0, we apply the first derivative test. To evaluate the sign of – – f'(x) = 5x (x − 3) for x ∈ (−√3, 0) and x ∈ (0, √3) , let x = −1 and x = 1 be the two test points. Since f'(−1) < 0 and f'(1) < 0 , we conclude that f is decreasing on both intervals and, therefore, f does not have a local extrema at x = 0 as shown in the following graph. 2 10/21/2020 2 8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10905 – – Figure 10 :The function f has a local maximum at x = −√3 and a local minimum at x = √3 EXERCISE 4 Consider the function f(x) = x − ( )x − 18x . The points c = 3, −2 satisfy f'(c) = 0 . Use the second derivative test to determine whether f has a local maximum or local minimum at those points. 3 3 2 2 Hint f ′′ (x) = 6x − 3 Answer f has a local maximum at −2 and a local minimum at 3. We have now developed the tools we need to determine where a function is increasing and decreasing, as well as acquired an understanding of the basic shape of the graph. In the next section we discuss what happens to a function as → ±∞. At that point, we have enough tools to provide accurate graphs of a large variety of functions. Key Concepts If c is a critical point of f and f '(x) > 0 for x < c and f '(x) < 0 for x > c , then f has a local maximum at c. If c is a critical point of f and f '(x) < 0 for x < c and f '(x) > 0 for x > c, then f has a local minimum at c. If f (x) > 0 over an interval I , then f is concave up over I . If f (x) < 0 over an interval I , then f is concave down over I . If f '(c) = 0 and f (c) > 0 , then f has a local minimum at c. If f '(c) = 0 and f (c) < 0 , then f has a local maximum at c. If f '(c) = 0 and f (c) = 0 , then evaluate f '(x) at a test point x to the left of c and a test point x to the right of c, to determine whether f has a local extremum at c. ′′ ′′ ′′ ′′ ′′ Glossary concave down if f is differentiable over an interval I and f' is decreasing over I , then f is concave down over I concave up if f is differentiable over an interval I and f' is increasing over I , then f is concave up over I concavity the upward or downward curve of the graph of a function concavity test 10/21/2020 9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10905 suppose f is twice differentiable over an interval I ; if f ′′ > 0 over I , then f is concave up over I ; if f ′′ < over I , then f is concave down over I first derivative test let f be a continuous function over an interval I containing a critical point c such that f is differentiable over I except possibly at c; if f' changes sign from positive to negative as x increases through c, then f has a local maximum at c; if f' changes sign from negative to positive as x increases through c, then f has a local minimum at c; if f' does not change sign as x increases through c, then f does not have a local extremum at c inflection point if f is continuous at c and f changes concavity at c, the point (c, f(c)) is an inflection point of f second derivative test suppose f'(c) = 0 and f ' is continuous over an interval containing c; if f maximum at c; if f (c) = 0 , then the test is inconclusive ′ ′′ (c) > 0 , then f has a local minimum at c; if f ′′ (c) < 0 , then f has a local ′′ Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BYSA-NC 4.0 license. Download for free at http://cnx.org. DS — DekiScript — No description Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10905 3.5E: Exercises Exercise 3.5E. 1 1) If c is a critical point of f(x), when is there no local maximum or minimum at c ? Explain. 2) For the function y = x , is x = 0 both an inflection point and a local maximum/minimum? 3 Answer It is not a local maximum/minimum because f' does not change sign 3) For the function y = x , is x = 0 an inflection point? 3 4) Is it possible for a point differentiable function? c to be both an inflection point and a local extrema of a twice Answer No 6) Why do you need continuity for the first derivative test? Come up with an example. 7) Explain whether a concave-down function has to cross y = 0 for some value of x . Answer False; for example, $y=\sqrt{x}$. 8) Explain whether a polynomial of degree 2 can have an inflection point. Answer Under Construction. Exercise 3.5E. 2 For the following exercises, analyze the graphs of f ', then list all intervals where f is increasing or decreasing. 1) 10/21/2020 3.5E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10930 Answer Increasing for −2 < x < −1 and x > 2 ; decreasing for x < −2 and −1 < x < 2 2) 3) 10/21/2020 3.5E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10930 Answer Decreasing for x < 1 , increasing for x > 1 4) 5) Answer Decreasing for −2 < x < −1 and 1 < x < 2 ; increasing for −1 < x < 1 and x < −2 and x > 2 Exercise 3.5E. 3 For the following exercises, analyze the graphs of f', then list all intervals where a. f is increasing and decreasing and b. the minima and maxima are located. 1) 10/21/2020 3.5E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10930 Answer a. Increasing over x < −2, −1 < x < 0, 1 < x < 2; x = 0 decreasing over b. maxima at x = −1 and x = 1 , minima at x = −2 and −2 < x < −1, 0 < x < 1, x > 2 , and x = 2 2) 3) Answer a. Increasing over x > 0 , decreasing over x < 0; b. Minimum at x = 0 4) 10/21/2020 3.5E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10930 Exercise 3.5E. 4 For the following exercises, analyze the graphs of f' , then list all inflection points and intervals f that are concave up and concave down. 1) Answer Concave up on all x , no in ection points 2) 10/21/2020 3.5E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10930 3) Answer Concave up on all x , no in ection points 4) 5) Answer 10/21/2020 3.5E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10930 Concave up for x < 0 and x > 1 , concave down for 0 < x < 1 , in ection points at x = 0 and x = 1 Exercise 3.5E. 5 For the following exercises, draw a graph that satisfies the given specifications for the domain x = [−3, 3]. The function does not have to be continuous or differentiable. 1) f(x) > 0, f'(x) > 0 over x > 1, −3 < x < 0, f'(x) = 0 over 0 < x < 1 2) f'(x) > 0 over x > 2, −3 < x < −1, f'(x) < 0 over −1 < x < 2, f 3) f ′′ over −1 < x < 1, f local minima at x = ±2 (x) < 0 x = 0, ′′ ′′ (x) < 0 (x) > 0, −3 < x < −1, 1 < x < 3, for all x local maximum at 4) There is a local maximum at x = 2, local minimum at x = 1, and the graph is neither concave up nor concave down. 5) There are local maxima at remains positive for all x. x = ±1, the function is concave up for all x , and the function Answer Under Construction Exercise 3.5E. 6 For the following exercises, determine a. intervals where f is increasing or decreasing and b. local minima and maxima of f . 1) f(x) = sinx + sin 3 x over −π<x<π Answer a. Increasing over − π 2 < x < π 2 , decreasing over x < −π b. Local maximum at x = ; local minimum at x = − 2) f(x) = x 2 π π 2 2 π 2 ,x > π 2 + cosx Answer Under Construction Exercise 3.5E. 7 For the following exercises, determine a. intervals where f is increasing or decreasing, 10/21/2020 3.5E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10930 b. local minima and maxima of f , c. intervals where f is concave up and concave down, and d. the inflection points of f . 1) f(x) = x 3 2 − 4x +x +2 Answer a. Concave up for x > 4 3 , b. Inflection point at x = 2) f(x) = x − 6x 3) f(x) = x − 6x 2 3 concave down for x < 34 3 4 3 2 Answer a. Increasing over x < 0 and x > 4, decreasing over 0 < x < 4 b. Maximum at x = 0 , minimum at x = 4 c. Concave up for x > 2 , concave down for x < 2 d. Infection point at x = 2 4) f(x) = x 4 3 − 6x 5) f(x) = x 11 10 − 6x Answer a. Increasing over x < 0 and x > b. Minimum at x = 7) f(x) = x 2 , decreasing over 0 < x < 60 11 11 d. Inflection point at x = 2 11 60 c. Concave down for x < 6) f(x) = x + x 60 54 11 , concave up for x > 54 11 54 11 3 −x +x +1 Answer a. Increasing over x > − , decreasing over x < − b. Minimum at x = − 1 1 2 2 1 2 c. Concave up for all x d. No inflection points 8) f(x) = x 3 4 +x Exercise 3.5E. 8 For the following exercises, determine 10/21/2020 3.5E.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10930 a. intervals where f is increasing or decreasing, b. local minima and maxima of f , c. intervals where f is concave up and concave down, and d. the inflection points of f . Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator. 1) f(x) = sin(πx) − cos(πx) over x = [−1, 1] Answer a. Increases over − 1 3 <x < 4 decreases over x > , 4 b. Minimum at x = − , maximum at x = 1 4 c. Concave up for − 3 4 d. Inflection points at x = − 4 3 4 4 ,x = 1 4 3 4 , concave down for x < − 1 <x < and x < − 3 3 4 and x > 1 4 1 4 2) f(x) = x + sin(2x) over x = [− 3) f(x) = sinx + tanx over (− π 2 , π 2 π 2 , π 2 ] ) Answer a. Increasing for all x b. No local minimum or maximum c. Concave up for x > 0 , concave down for x < 0 d. Inflection point at x = 0 4) f(x) = (x − 2) 2 5) f(x) = 1 1−x 2 (x − 4) ,x ≠ 1 Answer a. Increasing for all x where defined b. No local minima or maxima c. Concave up for x < 1 ; concave down for x > 1 d. No inflection points in domain 6) f(x) = sinx x over x = [−2π, 2π][2π, 0) ∪ (0, 2π] 7) f(x) = sin(x)e over x = [−π, π] x Answer a. Increasing over − π 4 <x < 3π 4 , decreasing over x > b. Minimum at x = − , maximum at x = π 4 c. Concave up for − π 2 <x < d. Infection points at x = ± 10/21/2020 π 2 3π 4 ,x <− π 4 3π 4 , concave down for x < − π 2 ,x > π 2 π 2 3.5E.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10930 8) f(x) = lnx √− x, x > 0 9) f(x) = 1 4 − √x + 1 x ,x > 0 Answer a. Increasing over x > 4, decreasing over 0 < x < 4 b. Minimum at x = 4 – – c. Concave up for 0 < x < 8√2 , concave down for x > 8√2 3 3 – d. Inflection point at x = 8√2 3 10) f(x) = 11) f(x) = e x x ,x ≠ 0 x ln(|x|) Answer: a. Increasing over x > e, decreasing over −1 < x < e b. Minimum at x = e c. Concave up for (−1, 0) ∪ (1, ∞) ), concavedownf or\((0, 1) d. Inflection point at x = 0 Exercise 3.5E. 9 For the following exercises, interpret the sentences in terms of f, f', and f ′′ . 1) The population is growing more slowly. Here f is the population. Answer: f > 0, f ' > 0, f ′′ <0 2) A bike accelerates faster, but a car goes faster. Here f = Bike’s position minus Car’s position. 3) The airplane lands smoothly. Here f is the plane’s altitude. Answer: f > 0, f ' < 0, f ′′ <0 4) Stock prices are at their peak. Here f is the stock price. 5) The economy is picking up speed. Here f is a measure of the economy, such as GDP. Answer f > 0, f ' > 0, f ′′ >0 Exercise 3.5E. 10 For the following exercises, consider a third-degree polynomial f′(1)=0,f′(3)=0. 10/21/2020 3.5E.10 CC-BY-NC-SA f(x), which has the properties https://math.libretexts.org/@go/page/10930 Determine whether the following statements are true or false. Justify your answer. 1) f(x) = 0 for some 1 ≤ x ≤ 3 2) f ′′ (x) = 0 for some 1 ≤ x ≤ 3 Answer: True, by the Mean Value Theorem 3) There is no absolute maximum at x = 3 4) If f(x) has three roots, then it has 1 inflection point. Answer: True, examine derivative 5) If f(x) has one inflection point, then it has three real roots. \begin{enumerate} \item {Consider the function 2 f (x) = x e −x . (3.5E.1) \begin{enumerate} \item Find the interval(s) on which $f(x)$ is increasing and \item Find the interval(s) on which $f(x)$ is decreasing and \item Find the local (relative) maximum and local (relative) minimum (if any). %\vspace{8in} \end{enumerate}} \item {Consider the function 2 f (x) = 3(x 2/3 − 4) . (3.5E.2) \begin{enumerate} \item Find the interval(s) on which $f(x)$ is increasing and \item Find the interval(s) on which $f(x)$ is decreasing and \item Find the local (relative) maximum and local (relative) minimum (if any). %\vspace{8in} \end{enumerate}} \item Consider the function 3 g(x) = 10 x 5 + 3x . (3.5E.3) Find \begin{enumerate} \item the interval(s) on which $g(x)$ is concave up and \item the interval(s) on which $g(x)$ is concave down and \item inflection points ( if any). 10/21/2020 3.5E.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10930 \end{enumerate} \item Consider the function 2 g(x) = e −x /2 . (3.5E.4) Find \begin{enumerate} \item the interval(s) on which $g(x)$ is concave up and \item the interval(s) on which $g(x)$ is concave down and \item inflection points ( if any). \end{enumerate} \item{ Find horizontal asymptote of the function −−−−−−−− − 2 y = f (x) = x − √ x + 2x − 6 . (3.5E.5) } %\vspace{8 in} \item { Consider the function f (x) = x e −2x . (3.5E.6) \begin{enumerate} \item Find the interval(s) on which $f(x)$ is increasing and \item Find the interval(s) on which $f(x)$ is decreasing and \item Find the local (relative) maximum and local (relative) minimum (if any). %\vspace{8in} \end{enumerate}} \item { Consider the function 2 f (x) = x e −x . (3.5E.7) \begin{enumerate} \item Find the interval(s) on which $f(x)$ is increasing and \item Find the interval(s) on which $f(x)$ is decreasing and \item Find the local (relative) maximum and local (relative) minimum (if any). %\vspace{8in} \end{enumerate}} \item { Consider the function 1 f (x) = x + . (3.5E.8) x \begin{enumerate} \item Find the interval(s) on which $f(x)$ is increasing and \item Find the interval(s) on which $f(x)$ is decreasing and \item Find the local (relative) maximum and local (relative) minimum (if any). %\vspace{8in} 10/21/2020 3.5E.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/10930 \end{enumerate}} \item { Consider the function 3 g(x) = 5 x 5 + 2x . (3.5E.9) Find \begin{enumerate} \item the interval(s) on which $g(x)$ is concave up and \item the interval(s) on which $g(x)$ is concave down and \item inflection points ( if any). %\vspace{8in} \end{enumerate}} \item { Consider the function 2 g(x) = ln(1 + x ). (3.5E.10) Find \begin{enumerate} \item the interval(s) on which $g(x)$ is concave up and \item the interval(s) on which $g(x)$ is concave down and \item inflection points ( if any). %\vspace{8in} \end{enumerate}} \item { Consider the function 2 2 g(x) = (3 − x ) . (3.5E.11) Find \begin{enumerate} \item the interval(s) on which $g(x)$ is concave up and \item the interval(s) on which $g(x)$ is concave down and \item inflection points ( if any). %\vspace{8in} \end{enumerate}} \item { Consider the function $$ \displaystyle{f(t)=\frac{1}{1-t^2}} $$ b) Find vertical and horizontal asymptotes (if any). c) Find the intervals when $f(t)$ is concave up and concave down. } \item {Consider the function $$ 10/21/2020 3.5E.13 CC-BY-NC-SA https://math.libretexts.org/@go/page/10930 \displaystyle{f(t)=e^{-t^2}} $$ a) Find the local maxima and minima for the function $f(t)$. b) Find the intervals when $f(t)$ is increasing and decreasing. c) Find the intervals when $f(t)$ is concave up and down. d) Find the inflection points for $f(t)$. } \item { Find all vertical and horizontal asymptotes for the function $$ \displaystyle{f(x)=\frac{x^2-5x+6}{x^2-4x+3}} $$ Justify your work using limits. } \item Suppose we have the following information, where $$y=\frac{x}{x^2-1},\qquad y^\prime=-\frac{x^2+1}{\left(x^2-1\right)^2}, \mbox{ and }y^{\prime\prime}=\frac{2x\left(x^2+3\right)}{\left(x^2-1 \right)^3}.$$ \noindent a) Identify any asymptotes for the function $y$ using limits. \noindent b) Find where the function $y$ is increasing/decreasing and any relative extrema. \noindent c) Find where the function $y$ is concave up/down and any points of inflection. \noindent d) Graph the function $y$. \item Let $f(x) = x^{5/4}-x^{1/4}.$ Determine the intervals on which $f$ is increasing and decreasing and find any local extrema. \item Let $g(x) = \ln (x^2+1).$ Determine the intervals of concavity of $g(x)$ and locate any inflection points. \item Let $f(x) = x^2e^{-16x^2}.$ Determine the intervals on which $f$ is increasing and decreasing and find any local extrema. \item Let $g(x) = (\ln x)^2.$ Determine the intervals of concavity of $g(x)$ and locate any inflection points. \item Given a function $\displaystyle{y=f(x)= x-2tan^{-1}(x)}$, find interval(s) on which the function is concave up and the interval(s) on which the function is concave down, and inflection point(s). \item Let $g(x) = x-2 \tan^{-1}(x).$ Determine the intervals of concavity of $g(x)$ and locate any inflection points. \end{enumerate} Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at 10/21/2020 3.5E.14 CC-BY-NC-SA https://math.libretexts.org/@go/page/10930 http://cnx.org. 10/21/2020 3.5E.15 CC-BY-NC-SA https://math.libretexts.org/@go/page/10930 3.6: Applied Optimization Problems In Section 3.3 we learned about extreme values -- the largest and smallest values a function attains on an interval. We motivated our interest in such values by discussing how it made sense to want to know the highest/lowest values of a stock or the fastest/slowest an object was moving. In this section, we apply the concepts of extreme values to solve "word problems," i.e., problems stated in terms of situations that require us to create the appropriate mathematical framework in which to solve the problem. We start with a classic example which is followed by a discussion of the topic of optimization. Solving Optimization Problems over a Closed, Bounded Interval Example 3.6.1 : Optimization: perimeter and area A man has 100 feet of fencing, a large yard, and a small dog. He wants to create a rectangular enclosure for his dog with the fencing that provides the maximal area. What dimensions provide the maximal area? Solution One can likely guess the correct answer -- that is great. We will proceed to show how calculus can provide this answer in a context that proves this answer is correct. It helps to make a sketch of the situation. Our enclosure is sketched twice in Figure 3.6.1, either with green grass and nice fence boards or as a simple rectangle. Either way, drawing a rectangle forces us to realize that we need to know the dimensions of this rectangle so we can create an area function -- after all, we are trying to maximize the area. Figure 3.6.1 : A sketch of the enclosure in Example 3.6.1. We let x and y denote the lengths of the sides of the rectangle. Clearly, Area = xy. (3.6.1) We do not yet know how to handle functions with 2 variables; we need to reduce this down to a single variable. We know more about the situation: the man has 100 feet of fencing. By knowing the perimeter of the rectangle must be 100, we can create another equation: Perimeter = 100 = 2x + 2y. (3.6.2) We now have 2 equations and 2 unknowns. In the latter equation, we solve for y : y = 50 − x. 10/17/2020 3.6.1 CC-BY-NC-SA (3.6.3) https://math.libretexts.org/@go/page/13711 Now substitute this expression for y in the area equation: Area = A(x) = x(50 − x). (3.6.4) Note we now have an equation of one variable; we can truly call the Area a function of x. This function only makes sense when 0 ≤ x ≤ 50 , otherwise we get negative values of area. So we find the extreme values of A(x) on the interval [0, 50]. To find the critical points, we take the derivative of A(x) and set it equal to 0, then solve for x. A(x) = x(50 − x) 2 (3.6.5) = 50x − x (3.6.6) A (x) = 50 − 2x (3.6.7) ′ We solve 50 − 2x = 0 to find x = 25; this is the only critical point. We evaluate A(x) at the endpoints of our interval and at this critical point to find the extreme values; in this case, all we care about is the maximum. Clearly A(0) = 0 and A(50) = 0 , whereas A(25) = 625ft . This is the maximum. Since we earlier found y = 50 − x , we find that y is also 25. Thus the dimensions of the rectangular enclosure with perimeter of 100 ft. with maximum area is a square, with sides of length 25 ft. 2 This example is very simplistic and a bit contrived. (After all, most people create a design then buy fencing to meet their needs, and not buy fencing and plan later.) But it models well the necessary process: create equations that describe a situation, reduce an equation to a single variable, then find the needed extreme value. "In real life," problems are much more complex. The equations are often not reducible to a single variable (hence multi-variable calculus is needed) and the equations themselves may be difficult to form. Understanding the principles here will provide a good foundation for the mathematics you will likely encounter later. We outline here the basic process of solving these optimization problems. Key Idea 6: Solving Optimization Problems 1. Understand the problem. Clearly identify what quantity is to be maximized or minimized. Make a sketch if helpful. 2. Create equations relevant to the context of the problem, using the information given. (One of these should describe the quantity to be optimized. We'll call this the fundamental equation.) 3. If the fundamental equation defines the quantity to be optimized as a function of more than one variable, reduce it to a single variable function using substitutions derived from the other equations. 4. Identify the domain of this function, keeping in mind the context of the problem. 5. Find the extreme values of this function on the determined domain. 6. Identify the values of all relevant quantities of the problem. We will use Key Idea 6 in a variety of examples. Example 3.6.2 : Optimization: perimeter and area Here is another classic calculus problem: A woman has a 100 feet of fencing, a small dog, and a large yard that contains a stream (that is mostly straight). She wants to create a rectangular enclosure with maximal area that uses the stream as one side. (Apparently, her dog won't swim away.) What dimensions provide the maximal area? Solution We will follow the steps outlined by Key Idea 6. 1. We are maximizing area. A sketch of the region will help; Figure 3.6.2 gives two sketches of the proposed enclosed area. A key feature of the sketches is to acknowledge that one side is not fenced. 10/17/2020 3.6.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/13711 Figure 3.6.2 : A sketch of the enclosure in Example 3.6.2. 2. We want to maximize the area; as in the example before, Area = xy. (3.6.8) This is our fundamental equation. This defines area as a function of two variables, so we need another equation to reduce it to one variable. We again appeal to the perimeter; here the perimeter is Perimeter = 100 = x + 2y. (3.6.9) Note how this is different than in our previous example. 3. We now reduce the fundamental equation to a single variable. In the perimeter equation, solve for y : y = 50 − x/2 . We can now write Area as 1 Area = A(x) = x(50 − x/2) = 50x − 2 x . (3.6.10) 2 Area is now defined as a function of one variable. 4. We want the area to be nonnegative. Since A(x) = x(50 − x/2), we want x ≥ 0 and 50 − x/2 ≥ 0 . The latter inequality implies that x ≤ 100, so 0 ≤ x ≤ 100 . 5. We now find the extreme values. At the endpoints, the minimum is found, giving an area of 0. Find the critical points. We have A (x) = 50 − x ; setting this equal to 0 and solving for x returns x = 50. This gives an area of ′ A(50) = 50(25) = 1250. (3.6.11) x A(x) 0 0 50 (50)(25) 100 0 2. We earlier set y = 50 − x/2 ; thus y = 25. Thus our rectangle will have two sides of length 25 and one side of length 50, with a total area of 1250 ft . 2 Keep in mind as we do these problems that we are practicing a process; that is, we are learning to turn a situation into a system of equations. These equations allow us to write a certain quantity as a function of one variable, which we then optimize. 10/17/2020 3.6.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/13711 Example 3.6.3 : Optimization: minimizing cost A power line needs to be run from a power station located on the beach to an offshore facility. Figure distances between the power station to the facility. 3.6.3 shows the It costs $50/ft. to run a power line along the land, and $130/ft. to run a power line under water. How much of the power line should be run along the land to minimize the overall cost? What is the minimal cost? Figure 3.6.3 : Running a power line from the power station to an offshore facility with minimal cost in Example 3.6.3 Solution We will follow the strategy of Key Idea 6 implicitly, without specifically numbering steps. There are two immediate solutions that we could consider, each of which we will reject through "common sense." First, we could minimize the distance by directly connecting the two locations with a straight line. However, this requires that all the wire be laid underwater, the most costly option. Second, we could minimize the underwater length by running a wire all 5000 ft. along the beach, directly across from the offshore facility. This has the undesired effect of having the longest distance of all, probably ensuring a non--minimal cost. The optimal solution likely has the line being run along the ground for a while, then underwater, as the figure implies. We need to label our unknown distances -- the distance run along the ground and the distance run underwater. Recognizing that the underwater distance can be measured as the hypotenuse of a right triangle, we choose to label the distances as shown in Figure 3.6.4. Figure 3.6.4 : Labeling unknown distances in Example 3.6.3. By choosing x as we did, we make the expression under the square root simple. We now create the cost function. Cost = land cost + $50× land distance 50(5000 − x) + + water cost $130× water distance − − − − − − − − − 2 2 c(x) = 50(5000 − x) + 130 √x + 1000 So we have . This function only makes sense on the interval we are fairly certain the endpoints will not give a minimal cost, we still evaluate c(x) at each to verify. c(0) = 380, 000 (3.6.12) − − − − − − − − − 2 2 130 √ x + 1000 . c(5000) ≈ 662, 873. . While [0, 5000] (3.6.13) We now find the critical values of c(x). We compute c (x) as ′ ′ c (x) = −50 + 10/17/2020 130x − − − − − − − − −. √ x2 + 10002 3.6.4 CC-BY-NC-SA (3.6.14) https://math.libretexts.org/@go/page/13711 Recognize that this is never undefined. Setting c (x) = 0 and solving for x, we have: ′ 130x −50 + − − − − − − − − − 2 √ x2 + 1000 =0 (3.6.15) = 50 (3.6.16) 130x − − − − − − − − − √ x2 + 10002 2 2 130 x 2 2 2 x 2 130 x 2 (130 (3.6.17) + 1000 2 2 2 2 130 x 2 = 50 − 50 x 2 2 − 50 )x 2 2 = 50 (x 2 = 50 2 + 1000 ) (3.6.18) 2 ⋅ 1000 (3.6.19) 2 = 50, 000 (3.6.20) 2 2 x 50, 000 = 2 130 (3.6.21) 2 − 50 50, 000 x = 50, 000 x = 2 = 416 120 Evaluating ≈ 416.67. (3.6.23) 3 gives a cost of about $370,000. The distance the power line is laid along land is − − − − − − − − − − − − − 5000 − 416.67 = 4583.33ft., and the underwater distance is √416.67 + 1000 ≈ 1083 ft. c(x) at (3.6.22) − − − − − − − − − √ 1302 − 502 x = 416.67 2 2 Example 3.6.4 : Maximizing the Volume of a Box An open-top box is to be made from a 24 in. by 36 in. the piece of cardboard by removing a square from each corner of the box and folding up the flaps on each side. What size square should be cut out of each corner to get a box with the maximum volume? Solution Step 1: Let x be the side length of the square to be removed from each corner (Figure 3.6.3). Then, the remaining four flaps can be folded up to form an open-top box. Let V be the volume of the resulting box. Figure 3.6.4 : A square with side length x inches is removed from each corner of the piece of cardboard. The remaining flaps are folded to form an open-top box. Step 2: We are trying to maximize the volume of a box. Therefore, the problem is to maximize V . Step 3: As mentioned in step 2, are trying to maximize the volume of a box. The volume of a box is V = L ⋅ W ⋅ H, where L, W ,and H are the length, width, and height, respectively. Step 4: From Figure 3.6.3, we see that the height of the box is 24 − 2x inches. Therefore, the volume of the box is 10/17/2020 x inches, the length is 3.6.5 CC-BY-NC-SA 36 − 2x inches, and the width is https://math.libretexts.org/@go/page/13711 V (x) = (36 − 2x)(24 − 2x)x . 3 = 4x 2 − 120 x + 864x Step 5: To determine the domain of consideration, let’s examine Figure 3.6.3. Certainly, we need x > 0. Furthermore, the side length of the square cannot be greater than or equal to half the length of the shorter side, 24 in.; otherwise, one of the flaps would be completely cut off. Therefore, we are trying to determine whether there is a maximum volume of the box for x over the open interval (0, 12). Since V is a continuous function over the closed interval [0, 12], we know V will have an absolute maximum over the closed interval. Therefore, we consider V over the closed interval [0, 12] and check whether the absolute maximum occurs at an interior point. Step 6: Since V (x) is a continuous function over the closed, bounded interval [0, 12], V must have an absolute maximum (and an absolute minimum). Since V (x) = 0 at the endpoints and V (x) > 0 for 0 < x < 12, the maximum must occur at a critical point. The derivative is 2 V '(x) = 12 x − 240x + 864. To find the critical points, we need to solve the equation 2 12 x − 240x + 864 = 0. Dividing both sides of this equation by 12, the problem simplifies to solving the equation 2 x − 20x + 72 = 0. Using the quadratic formula, we find that the critical points are −−−−−−−−−−−−− − 20 ± √ (−20 )2 − 4(1)(72) x = 2 − − − 20 ± √112 = . 2 – 20 ± 4 √7 = 2 – = 10 ± 2 √7 – – Since 10 + 2√7 is not in the domain of consideration, the only critical point we need to consider is 10 − 2√7 . Therefore, – the volume is maximized if we let x = 10 − 2√7 in. The maximum volume is – – 3 V (10 − 2 √7) = 640 + 448 √7 ≈ 1825 in. as shown in the following graph. Figure 3.6.4 : Maximizing the volume of the box leads to finding the maximum value of a cubic polynomial. Watch a video about optimizing the volume of a box. Exercise 3.6.2 10/17/2020 3.6.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/13711 Suppose the dimensions of the cardboard in Example 3.6.2 are 20 in. by 30 in. Let x be the side length of each square and write the volume of the open-top box as a function of x. Determine the domain of consideration for x. Hint The volume of the box is L ⋅ W ⋅ H . Answer V (x) = x(20 − 2x)(30 − 2x). The domain is [0, 10]. Example 3.6.5 : Minimizing Travel Time An island is 2 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that is 6 mi west of that point. The visitor is planning to go from the cabin to the island. Suppose the visitor runs at a rate of 8 mph and swims at a rate of 3 mph. How far should the visitor run before swimming to minimize the time it takes to reach the island? Solution Step 1: Let x be the distance running and let y be the distance swimming (Figure from the cabin to the island. ). Let T be the time it takes to get 3.6.5 Figure 3.6.5 : How can we choose x and y to minimize the travel time from the cabin to the island? Step 2: The problem is to minimize T . Step 3: To find the time spent traveling from the cabin to the island, add the time spent running and the time spent swimming. Since Distance = Rate × Time (D = R × T ), the time spent running is Drunning Trunning = x = Rrunning 8 , and the time spent swimming is Dswimming Tswimming = y = Rswimming 3 . Therefore, the total time spent traveling is x T = y + 8 . 3 Step 4: From Figure 3.6.5, the line segment of y miles forms the hypotenuse of a right triangle with legs of length 2mi and − − − − − − − − − − 6 − xmi . Therefore, by the Pythagorean theorem, 2 + (6 − x ) = y , and we obtain y = √(6 − x ) + 4 . Thus, the total time spent traveling is given by the function 2 2 − − − − − − − − − − 2 √(6 − x ) + 4 x T (x) = + 8 10/17/2020 2 3 2 . 3.6.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/13711 Step 5: From Figure 3.6.5, we see that 0 ≤ x ≤ 6 . Therefore, [0, 6] is the domain of consideration. Step 6: Since T (x) is a continuous function over a closed, bounded interval, it has a maximum and a minimum. Let’s begin by looking for any critical points of T over the interval [0, 6]. The derivative is 1 T '(x) = 1 2 −1/2 [(6 − x ) + 4] − 8 ⋅ 2(6 − x) 2 3 1 = (6 − x) − 8 − − − − − − − − − − 2 3 √ (6 − x ) + 4 If T '(x) = 0,, then 1 6 −x = 8 − − − − − − − − − − 2 3 √ (6 − x ) + 4 Therefore, − − − − − − − − − − 2 3 √ (6 − x ) + 4 = 8(6 − x). Squaring both sides of this equation, we see that if x satisfies this equation, then x must satisfy 2 9[(6 − x ) 2 + 4] = 64(6 − x ) , which implies 2 55(6 − x ) = 36. We conclude that if x is a critical point, then x satisfies 2 (x − 6 ) 36 = . 55 Therefore, the possibilities for critical points are 6 (x = 6 ± − − √55 . − − − − Since x = 6 + 6/√55 is not in the domain, it is not a possibility for a critical point. On the other hand, x = 6 − 6/√55 is in the domain. Since we squared both sides of Equation to arrive at the possible critical points, it remains to verify that − − − − − − x = 6 − 6/ √55 satisfies Equation. Since x = 6 − 6/ √55 does satisfy that equation, we conclude that x = 6 − 6/ √55 is a critical point, and it is the only one. To justify that the time is minimized for this value of x, we just need to check the values of T (x) at the endpoints x = 0 and x = 6 , and compare them with the value of T (x) at the critical point − − x = 6 − 6/ √55 . We find that T (0) ≈ 2.108h and T (6) ≈ 1.417h, whereas − − T (6 − 6/ √55) ≈ 1.368h. Therefore, we conclude that T has a local minimum at x ≈ 5.19mi. Exercise 3.6.3 Suppose the island is 1 mi from shore, and the distance from the cabin to the point on the shore closest to the island is 15 mi. Suppose a visitor swims at the rate of 2.5 mph and runs at a rate of 6 mph. Let x denote the distance the visitor will run before swimming, and find a function for the time it takes the visitor to get from the cabin to the island. Hint The time T = Trunning + Tswimming . Answer 10/17/2020 3.6.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/13711 − − − − − − − − − − − 2 √ (15 − x ) + 1 x T (x) = + 6 2.5 In business, companies are interested in maximizing revenue. In the following example, we consider a scenario in which a company has collected data on how many cars it is able to lease, depending on the price it charges its customers to rent a car. Let’s use these data to determine the price the company should charge to maximize the amount of money it brings in. Example 3.6.6 : Maximizing Revenue Owners of a car rental company have determined that if they charge customers p dollars per day to rent a car, where 50 ≤ p ≤ 200 , the number of cars n they rent per day can be modeled by the linear function n(p) = 1000 − 5p . If they charge $50 per day or less, they will rent all their cars. If they charge $200 per day or more, they will not rent any cars. Assuming the owners plan to charge customers between $50 per day and $200 per day to rent a car, how much should they charge to maximize their revenue? Solution Step 1: Let p be the price charged per car per day and let n be the number of cars rented per day. Let R be the revenue per day. Step 2: The problem is to maximize R. Step 3: The revenue (per day) is equal to the number of cars rented per day times the price charged per car per day—that is, R = n × p. Step 4: Since the number of cars rented per day is modeled by the linear function n(p) = 1000 − 5p, the revenue R can be represented by the function R(p) = n × p = (1000 − 5p)p 2 = −5 p + 1000p. Step 5: Since the owners plan to charge between $50 per car per day and maximum revenue R(p) for p in the closed interval [50, 200]. $200 per car per day, the problem is to find the Step 6: Since R is a continuous function over the closed, bounded interval [50, 200], it has an absolute maximum (and an absolute minimum) in that interval. To find the maximum value, look for critical points. The derivative is R'(p) = −10p + 1000. Therefore, the critical point is p = 100 When p = 100, R(100) = $50, 000. When p = 50, R(p) = $37, 500. When p = 200, R(p) = $0 . Therefore, the absolute maximum occurs at p = $100. The car rental company should charge maximize revenue as shown in the following figure. $100 per day per car to Figure 3.6.6 : To maximize revenue, a car rental company has to balance the price of a rental against the number of cars people will rent at that price. Exercise 3.6.4 10/17/2020 3.6.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/13711 A car rental company charges its customers p dollars per day, where 60 ≤ p ≤ 150 . It has found that the number of cars rented per day can be modeled by the linear function n(p) = 750 − 5p. How much should the company charge each customer to maximize revenue? Hint R(p) = n × p, where n is the number of cars rented and p is the price charged per car. Answer The company should charge $75 per car per day. Example 3.6.7 : Maximizing the Area of an Inscribed Rectangle A rectangle is to be inscribed in the ellipse 2 x +y 2 = 1. 4 What should the dimensions of the rectangle be to maximize its area? What is the maximum area? Solution: Step 1: For a rectangle to be inscribed in the ellipse, the sides of the rectangle must be parallel to the axes. Let length of the rectangle and W be its width. Let A be the area of the rectangle. L be the Figure 3.6.7 : We want to maximize the area of a rectangle inscribed in an ellipse. Step 2: The problem is to maximize A . Step 3: The area of the rectangle is A = LW . Step 4: Let (x, y) be the corner of the rectangle that lies in the first quadrant, as shown in Figure 3.6.7. We can write length − − − − − − 2 1 −x 2 L = 2x and width W = 2y . Since x 4 +y 2 and y > 0 , we have y = √ =1 − − − − − − 2 1 −x A = LW = (2x)(2y) = 4x √ 4 . Therefore, the area is − −−− − 2 = 2x √4 − x 4 Step 5: From Figure 3.6.7, we see that to inscribe a rectangle in the ellipse, the x-coordinate of the corner in the first quadrant must satisfy 0 < x < 2 . Therefore, the problem reduces to looking for the maximum value of A(x) over the open interval (0, 2). Since A(x) will have an absolute maximum (and absolute minimum) over the closed interval [0, 2], we − −−− − consider A(x) = 2x √4 − x over the interval [0, 2]. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval. 2 Step 6: As mentioned earlier, A(x) is a continuous function over the closed, bounded interval [0, 2]. Therefore, it has an absolute maximum (and absolute minimum). At the endpoints x = 0 and x = 2, A(x) = 0. For 0 < x < 2, A(x) > 0 . Therefore, the maximum must occur at a critical point. Taking the derivative of A(x), we obtain 10/17/2020 3.6.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/13711 − −−− − ′ 2 A (x) = 2 √ 4 − x 1 + 2x ⋅ 2 − −−− − 2 = 2√ 4 − x − −−− − (−2x) 2 2√ 4 − x 2x − − −−− − √ 4 − x2 2 8 − 4x = − −−− − √ 4 − x2 . To find critical points, we need to find where A (x) = 0. We can see that if x is a solution of ′ 2 8 − 4x − −−− − =0 √4 − x2 , then x must satisfy 2 8 − 4x = 0. (3.6.24) – x = ±√2 Therefore, x = 2. Thus, are the possible solutions of Equation. Since we are considering x over the interval – – – [0, 2], x = √2 is a possibility for a critical point, but x = −√2 is not. Therefore, we check whether √2 is a solution of – – Equation. Since x = √2 is a solution of Equation 3.6.24, we conclude that √2 is the only critical point of A(x) in the interval [0, 2]. 2 Therefore, A(x) must have an absolute maximum at the critical point – rectangle, we need to find the length L and the width W . If x = √2 then − − − − − − − − − – 2 (√2) y = √1 − − − − − − 1 = √1 − 4 Therefore, the dimensions of the rectangle are and . To determine the dimensions of the 1 = 2 – L = 2x = 2 √2 – x = √2 –. √2 (3.6.25) 2 W = 2y = – – A = LW = (2 √2)(√2) = 4. – √2 – = √2 . The area of this rectangle is Exercise 3.6.5 Modify the area function consideration? A if the rectangle is to be inscribed in the unit circle 2 x = + y2 1 . What is the domain of Hint If (x, y) is the vertex of the square that lies in the first quadrant, then the area of the square is A = (2x)(2y) = 4xy. Answer − −−− − A(x) = 4x √1 − x2 . The domain of consideration is [0, 1]. Solving Optimization Problems when the Interval Is Not Closed or Is Unbounded In the previous examples, we considered functions on closed, bounded domains. Consequently, by the extreme value theorem, we were guaranteed that the functions had absolute extrema. Let’s now consider functions for which the domain is neither closed nor bounded. Many functions still have at least one absolute extrema, even if the domain is not closed or the domain is unbounded. For example, the function f (x) = x + 4 over (−∞, ∞) has an absolute minimum of 4 at x = 0 . Therefore, we can still consider functions over unbounded domains or open intervals and determine whether they have any absolute extrema. In the next example, we try to minimize a function over an unbounded domain. We will see that, although the domain of consideration is (0, ∞), the function has an absolute minimum. 2 In the following example, we look at constructing a box of least surface area with a prescribed volume. It is not difficult to show that for a closed-top box, by symmetry, among all boxes with a specified volume, a cube will have the smallest surface 10/17/2020 3.6.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/13711 area. Consequently, we consider the modified problem of determining which open-topped box with a specified volume has the smallest surface area. Example 3.6.8 : Minimizing Surface Area A rectangular box with a square base, an open top, and a volume of 216in. is to be constructed. What should the dimensions of the box be to minimize the surface area of the box? What is the minimum surface area? 3 Solution Step 1: Draw a rectangular box and introduce the variable x to represent the length of each side of the square base; let represent the height of the box. Let S denote the surface area of the open-top box. y Figure 3.6.8.1 : We want to minimize the surface area of a square-based box with a given volume. Step 2: We need to minimize the surface area. Therefore, we need to minimize S . Step 3: Since the box has an open top, we need only determine the area of the four vertical sides and the base. The area of each of the four vertical sides is x ⋅ y. The area of the base is x . Therefore, the surface area of the box is 2 2 S = 4xy + x Step 4: Since the volume of this box is x 2 y . and the volume is given as 216in. , the constraint equation is 3 2 x y = 216 Solving the constraint equation for y , we have y = 216 2 . . Therefore, we can write the surface area as a function of x only: x 216 S(x) = 4x( 2 2 )+x . (3.6.26) x Therefore, S(x) = 864 . 2 +x x Step 5: Since we are requiring that x y = 216, we cannot have x = 0 . Therefore, we need x > 0 . On the other hand, x is allowed to have any positive value. Note that as x becomes large, the height of the box y becomes correspondingly small so that x y = 216. Similarly, as x becomes small, the height of the box becomes correspondingly large. We conclude that the domain is the open, unbounded interval (0, ∞). Note that, unlike the previous examples, we cannot reduce our problem to looking for an absolute maximum or absolute minimum over a closed, bounded interval. However, in the next step, we discover why this function must have an absolute minimum over the interval (0, ∞). 2 2 Step 6: Note that as x → 0 + S(x) → ∞. Also, as x → ∞, S(x) → ∞ . Since S is a continuous function that approaches infinity at the ends, it must have an absolute minimum at some x ∈ (0, ∞). This minimum must occur at a critical point of S . The derivative is , 864 S'(x) = − Therefore, S'(x) = 0 when 2x = 864 x2 x2 + 2x. (3.6.27) . Solving this equation for x, we obtain x 3 = 432 − − − – , so x = √432 = 6√2. Since this 3 3 – is the only critical point of S , the absolute minimum must occur at x = 6√2 (see Figure 3.6.9). 3 When 3 – x = 6 √2 , 216 y = – 2 (6 √2) 3 3 – = 3 √2in . Therefore, the dimensions of the box should be 3 – x = 6 √2in. and 3 – y = 3 √2in. With these dimensions, the surface area is 10/17/2020 3.6.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/13711 3 – S(6 √2) = 864 3 – 2 3 – 2 + (6 √2) = 108 √4in. 3 – 6 √2 (3.6.28) Figure 3.6.9.2 : We can use a graph to determine the dimensions of a box of given the volume and the minimum surface area. Exercise 3.6.8 Consider the same open-top box, which is to have volume 216in. . Suppose the cost of the material for the base is 20 ¢/in. and the cost of the material for the sides is 30 ¢/in. and we are trying to minimize the cost of this box. Write the cost as a function of the side lengths of the base. (Let x be the side length of the base and y be the height of the box.) 3 2 2 Hint If the cost of one of the sides is 30¢/in. , the cost of that side is 0.30xy. 2 Answer 259.2 c(x) = 2 + 0.2 x x dollars Example 3.6.9 : Minimizing Cost An open-topped cylindrical can is to have volume 300 cm . The material for the bottom of the can costs for its curved side costs 5 cents/cm . Find the dimensions of the can that minimize the cost of the can. 3 10 cents/cm 2 , 2 Solution Step 1: Draw a cylindrical can and introduce the variable height of the can. Let C denote the cost to produce a can. r to represent the radius of the circular base; let h represent the Figure 3.6.9.1 : We want to minimize the cost of the open-topped cylindrical can with a given volume Step 2: Given: Volume of the can V = 300 . We need to minimize the cost. Therefore, we need to minimize C . Step 3: Since the can has an open top, we need only determine the cost to produce the bottom and the cost to produce the side. 10/17/2020 3.6.13 CC-BY-NC-SA https://math.libretexts.org/@go/page/13711 Figure 3.6.9.2 : Surface area of the curved side and the area of the bottom The surface area of the curved side is 2πrh The area of the bottom is πr . Therefore, the cost to produce the can is 2 2 C = (5)2πrh + (10)πr Step 4: Since the volume of this can π r 2 h , with r ≥ 0 . and the volume is given as 300, the constraint equation is . 2 π r h = 300 Solving the constraint equation for h , we have h = 300 C = (5)2πr 2 2 + (10)π r 300 2 πr . Therefore, we can write the cost as a function of \ (r\) only: . πr Therefore, C 3000 = 2 + (10)π r r . Step 5: Since we are requiring that π r h = 300, we cannot have r = 0 . Therefore, we need r > 0 . On the other hand, r is allowed to have any positive value. Note that as r becomes large, the height of the can h becomes correspondingly small so that π r h = 300. Similarly, as r becomes small, the height of the can becomes correspondingly large. We conclude that the domain is the open, unbounded interval (0, ∞). In the next step, we discover why this function must have an absolute minimum over the interval (0, ∞). 2 2 Step 6: Note that as x → 0 + C (r) → ∞. Also, as r → ∞, C (r) → ∞ . Since C is a continuous function that approaches infinity at the ends, it must have an absolute minimum at some r ∈ (0, ∞). This minimum must occur at a critical point of C . The derivative is , 3000 C '(x) = − + 2r. 2 (3.6.29) r Therefore, C '(r) = 0 when 2r = 3000 2 . Solving this equation for r, we obtain r 3 r − − − − , so r = √1500. Since this is the 3 = 1500 − − − − only critical point of C , the absolute minimum must occur at r = √1500 . 3 3 300 − − − − When r = √1500 , h = − − − −2 π √1500 3 300 − − − − . Therefore, the dimensions of the can sho uld be r = √1500in. and h = 3 − − − −2 π √1500 . 3 With these dimensions, the cost is Exercise 3.6.9 Pop cans to hold 300 ml are made in the shape of right circular cylinders. Find the dimensions of the can that minimize the surface area. Answer Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page. 10/17/2020 3.6.14 CC-BY-NC-SA https://math.libretexts.org/@go/page/13711 Example 3.6.10 : Closest point Find the point on the curve y = 2 −x 3 that is closest to the point P (4, −1). In the exercises, you will see a variety of situations that require you to combine problem--solving skills with calculus. Focus on the process; learn how to form equations from situations that can be manipulated into what you need. Eschew memorizing how to do "this kind of problem" as opposed to "that kind of problem." Learning a process will benefit one far longer than memorizing a specific technique. The next section introduces our final application of the derivative: differentials. Given approximating the change in y after x changes by a small amount. , they offer a method of y = f (x) Key Concepts To solve an optimization problem, begin by drawing a picture and introducing variables. Find an equation relating the variables. Find a function of one variable to describe the quantity that is to be minimized or maximized. Look for critical points to locate local extrema. Glossary optimization problems problems that are solved by finding the maximum or minimum value of a function Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/ Integrated by Justin Marshall. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/17/2020 3.6.15 CC-BY-NC-SA https://math.libretexts.org/@go/page/13711 3.6E: Exercises Exercise 3.6E . 1 For the following exercises, answer by proof, counterexample, or explanation. 1) When you find the maximum for an optimization problem, why do you need to check the sign of the derivative around the critical points? Answer The critical points can be the minima, maxima, or neither. 2) Why do you need to check the endpoints for optimization problems? 3) True or False. For every continuous nonlinear function, you can find the value xx that maximizes the function. Answer False; y = −x has a minimum only 2 4) True or False. For every continuous nonconstant function on a closed, finite domain, there exists at least one xx that minimizes or maximizes the function Exercise 3.6E . 2 To carry a suitcase on an airplane, the length+width+height of the box must be less than or equal to 62in. Assuming the height is fixed, show that the maximum volume is V = h(31 − (12)h) What height allows you to have the largest volume? Answer Under Construction Exercise 3.6E . 3 You are constructing a cardboard box with the dimensions 2m by 4m. You then cut equal-size squares from each corner so you may fold the edges. What are the dimensions of the box with the largest volume? Answer Under Construction Exercise 3.6E . 4 Find the positive integer that minimizes the sum of the number and its reciprocal. Answer 10/21/2020 3.6E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/13713 11 Exercise 3.6E . 5 For the following exercises, consider the construction of a pen to enclose an area. 1) You have 400ft of fencing to construct a rectangular pen for cattle. What are the dimensions of the pen that maximize the area? Answer 100ft x 100ft 2) You have 800ft of fencing to make a pen for hogs. If you have a river on one side of your property, what is the dimension of the rectangular pen that maximizes the area? 3) You need to construct a fence around an area of 1600ft. 1600ft. What are the dimensions of the rectangular pen to minimize the amount of material needed? Answer 40ft x 40ft Exercise 3.6E . 6 Two poles are connected by a wire that is also connected to the ground. The first pole is 20ft tall and the second pole is 10ft tall. There is a distance of 30ft between the two poles. Where should the wire be anchored to the ground to minimize the amount of wire needed? Answer Under Construction Exercise 3.6E . 7 You are moving into a new apartment and notice there is a corner where the hallway narrows from 8ft to 6ft. What is the length of the longest item that can be carried horizontally around the corner? 10/21/2020 3.6E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/13713 Answer 19.73 ft Exercise 3.6E . 8 1) A patient’s pulse measures 70bpm,80bpm, then 120bpm. To determine an accurate measurement of pulse, the doctor wants to know what value minimizes the expression (x − 70) + (x − 80) + (x − 120) ? 2 2 2 2) In the previous problem, assume the patient was nervous during the third measurement, so we only weight that value half as much as the others. What is the value that minimizes (x − 70) + (x − 80) + 12(x − 120) ? 2 2 2 Answer 84bpm Exercise 3.6E . 9 You can run at a speed of 66 mph and swim at a speed of 33 mph and are located on the shore, 44 miles east of an island that is 11 mile north of the shoreline. How far should you run west to minimize the time needed to reach the island? Answer Under Construction Exercise 3.6E . 10 For the following problems, consider a lifeguard at a circular pool with diameter 40m. He must reach someone who is drowning on the exact opposite side of the pool, at position C. The lifeguard swims with a speed v and runs around the pool at speed w=3v. 10/21/2020 3.6E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/13713 1) Find a function that measures the total amount of time it takes to reach the drowning person as a function of the swim angle, θ. 2) Find at what angle θ the lifeguard should swim to reach the drowning person in the least amount of time. Answer Under Construction Exercise 3.6E . 11 For the following exercises, consider a limousine that gets $15/h, and gas is $3.5/gal. m(v) = (120 − 2v)5mi/gal at speed v, the chauffeur costs 1) Find the cost per mile at speed v. 2) Find the cheapest driving speed. Answer Approximately 34.02mph Exercise 3.6E . 12 For the following exercises, consider a pizzeria that sell pizzas for a revenue of C (x) = b + cx + dx , where x represents the number of pizzas. R(x) = ax and costs 2 1) Find the profit function for the number of pizzas. How many pizzas gives the largest profit per pizza? 2) Assume that R(x) = 10x and C (x) = 2x + x . How many pizzas sold maximizes the profit? 2 Answer 44 3) Assume that R(x) = 15x, and C (x) = 60 + 3x + 12x . How many pizzas sold maximizes the profit? 2 Exercise 3.6E . 13 Consider a wire 4ft long cut into two pieces. One piece forms a circle with radius r and the other forms a square of side x. Choose x to maximize the sum of their areas. Answer Under Construction Exercise 3.6E . 14 For the following exercises, consider two non-negative numbers x and y such that x + y = 10 . Maximize and minimize the quantities. 10/21/2020 3.6E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/13713 1) xy Answer Maximal: x = 5, y = 5 ; minimal:x = 0, y = 10x = 0, y = 10 and y = 0, x = 10y = 0, x = 10 2) x 2 y 2 3) y − 1x Answer Maximal: x = 1, y = 9 ; minimal: none 4) x 2 −y Exercise 3.6E . 15 For the following exercises, draw the given optimization problem and solve. 1) Find the volume of the largest right circular cylinder that fits in a sphere of radius 11. Answer Under Construction 2) Find the volume of the largest right cone that fits in a sphere of radius 11. 3) Find the area of the largest rectangle that fits into the triangle with sides x=0,y=0 and x 4 +y 6 =1 Answer Under Construction 4) Find the largest volume of a cylinder that fits into a cone that has base radius R and height h. 5) Find the dimensions of the closed cylinder volume V=16π that has the least amount of surface area. Answer r=2,h=4 6) Find the dimensions of a right cone with surface area S=4π that has the largest volume. Exercise 3.6E . 16 For the following exercises, consider the points on the given graphs. Use a calculator to graph the functions. 1) Where is the line y = 5 − 2x closest to the origin? Answer (2,1) 2) Where is the line y=5−2x closest to point (1,1)? 3) Where is the parabola y = x closest to point (2,0)? 2 Answer (0.8351,0.6974) 4) Where is the parabola y = x closest to point (0,3)? 2 10/21/2020 3.6E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/13713 Exercise 3.6E . 17 For the following exercises, set up, but do not evaluate, each optimization problem. 1) A window is composed of a semicircle placed on top of a rectangle. If you have 20ft of window-framing materials for the outer frame, what is the maximum size of the window you can create? Use r to represent the radius of the semicircle. Answer 2 A = 20r − 2 r 2 − 12π r 2) You have a garden row of 20 watermelon plants that produce an average of 30 watermelons apiece. For any additional watermelon plants planted, the output per watermelon plant drops by one watermelon. How many extra watermelon plants should you plant? 3) You are constructing a box for your cat to sleep in. The plush material for the square bottom of the box costs $5/f t and the material for the sides costs $2/f t . You need a box with volume 4f t . Find the dimensions of the box that minimize cost. Use x to represent the length of the side of the box. 2 2 2 Answer: 2 C (x) = 5 x + 32x 4) You are building five identical pens adjacent to each other with a total area of 1000m , as shown in the following figure. What dimensions should you use to minimize the amount of fencing? 2 5) You are the manager of an apartment complex with 50 units. When you set rent at $800/month, all apartments are rented. As you increase rent by $25/month, one fewer apartment is rented. Maintenance costs run $50/month for each occupied unit. What is the rent that maximizes the total amount of profit? Answer P (x) = (50 − x)(800 + 25x − 50) 10/21/2020 3.6E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/13713 3.7: Curve skectching Guidelines for Drawing the Graph of a Function We now have enough analytical tools to draw graphs of a wide variety of algebraic and transcendental functions. Before showing how to graph specific functions, let’s look at a general strategy to use when graphing any function. PROBLEM-SOLVING STRATEGY: DRAWING THE GRAPH OF A FUNCTION Given a function f , use the following steps to sketch a graph of f : 1. Domain: Determine the domain of the function. 2. Intercepts: Locate the x- and y -intercepts. 3. Symmetry: If f(−x) = f(x) , then f is symmetric about y -axis, also called even function. If f (−x) = −f (x) , then f is symmetric about the origin, also called odd function. 4. Asymptote(s): Evaluate lim f(x) and lim f(x) to determine the end behavior. If either of these limits is a finite number L, then y = L is a x→∞ x→−∞ horizontal asymptote. If either of these limits is ∞ or −∞, determine whether f has an oblique asymptote. If f is a rational function such that p(x) f(x) = q(x) , where the degree of the numerator is greater than the degree of the denominator, then f can be written as p(x) f(x) = r(x) = g(x) + q(x) (3.7.1) q(x), where the degree of r(x) is less than the degree of q(x). The values of f(x) approach the values of g(x) as x → ±∞. If g(x) is a linear function, it is known as an oblique asymptote. Determine whether f has any vertical asymptotes. 5. Local extrema: Calculate f'. Find all critical points and determine the intervals where f is increasing and where f is decreasing. Determine whether f has any local extrema. 6. Concavity: Calculate f . Determine the intervals where f is concave up and where f is concave down. Use this information to determine whether f has any inflection points. The second derivative can also be used as an alternate means to determine or verify that f has a local extremum at a critical point. ′′ Now let’s use this strategy to graph several different functions. We start by graphing a polynomial function. EXAMPLE 3.7.1: SKETCHING A GRAPH OF A POLYNOMIAL Sketch a graph of f(x) = (x − 1) 2 (x + 2). Solution Step 1: Since f is a polynomial, the domain is the set of all real numbers. Step 2: When x = 0, f(x) = 2. Therefore, the y-intercept is gives us the x -intercepts (1, 0) and (−2, 0) . (0, 2) . To find the x -intercepts, we need to solve the equation 2 (x − 1) (x + 2) = 0 , Step 3: Consider $f(-x)= ((-x)−1)^2((-x)+2)=(x+1)^2(2-x),$ which is neither equal to −f (x) nor f (x). There fore there is no symmetry. Step 4: We need to evaluate the end behavior of f. As 2 x → ∞, (x − 1) and → ∞ (x + 2) → ∞ . Therefore, lim f(x) = ∞ As x→∞ 2 x → −∞, (x − 1) → ∞ and (x + 2) → −∞ . Therefore, lim f(x) = −∞ . To get even more information about the end behavior of f , we can x→∞ multiply the factors of f .When doing so, we see that 2 3 f(x) = (x − 1) (x + 2) = x − 3x + 2. (3.7.2) Since the leading term of f is x , we conclude that f behaves like y = x as x → ±∞. 3 3 Step 5: Since f is a polynomial function, it does not have any vertical asymptotes. Step 6: The first derivative of f is 2 f'(x) = 3x − 3. (3.7.3) Therefore, f has two critical points: x = 1, −1. Divide the interval (−∞, ∞) into the three smaller intervals: (−∞, −1), (−1, 1) , and (1, ∞) . Then, choose test points x = −2, x = 0 , and x = 2 from these intervals and evaluate the sign of f'(x) at each of these test points, as shown in the following table. Interval Test point Sign of Derivative ′ 2 f (x) = 3x (−∞, −1) 10/21/2020 x = −2 Conclusion − 3 = 3(x − 1)(x + 1) (+)(−)(−) = + 3.7.1 CC-BY-NC-SA f is increasing https://math.libretexts.org/@go/page/13717 Interval Test point Sign of Derivative ′ 2 f (x) = 3x Conclusion − 3 = 3(x − 1)(x + 1) (−1, 1) x = 0 (+)(−)(+) = − (1, ∞) (x=2\) (+)(+)(+) = + f f decreasing is increasing From the table, we see that f has a local maximum at x = −1 and a local minimum at x = 1 . Evaluating f(x) at those two points, we find that the local maximum value is f(−1) = 4 and the local minimum value is f(1) = 0. Step 7: The second derivative of f is ′′ f (x) = 6x. (3.7.4) The second derivative is zero at x = 0. Therefore, to determine the concavity of f , divide the interval (−∞, ∞) into the smaller intervals (−∞, 0) and (0, ∞) , and choose test points x = −1 and x = 1 to determine the concavity of f on each of these smaller intervals as shown in the following table. Interval Test Point Sign of (−∞, 0) x = −1 − (0, ∞) x = 1 + ′′ Conclusion (x) = 6x f is concave down.. f is concave up. We note that the information in the preceding table confirms the fact, found in step 5, that f has a local maximum at x = −1 and a local minimum at x = 1 . In addition, the information found in step 5 —namely, f has a local maximum at x = −1 and a local minimum at x = 1 , and f'(x) = 0 at those points—combined with the fact that f changes sign only at x = 0 confirms the results found in step 6 on the concavity of f . ′′ Combining this information, we arrive at the graph of f(x) = (x − 1) 2 (x + 2) shown in the following graph. EXERCISE 3.7.1 Sketch a graph of f(x) = (x − 1) 3 (x + 2). Answer f is a fourth-degree polynomial. Answer 10/21/2020 3.7.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/13717 EXAMPLE 3.7.2: SKETCHING A RATIONAL FUNCTION Sketch the graph of f(x) = 2 x 2 (1−x ) . Solution Step 1: The function f is defined as long as the denominator is not zero. Therefore, the domain is the set of all real numbers x except x = ±1. Step 2: Find the intercepts. If x = 0, then f(x) = 0 , so 0 is an intercept. If y = 0 2 , then x 2 (1−x ) = 0, which implies x=0. Therefore, (0,0) is the only intercept. Step 3: Consider $f(-x) =\frac{(-x)^2}{(1−(-x)^2)}=\frac{x^2}{(1−x^2)}$=f(x)\). Therefore f is symmetric about y-axis. Step 4: Evaluate the limits at infinity. Since f is a rational function, divide the numerator and denominator by the highest power in the denominator: x .We obtain 2 2 x lim 1 = 2 1−x x→±∞ lim x→±∞ = −1. 1 −1 2 x Therefore, f has a horizontal asymptote of y = −1 as x → ∞ and x → −∞. Step 5: To determine whether f has any vertical asymptotes, first check to see whether the denominator has any zeroes. We find the denominator is zero when x = ±1 . To determine whether the lines x = 1 or x = −1 are vertical asymptotes of f , evaluate lim f(x) and lim f(x). By looking at x→1 x→−1 each one-sided limit as x → 1, we see that 2 2 x lim + x→1 2 = −∞ and x lim − 1−x x→1 = ∞. 2 1−x In addition, by looking at each one-sided limit as x → −1, we find that 2 2 x lim + x→−1 2 = ∞ x and 1−x lim − x→−1 = −∞. 2 1−x Step 6: Calculate the first derivative: 2 2 (1−x )(2x)−x (−2x) f'(x) = 2 2 = 2x 2 (1−x ) 2 . (1−x ) Critical points occur at points x where f'(x) = 0 or f'(x) is undefined. We see that f'(x) = 0 when x = 0. The derivative f' is not undefined at any point in the domain of f . However, x = ±1 are not in the domain of f . Therefore, to determine where f is increasing and where f is decreasing, divide the interval (−∞, ∞) into four smaller intervals: (−∞, −1), (−1, 0), (0, 1), and (1, ∞) , and choose a test point in each interval to determine the sign of f'(x) in each of these intervals. The values x = −2, x = − , x = , and x = 2 are good choices for test points as shown in the following table. 10/21/2020 1 1 2 2 Interval Test point Sign of f'(x) = (−∞, −1) x = −2 −/+ = − f is decreasing (−1, 0) x = −/2 −/+ = − f is decreasing 3.7.3 CC-BY-NC-SA 2x 2 2 (1−x ) Conclusion https://math.libretexts.org/@go/page/13717 Interval Test point Sign of f'(x) = Conclusion 2x 2 2 (1−x ) (0, 1) x = 1/2 +/+ = + f is increasing (1, ∞) x = 2 +/+ = + f is increasing From this analysis, we conclude that f has a local minimum at x = 0 but no local maximum. Step 7: Calculate the second derivative: 2 2 f ′′ 2 (1−x ) (2)−2x(2(1−x )(−2x)) (x) = 2 4 (1−x ) 2 = 2 2 (1−x )[2(1−x )+8x ] 2 4 (1−x ) 2 2 2(1−x )+8x = 2 3 (1−x ) 2 = 6x +2 2 3 . (1−x ) To determine the intervals where f is concave up and where f is concave down, we first need to find all points x where f (x) = 0 or f (x) is undefined. Since the numerator 6x + 2 ≠ 0 for any x, f (x) is never zero. Furthermore, f is not undefined for any x in the domain of f . However, as discussed earlier, x = ±1 are not in the domain of f . Therefore, to determine the concavity of f , we divide the interval (−∞, ∞) into the three smaller intervals (−∞, −1), (−1, −1) , and (1, ∞) , and choose a test point in each of these intervals to evaluate the sign of f (x). in each of these intervals. The values x = −2, x = 0 , and x = 2 are possible test points as shown in the following table. ′′ 2 ′′ ′′ ′′ ′′ Interval Test Point Sign of f Conclusion 2 ′′ (x) = 6x +2 2 3 (1−x ) (−∞, −1) x = −2 +/− = − (−1, −1) x = 0 +/+ = + (1, ∞) x = 2 +/− = − f is concave dow f f is concave up is concave dow Combining all this information, we arrive at the graph of f shown below. Note that, although f changes concavity at x = −1 and x = 1 , there are no inflection points at either of these places because f is not continuous at x = −1 or x = 1. 10/21/2020 3.7.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/13717 EXERCISE 3.7.2 Sketch a graph of f(x) = (3x+5) (8+4x). Hint A line y = L is a horizontal asymptote of f if the limit as x → ∞ or the limit as x → −∞ of f(x) is L. A line x = a is a vertical asymptote if at least one of the one-sided limits of f as x → a is ∞ or −∞. Answer EXAMPLE 3.7.3: SKETCHING A RATIONAL FUNCTION WITH AN OBLIQUE ASYMPTOTE 2 Sketch the graph of f(x) = x (x−1) Solution Step 1: The domain of f is the set of all real numbers x except x = 1. Step 2: Find the intercepts. We can see that when x = 0, f(x) = 0, so (0, 0) is the only intercept. Step 3: Consider $f(-x)=\frac{(-x)^2}{((-x)−1)}= \frac{x^2}{(-x−1)}$, which is neither equal to −f (x) nor f (x). There fore there is no symmetry. Step 4: Evaluate the limits at infinity. Since the degree of the numerator is one more than the degree of the denominator, asymptote. To find the oblique asymptote, use long division of polynomials to write 2 f(x) = x x−1 = x +1+ 1 x−1 f must have an oblique . Since 1/(x − 1) → 0 as x → ±∞, f(x) approaches the line y = x + 1 as x → ±∞ . The line y = x + 1 is an oblique asymptote for f . Step 5: To check for vertical asymptotes, look at where the denominator is zero. Here the denominator is zero at limits as x → 1, we find 2 + Looking at both one-sided 2 x lim x→1 x = 1. = ∞ x −1 x and lim − x→1 = −∞. x −1 Therefore, x = 1 is a vertical asymptote, and we have determined the behavior of f as x approaches 1 from the right and the left. Step 6: Calculate the first derivative: 2 2 (x−1)(2x)−x (1) f'(x) = = 2 (x−1) x −2x 2 . (x−1) We have f'(x) = 0 when x − 2x = x(x − 2) = 0 . Therefore, x = 0 and x = 2 are critical points. Since f is undefined at x = 1 , we need to divide the interval (−∞, ∞) into the smaller intervals (−∞, 0), (0, 1), (1, 2), and (2, ∞) , and choose a test point from each interval to evaluate the sign of f'(x) in each of these smaller intervals. For example, let x = −1, x = ,x = , and x = 3 be the test points as shown in the following table. 2 Interval Test Point 1 3 2 2 Sign of f ′′ (x) = Conclusion 2 3 (x−1) (−∞, 1) 10/21/2020 x = 0 +/− = − 3.7.5 CC-BY-NC-SA f is concave dow https://math.libretexts.org/@go/page/13717 Interval Test Point Sign of f ′′ (x) = Conclusion 2 3 (x−1) (x=2\) (1, ∞) f +/+ = + is concave up From the information gathered, we arrive at the following graph for f. EXERCISE 3.7.3 Find the oblique asymptote for f(x) = 3 (3x −2x+1) 2 (2x −4) . Hint Use long division of polynomials. Answer y = 3 2 x EXAMPLE 3.7.4: SKETCHING THE GRAPH OF A FUNCTION WITH A CUSP Sketch a graph of f(x) = (x − 1) 2/3 Solution Step 1: Since the cube-root function is defined for all real numbers x and (x − 1) 2/3 − − − − 2 3 − = (√x − 1 ) , the domain of f is all real numbers. Step 2: To find the y -intercept, evaluate f(0). Since f(0) = 1, the y -intercept is (0, 1) . To find the x -intercept, solve (x − 1) this equation is x = 1 , so the x -intercept is (1, 0). 2/3 = 0 . The solution of Step 3: Consider $f(-x)=((-x)−1)^{2/3}=(x+1)^{2/3}$, which is neither equal to −f (x) nor f (x). There fore there is no symmetry. Step 4: Since 2/3 lim (x − 1) = ∞, the function continues to grow without bound as x → ∞ and x → −∞. x→±∞ Step 5: The function has no vertical asymptotes. Step 6: To determine where f is increasing or decreasing, calculate f'. We find 2 f'(x) = −1/3 (x − 1) 3 2 = (3.7.5) 3(x − 1)1/3 This function is not zero anywhere, but it is undefined when x = 1. Therefore, the only critical point is x = 1. Divide the interval (−∞, ∞) into the smaller intervals (−∞, 1) and (1, ∞) , and choose test points in each of these intervals to determine the sign of f'(x) in each of these smaller intervals. Let x = 0 and x = 2 be the test points as shown in the following table. Interval Test Point Sign of f'(x) = 2 1/3 Conclusion 3(x−1) (−∞, 1) 10/21/2020 x = 0 +/− = − 3.7.6 CC-BY-NC-SA f is decreasing https://math.libretexts.org/@go/page/13717 Interval Test Point Sign of f'(x) = Conclusion 2 1/3 3(x−1) (1, ∞) x = 2 f +/+ = + is increasing We conclude that h as a local minimum at x = 1 . Evaluating f at x = 1 , we find that the value of f at the local minimum is zero. Note that f'(1) is undefined, so to determine the behavior of the function at this critical point, we need to examine lim f'(x). Looking at the one-sided limits, we have x→1 2 2 lim + x→1 1/3 = ∞\)and$ lim − 3(x − 1) x→1 = −∞. 1/3 (3.7.6) 3(x − 1) Therefore, f has a cusp at x = 1. Step 7: To determine concavity, we calculate the second derivative of f f ′′ 2 (x) = − : −4/3 (x − 1) 9 −2 = . (3.7.7) 9(x − 1)4/3 We find that f (x) is defined for all x , but is undefined when x = 1 . Therefore, divide the interval (−∞, ∞) into the smaller intervals (−∞, 1) and (1, ∞) , and choose test points to evaluate the sign of f (x) in each of these intervals. As we did earlier, let x = 0 and x = 2 be test points as shown in the following table. ′′ ′′ Interval Test Point Sign of f ′′ (x) = Conclusion −2 4/3 3(x−1) (−∞, 1) x = 0 −/+ = − f is concave dow (1, ∞) x = 2 \−(+/+=−$ f is concave dow From this table, we conclude that i s concave down everywhere. Combining all of this information, we arrive at the following graph for f . EXERCISE 3.7.4 Consider the function f(x) = 5 − x 2/3 . Determine the point on the graph where a cusp is located. Determine the end behavior of f . Hint A function f has a cusp at a point a if f(a) exists, f sided limit is −∞. ′ (a) is undefined, one of the one-sided limits as x → a of f ′ (x)is + ∞ , and the other one- Answer The function f has a cusp at (0, 5) lim − x→0 f'(x) = ∞ , lim f'(x) = −∞ . For end behavior, + lim f(x) = −∞. x→±∞ x→0 Example 3.7.5 : Exponential function Sketch a graph of f (x) = e 10/21/2020 2 −x 2 . 3.7.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/13717 Key Concepts The limit of f (x) is L as x → ∞ (or as x → −∞) if the values f (x) become arbitrarily close to L as xbecomes sufficiently large. The limit of f (x) is ∞ as x → ∞ if f (x) becomes arbitrarily large as x becomes sufficiently large. The limit of f (x) is −∞ as x → ∞ if f (x) < 0 and |f (x)| becomes arbitrarily large as x becomes sufficiently large. We can define the limit of f (x) as x approaches −∞ similarly. For a polynomial function p(x) = a x + a x + … + a x + a , where a ≠ 0 , the end behavior is determined by the leading term a x . If n ≠ 0, p(x) approaches ∞ or −∞ at each end. n n−1 n For a rational function f (x) = p(x) q(x), n−1 n 1 0 n n the end behavior is determined by the relationship between the degree of p and the degree of q. If the degree of p is less than the degree of q, the line y = 0 is a horizontal asymptote for f . If the degree of p is equal to the degree of q, then the line y = an bn is a horizontal asymptote, where a and b are the leading coefficients of p and q, respectively. If the degree of p is greater than the degree of q, then f approaches ∞ or −∞ at each end. n n Glossary end behavior the behavior of a function as x → ∞ and x → −∞ horizontal asymptote if lim f(x) = L or x→∞ lim f(x) = L , then y = L is a horizontal asymptote of f x→−∞ infinite limit at infinity a function that becomes arbitrarily large as x becomes large limit at infinity a function that becomes arbitrarily large as x becomes large oblique asymptote the line y = mx + b if f(x) approaches it as x → ∞ orx → −∞ Contributors {{template.ContribOpenStaxCalc()} Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 3.7.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/13717 Welcome to the Mathematics Library. This Living Library is a principal hub of the LibreTexts project, which is a multiinstitutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning. The LibreTexts approach is highly collaborative where an Open Access textbook environment is under constant revision by students, faculty, and outside experts to supplant conventional paper-based books. Courses 10/21/2020 Bookshelves 1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10969 3.8: Newton's Method In many areas of pure and applied mathematics, we are interested in finding solutions to an equation of the form f(x) = 0. For most functions, however, it is difficult—if not impossible—to calculate their zeroes explicitly. In this section, we take a look at a technique that provides a very efficient way of approximating the zeroes of functions. This technique makes use of tangent line approximations and is behind the method used often by calculators and computers to find zeroes. Describing Newton’s Method Consider the task of finding the solutions of f(x) = 0. If f is the first-degree polynomial f(x) = ax + b , then the solution of f(x) = 0 is given by the formula x = − . If f is the seconddegree polynomial f(x) = ax + bx + c , the solutions of f(x) = 0 can be found by using the quadratic formula. However, for polynomials of degree 3 or more, finding roots of f becomes more complicated. Although formulas exist for third- and fourth-degree polynomials, they are quite complicated. Also, if f is a polynomial of degree 5 or greater, it is known that no such formulas exist. For example, consider the function b a 2 5 f(x) = x 4 + 8x 3 + 4x − 2x − 7. (3.8.1) No formula exists that allows us to find the solutions of f(x) = 0. Similar difficulties exist for nonpolynomial functions. For example, consider the task of finding solutions of tan(x) − x = 0. No simple formula exists for the solutions of this equation. In cases such as these, we can use Newton’s method to approximate the roots. Newton’s method makes use of the following idea to approximate the solutions of f(x) = 0. By sketching a graph of f , we can estimate a root of f(x) = 0 . Let’s call this estimate x . We then draw the tangent line to f at x . If f'(x0) ≠ 0 , this tangent line intersects the x -axis at some point (x1, 0) . Now let x be the next approximation to the actual root. Typically, x is closer than x to an actual root. Next we draw the tangent line to f at x . If f'(x1) ≠ 0 , this tangent line also intersects the x axis, producing another approximation, x . We continue in this way, deriving a list of approximations: x , x , x , … . Typically, the numbers x , x , x , … quickly approach an actual root x∗, as shown in the following figure. 0 0 1 1 0 1 2 0 10/21/2020 1 2 0 1 3.8.1 CC-BY-NC-SA 2 https://math.libretexts.org/@go/page/10307 Figure 3.8.1 :The approximations x , x , x , … approach the actual root x∗. The approximations are derived by looking at tangent lines to the graph of f . 0 1 2 Now let’s look at how to calculate the approximations approximation, the approximation x is defined by letting (x line to f at x . The equation of this tangent line is given by 1 , 0) 1 If x is our first be the x -intercept of the tangent x0 , x1 , x2 , … . 0 0 y = f(x0 ) + f'(x0 )(x − x0 ). (3.8.2) f(x0 ) + f'(x0 )(x1 − x0 ) = 0. (3.8.3) Therefore, x must satisfy 1 Solving this equation for x , we conclude that 1 f(x0) x1 = x0 − ′ . (3.8.4) f (x0) Similarly, the point (x equation 2, 0) is the x -intercept of the tangent line to f at x . Therefore, x satisfies the 1 2 f(x1) x2 = x1 − ′ . (3.8.5) f (x1) In general, for n > 0, x satisfies n f(xn−1 ) xn = xn−1 − ′ . (3.8.6) f (xn−1 ) 10/21/2020 3.8.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10307 Next we see how to make use of this technique to approximate the root of the polynomial 3 f(x) = x − 3x + 1. EXAMPLE 3.8.1: FINDING A ROOT OF A POLYNOMIAL Use Newton’s method to approximate a root of x = 2 and find x , x , x , x , and x . 0 1 2 3 4 3 f(x) = x − 3x + 1 in the interval [1, 2] . Let 5 Solution From Figure, we see that f has one root over the interval (1, 2) . Therefore x = 2 seems like a reasonable first approximation. To find the next approximation, we use Equation. Since f(x) = x − 3x + 1 , the derivative is f'(x) = 3x − 3 . Using Equation with n = 1 (and a calculator that displays 10 digits), we obtain 0 3 2 f(x0 ) x1 = x0 − f(2) = 2− ′ f (x0 ) To find the next approximation, the calculator. We find that x2 ′ 3 = 2− ≈ 1.666666667. (3.8.7) 9 f (2) , we use Equation with n = 2 and the value of x1 stored on f(x1) x2 = x1 = ≈ 1.548611111. ′ (3.8.8) f (x1) Continuing in this way, we obtain the following results: x1 ≈ 1.666666667 (3.8.9) x2 ≈ 1.548611111 (3.8.10) x3 ≈ 1.532390162 (3.8.11) x4 ≈ 1.532088989 (3.8.12) x5 ≈ 1.532088886 (3.8.13) x6 ≈ 1.532088886. (3.8.14) We note that we obtained the same value for x and x . Therefore, any subsequent application of Newton’s method will most likely give the same value for x . 5 6 n 10/21/2020 3.8.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10307 EXERCISE 3.8.1 Figure 3.8.2 : The function f(x) = x − 3x + 1 has one root over the interval [1, 2]. Letting x = 0 , let’s use Newton’s method to approximate the root of f(x) = x − 3x + 1 over the interval [0, 1] by calculating x and x . 3 3 0 1 2 Hint Use Equation Answer x1 ≈ 0.33333333, x2 ≈ 0.347222222 (3.8.15) Newton’s method can also be used to approximate square roots. Here we show how to approximate – √2. This method can be modified to approximate the square root of any positive number. EXAMPLE 3.8.2: FINDING A SQUARE ROOT – Use Newton’s method to approximate √2 (Figure). Let f(x) = x − 2 , let x = 2 , and calculate – x , x , x , x , x . (We note that since f(x) = x − 2 has a zero at √2, the initial value x = 2 – is a reasonable choice to approximate √2.) 2 0 2 1 2 3 4 5 0 Solution For f(x) = x 2 − 2, f'(x) = 2x. xn = xn−1 − From Equation, we know that f ( xn−1 ) ′ f ( xn−1 ) 2 x = xn−1 − = = 1 2 1 2 n−1 −2 2xn−1 1 xn−1 + xn−1 (xn−1 + 2 xn−1 ). Therefore, x1 = 10/21/2020 1 2 (x0 + 2 x0 ) = 1 2 (2 + 2 2 ) = 1.5 3.8.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10307 x2 = 1 2 (x1 + 2 x1 ) = 1 2 (1.5 + 2 1.5 ) ≈ 1.416666667. Continuing in this way, we find that x1 = 1.5 x2 ≈ 1.416666667 x3 ≈ 1.414215686 x4 ≈ 1.414213562 x5 ≈ 1.414213562. Since we obtained the same value for x and x , it is unlikely that the value xn will change on – any subsequent application of Newton’s method. We conclude that √2 ≈ 1.414213562. 4 5 – Figure 3.8.3 : We can use Newton’s method to find √2. 10/21/2020 3.8.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10307 EXERCISE 3.8.2 – Use Newton’s method to approximate √3 by letting f(x) = x 2 −3 and x 0 = 3 . Find x and x . 1 2 Hint For f(x) = x 2 −3 , Equation reduces to x n = xn−1 2 + 3 2xn−1 . Answer x1 = 2, x2 = 1.75 When using Newton’s method, each approximation after the initial guess is defined in terms of the previous approximation by using the same formula. In particular, by defining the function f (x) F (x) = x − [ f '(x) ] , we can rewrite Equation as xn = F (xn−1 ) . This type of process, where each is defined in terms of x by repeating the same function, is an example of an iterative process. Shortly, we examine other iterative processes. First, let’s look at the reasons why Newton’s method could fail to find a root. xn n−1 Failures of Newton’s Method Typically, Newton’s method is used to find roots fairly quickly. However, things can go wrong. Some reasons why Newton’s method might fail include the following: 1. At one of the approximations x , the derivative f' is zero at x , but f(x ) ≠ 0 . As a result, the tangent line of f at x does not intersect the x -axis. Therefore, we cannot continue the iterative process. 2. The approximations x , x , x , … may approach a different root. If the function f has more than one root, it is possible that our approximations do not approach the one for which we are looking, but approach a different root (see Figure). This event most often occurs when we do not choose the approximation x close enough to the desired root. 3. The approximations may fail to approach a root entirely. In Example, we provide an example of a function and an initial guess x such that the successive approximations never approach a root because the successive approximations continue to alternate back and forth between two values. n n n n 0 1 2 0 0 10/21/2020 3.8.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10307 Figure 3.8.4 : If the initial guess x is too far from the root sought, it may lead to approximations that approach a different root. 0 EXAMPLE 3.8.3: WHEN NEWTON’S METHOD FAILS Consider the function f(x) = x to approach a root of f . 3 − 2x + 2 . Let x 0 = 0 . Show that the sequence x 1 , x2 , … fails Solution For f(x) = x 3 − 2x + 2, the derivative is f'(x) = 3x 2 f(x0) x1 = x0 − −2 .Therefore, f(0) = 0− f'(x0) 2 = − = 1. (3.8.16) = 0. (3.8.17) −2 f'(0) In the next step, f(x1) x2 = x1 − ′ f (x1) f(1) = 1− 1 = 1− f'(1) 1 Consequently, the numbers x , x , x , … continue to bounce back and forth between 0 and 1 and never get closer to the root of f which is over the interval [−2, −1] (see Figure). Fortunately, if we choose an initial approximation x closer to the actual root, we can avoid this situation. 0 1 2 0 10/21/2020 3.8.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10307 Figure 3.8.5 : The approximations continue to alternate between 0 and 1 and never approach the root of f . EXERCISE 3.8.3 For f(x) = x 3 − 2x + 2, let x 0 = −1.5 and find x and x . 1 2 Hint Use Equation. Answer x1 ≈ −1.842105263, x2 ≈ −1.772826920 From Example, we see that Newton’s method does not always work. However, when it does work, the sequence of approximations approaches the root very quickly. Discussions of how quickly the sequence of approximations approach a root found using Newton’s method are included in texts on numerical analysis. Other Iterative Processes As mentioned earlier, Newton’s method is a type of iterative process. We now look at an example of a different type of iterative process. Consider a function F and an initial number x . Define the subsequent numbers x by the formula x = F (x ). This process is an iterative process that creates a list of numbers x , x , x , … , x , … . This list of numbers may approach a finite number x∗ as n gets larger, or it may not. In Example, we see an example of a function F and an initial guess x such that the resulting list of numbers approaches a finite value. 0 n n n−1 0 1 2 n 0 EXAMPLE 3.8.4: FINDING A LIMIT FOR AN ITERATIVE PROCESS Let and let x = 0 . For all n ≥ 1 , let x = F (x ) . Find the values x ,x ,x ,x ,x . Make a conjecture about what happens to this list of numbers x , x , x … , x , … as n → ∞ . If the list of numbers x , x , x , … approaches a finite number x∗, then x∗ satisfies x∗ = F (x∗) , and (x*\) is called a fixed point of F . 1 F (x) = 1 2 3 1 2 3 2 4 0 n n−1 5 n Solution: If x 0 10/21/2020 x +4 = 0 1 2 3 , then 3.8.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10307 x1 = x2 = x3 = x4 = x5 = x6 = x7 = x8 = x9 = 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 (0) + 4 = 4 (4) + 4 = 6 (6) + 4 = 7 (7) + 4 = 7.5 (7.5) + 4 = 7.75 (7.75) + 4 = 7.875 (7.875) + 4 = 7.9375 (7.9375) + 4 = 7.96875 (7.96875) + 4 = 7.984375. From this list, we conjecture that the values x approach 8. n Figure provides a graphical argument that the values approach 8 as n → ∞ . Starting at the point (x , x ), we draw a vertical line to the point (x , F (x )) . The next number in our list is x = F (x ) . We use x to calculate x . Therefore, we draw a horizontal line connecting (x , x ) to the point (x , x ) on the line y = x , and then draw a vertical line connecting (x , x ) to the point (x , F (x )). The output F (x ) becomes x . Continuing in this way, we could create an infinite number of line segments. These line segments are trapped between the lines F (x) = + 4 and y = x . The line segments get closer to the intersection point of these two lines, which occurs when x = F (x) . Solving the equation x = + 4, we conclude they intersect at x = 8 . Therefore, our graphical evidence agrees with our numerical evidence that the list of numbers x , x , x , … approaches x∗ = 8 as n → ∞ . 0 0 0 1 0 1 1 1 0 2 0 1 1 1 1 1 1 2 x 2 x 2 0 1 2 Figure 3.8.6 : This iterative process approaches the value x∗ = 8. 10/21/2020 3.8.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10307 EXERCISE 3.8.4 Consider the function F (x) = x + 6 . Let x = 0 and let x = F (x ) for n ≥ 2 . Find x ,x ,x ,x ,x . Make a conjecture about what happens to the list of numbers x 1 x , x , … x , … as n → ∞. 1 0 3 1 , 2 3 2 3 4 n n−1 5 n Hint Consider the point where the lines y = x and y = F (x) intersect. Answer x1 = 6, x2 = 8, x3 = 26 3 , x4 = 80 9 , x5 = 242 27 ; x∗ = 9 ITERATIVE PROCESSES AND CHAOS Iterative processes can yield some very interesting behavior. In this section, we have seen several examples of iterative processes that converge to a fixed point. We also saw in Example that the iterative process bounced back and forth between two values. We call this kind of behavior a 2cycle. Iterative processes can converge to cycles with various periodicities, such as 2−cycles, 4−cycles (where the iterative process repeats a sequence of four values), 8-cycles, and so on. Some iterative processes yield what mathematicians call chaos. In this case, the iterative process jumps from value to value in a seemingly random fashion and never converges or settles into a cycle. Although a complete exploration of chaos is beyond the scope of this text, in this project we look at one of the key properties of a chaotic iterative process: sensitive dependence on initial conditions. This property refers to the concept that small changes in initial conditions can generate drastically different behavior in the iterative process. Probably the best-known example of chaos is the Mandelbrot set (see Figure), named after Benoit Mandelbrot (1924–2010), who investigated its properties and helped popularize the field of chaos theory. The Mandelbrot set is usually generated by computer and shows fascinating details on enlargement, including self-replication of the set. Several colorized versions of the set have been shown in museums and can be found online and in popular books on the subject. 10/21/2020 3.8.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10307 Figure 3.8.7 : The Mandelbrot set is a well-known example of a set of points generated by the iterative chaotic behavior of a relatively simple function. In this project we use the logistic map f(x) = rx(1 − x) (3.8.18) where x ∈ [0, 1] and r > 0 as the function in our iterative process. The logistic map is a deceptively simple function; but, depending on the value of r , the resulting iterative process displays some very interesting behavior. It can lead to fixed points, cycles, and even chaos. To visualize the long-term behavior of the iterative process associated with the logistic map, we will use a tool called a cobweb diagram. As we did with the iterative process we examined earlier in this section, we first draw a vertical line from the point (x , 0) to the point (x , f(x0)) = (x , x ) . We then draw a horizontal line from that point to the point (x , x ), then draw a vertical line to (x , f(x )) = (x , x ) , and continue the process until the long-term behavior of the system becomes apparent. Figure shows the long-term behavior of the logistic map when r = 3.55 and x = 0.2 . (The first 100 iterations are not plotted.) The long-term behavior of this iterative process is an 8-cycle. 0 0 0 1 1 1 1 1 1 2 0 10/21/2020 3.8.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10307 Figure 3.8.8 : A cobweb diagram for f(x) = 3.55x(1 − x) is presented here. The sequence of values results in an 8-cycle. 1. Let r = 0.5 and choose x = 0.2 . Either by hand or by using a computer, calculate the first 10 values in the sequence. Does the sequence appear to converge? If so, to what value? Does it result in a cycle? If so, what kind of cycle (for example, 2 − cycle, 4 − cycle. )? 2. What happens when r = 2? 3. For r = 3.2 and r = 3.5, calculate the first 100 sequence values. Generate a cobweb diagram for each iterative process. (Several free applets are available online that generate cobweb diagrams for the logistic map.) What is the long-term behavior in each of these cases? 4. Now let r = 4. Calculate the first 100 sequence values and generate a cobweb diagram. What is the long-term behavior in this case? 5. Repeat the process for r = 4, but let x = 0.201. How does this behavior compare with the behavior for x = 0.2 ? 0 0 0 Key Concepts Newton’s method approximates roots of f (x) = 0 by starting with an initial approximation x , then uses tangent lines to the graph of f to create a sequence of approximations 0 x1 , x2 , x3 , … . Typically, Newton’s method is an efficient method for finding a particular root. In certain cases, Newton’s method fails to work because the list of numbers x , x , x , … does not approach a finite value or it approaches a value other than the root sought. Any process in which a list of numbers x , x , x , … is generated by defining an initial number x and defining the subsequent numbers by the equation x = F (x ) for some function F is an iterative process. Newton’s method is an example of an iterative process, 0 0 1 10/21/2020 n f (x) f '(x) ] 2 2 0 where the function F (x) = x − [ 1 n−1 for a given function f . 3.8.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/10307 Glossary iterative process process in which a list of numbers x , x and defining x = F (x ) for n ≥ 1 0 n 1 , x2 , x3 … is generated by starting with a number x 0 n−1 Newton’s method method for approximating roots of f(x) = 0; using an initial guess x0; each subsequent approximation is defined by the equation x n = xn−1 − f ( xn−1 ) ′ f ( xn−1 ) Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. DS 10/21/2020 — DekiScript — No description 3.8.13 CC-BY-NC-SA https://math.libretexts.org/@go/page/10307 3.8E: Exercises Exercise 3.8E . 1 For the following exercises, write Newton’s formula as x n+1 1) f(x) = x +1 2) f(x) = x + 2x + 1 2 3 = F (xn ) for solving f(x) = 0 . Answer 3 F (xn ) = xn − xn +2 xn +1 2 3 xn +2 3) f(x) = sinx 4) f(x) = e x Answer F (xn ) = xn − 5) f(x) = x 3 e e xn xn x + 3x e Exercise 3.8E . 2 For the following exercises, solve f(x) = 0 using the iteration x = x f(x ) , which differs slightly from Newton’s method. Find a c that works and a c that fails to converge, with the exception of c = 0. n+1 1) f(x) = x 2 − 4, with x 0 n−c n = 0 Answer |c| > 0.5 fails, |c| ≤ 0.5 works 2) f(x) = x 2 − 4x + 3, with x 0 = 2 3) What is the value of “c” for Newton’s method? Answer c = 1 f '( xn ) Exercise 3.8E . 3 For the following exercises, start at a. x 0 b. x 0 = 0.6 and = 2. Compute x and x using the specified iterative method. 1 10/21/2020 2 3.8E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10937 1) x = xn − 2) x = 2xn (1 − xn ) 2 n+1 n+1 1 2 Answer a. x 1 b. x 1 = 12 25 , x2 = 312 625 = −4, x2 = −40 3) x − − = √xn 4) x = n+1 n+1 1 √xn Answer a. x 1 b. x 1 = 1.291, x2 = 0.8801 = 0.7071, x2 = 1.189 5) x = 3xn (1 − xn ) 6) x = xn + xn−2 n+1 2 n+1 Answer a. x 1 b. x 1 =− 26 25 , x2 = − 1224 625 = 4, x2 = 18 7) x = 8) x = | xn | n+1 n+1 1 2 xn − 1 Answer a. x 1 b. x 1 = 6 10 , x2 = 6 10 = 2, x2 = 2 Exercise 3.8E . 4 For the following exercises, solve to four decimal places using Newton’s method and a computer or calculator. Choose any initial guess x that is not the exact root. 0 1) x − 10 = 0 2) x − 100 = 0 2 4 Answer 3.1623or − 3.1623 3) x −x = 0 4) x −x = 0 2 3 Answer 10/21/2020 3.8E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10937 0, −1or1 5) x + 5cos(x) = 0 6) x + tan(x) = 0, choose x 0 ∈ (− π 2 , π 2 ) Answer 0 7) 1 1−x = 2 8) 1 + x + x 2 3 +x 4 +x = 2 Answer or −1.2906 0.5188 9) x 3 3 + (x + 1) 10) x = sin2 ( = 10 3 x) Answer 0 Exercise 3.8E . 5 For the following exercises, use Newton’s method to find the fixed points of the function where f(x) = x ; round to three decimals. 1) sinx 2) tan(x) on x = ( π 2 , 3π 2 ) Answer 4.493 3) e x −2 4) ln(x) + 2 Answer 0.159, 3.146 Exercise 3.8E . 6 Newton’s method can be used to find maxima and minima of functions in addition to the roots. In this case apply Newton’s method to the derivative function f'(x) to find its roots, instead of the original function. For the following exercises, consider the formulation of the method. To find candidates for maxima and minima, we need to find the critical points f '(x) = 0. Show that to solve for the critical points of a function f (x), Newton’s method is given by 10/21/2020 3.8E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10937 xn+1 = xn − f'(xn ) f ′′ . (xn ) What additional restrictions are necessary on the function f ? Answer We need f to be twice continuously differentiable. Exercise 3.8E . 7 For the following exercises, use Newton’s method to find the location of the local minima and/or maxima of the following functions; round to three decimals. 1) Minimum of f(x) = x 2 + 2x + 4 2) Minimum of f(x) = 3x 3 2 + 2x − 16 Answer x =0 3) Minimum of f(x) = x 2 x e 4) Maximum of f(x) = x + 1 x Answer x = −1 5) Maximum of f(x) = x 3 6) Maximum of f(x) = 2 + 10x + 15x − 2 3 √x−√x x Answer x = 5.619 7) Minimum of f(x) = x 2 8) Minimum of f(x) = x 4 closest non-zero minimum to x = 0 sinx, 3 +x 2 + 3x + 12x + 6 Answer x = −1.326 Exercise 3.8E . 8 For the following exercises, use the specified method to solve the equation. If it does not work, explain why it does not work. 1) Newton’s method, x 2 +2 = 0 2) Newton’s method, 0 = e x Answer There is no solution to the equation. 10/21/2020 3.8E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10937 3) Newton’s method, 0 = 1 + x starting at x 2 0 4) Solving x starting at x 3 n+1 = −xn 0 = 0 = −1 Answer It enters a cycle. Exercise 3.8E . 9 For the following exercises, use the secant method, an alternative iterative method to Newton’s method. The formula is given by xn = xn−1 − f(xn−1 ) 1) a root to 0 = x 2 −x −3 xn−1 −xn−2 f ( xn−1 )−f ( xn−2 ) . accurate to three decimal places. 2) Find a root to 0 = sinx + 3x accurate to four decimal places. Answer 0 3) Find a root to 0 = e x −2 accurate to four decimal places. 4) Find a root to ln(x + 2) = 1 2 accurate to four decimal places. Answer −0.3513 5) Why would you use the secant method over Newton’s method? What are the necessary restrictions on f ? Exercise 3.8E . 10 For the following exercises, use both Newton’s method and the secant method to calculate a root for the following equations. Use a calculator or computer to calculate how many iterations of each are needed to reach within three decimal places of the exact answer. For the secant method, use the first guess from Newton’s method. 1) f(x) = x 2 + 2x + 1, x0 = 1 Answer Newton: 11 iterations, secant: 16 iterations 2) f(x) = x 2 , x0 = 1 3) f(x) = sinx, x 0 = 1 Answer Newton: three iterations, secant: six iterations 10/21/2020 3.8E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10937 4) f(x) = e x 5) f(x) = x 3 − 1, x0 = 2 + 2x + 4, x0 = 0 Answer Newton: five iterations, secant: eight iterations Exercise 3.8E . 11 In the following exercises, consider Kepler’s equation regarding planetary orbits, M = E − εsin(E) , where M is the mean anomaly, E is eccentric anomaly, and ε measures eccentricity. 1) Use Newton’s method to solve for the eccentric anomaly E when the mean anomaly M = π 3 and the eccentricity of the orbit ε = 0.25; round to three decimals. 2) Use Newton’s method to solve for the eccentric anomaly E when the mean anomaly and the eccentricity of the orbit ε = 0.8; round to three decimals. M = 3π 2 Answer E = 4.071 Exercise 3.8E . 12 The following exercises consider a bank investment. The initial investment is years, the investment has tripled to $30, 000. $10, 000 . After 25 1) Use Newton’s method to determine the interest rate if the interest was compounded annually. 2) Use Newton’s method to determine the interest rate if the interest was compounded continuously. Answer 4.394 Exercise 3.8E . 13 The cost for printing a book can be given by the equation the break-even point if the printer sells each book for $20. C (x) = 1000 + 12x + ( 1 2 2/3 )x . Use Newton’s method to find Answer Under Construction 10/21/2020 3.8E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10937 3.9: Anti derivatives and Rectilinear Motion At this point, we have seen how to calculate derivatives of many functions and have been introduced to a variety of their applications. We now ask a question that turns this process around: Given a function f , how do we find a function with the derivative f and why would we be interested in such a function? We answer the first part of this question by defining antiderivatives. The antiderivative of a function f is a function with a derivative f . Why are we interested in antiderivatives? The need for antiderivatives arises in many situations, and we look at various examples throughout the remainder of the text. Here we examine one specific example that involves rectilinear motion. In our examination in Derivatives of rectilinear motion, we showed that given a position function s(t) of an object, then its velocity function v(t) is the derivative of s(t) —that is, v(t) = s'(t) . Furthermore, the acceleration a(t) is the derivative of the velocity v(t) —that is, a(t) = v'(t) = s (t) . Now suppose we are given an acceleration function a , but not the velocity function v or the position function s . Since a(t) = v'(t) , determining the velocity function requires us to find an antiderivative of the acceleration function. Then, since v(t) = s'(t), determining the position function requires us to find an antiderivative of the velocity function. Rectilinear motion is just one case in which the need for antiderivatives arises. We will see many more examples throughout the remainder of the text. For now, let’s look at the terminology and notation for antiderivatives, and determine the antiderivatives for several types of functions. We examine various techniques for finding antiderivatives of more complicated functions later in the text (Introduction to Techniques of Integration). ′′ The Reverse of Differentiation At this point, we know how to find derivatives of various functions. We now ask the opposite question. Given a function f , how can we find a function with derivative f ? If we can find a function F derivative f , we call F an antiderivative of f . Definition: Antiderivative A function F is an antiderivative of the function f if F '(x) = f (x) (3.9.1) for all x in the domain of f . Consider the function f (x) = 2x. Knowing the power rule of differentiation, we conclude that F (x) = x is an antiderivative of f since F '(x) = 2x. Are there any other antiderivatives of f ? Yes; since the derivative of any constant C is zero, x + C is also an antiderivative of – 2x. Therefore, x + 5 and x − √2 are also antiderivatives. Are there any others that are not of the form x + C for some constant C ? The answer is no. From Corollary 2 of the Mean Value Theorem, we know that if F and G are differentiable functions such that F '(x) = G'(x), then F (x) − G(x) = C for some constant C . This fact leads to the following important theorem. 2 2 2 2 2 General Form of an Antiderivative Let F be an antiderivative of f over an interval I . Then, I. for each constant C , the function F (x) + C is also an antiderivative of f over I ; II. if G is an antiderivative of f over I , there is a constant C for which G(x) = F (x) + C over I . In other words, the most general form of the antiderivative of f over I is F (x) + C . We use this fact and our knowledge of derivatives to find all the antiderivatives for several functions. Example 3.9.1 : Finding Antiderivatives For each of the following functions, find all antiderivatives. a. 2 f (x) = 3x b. f (x) = 1 x c. f (x) = cos x d. f (x) = e x Solution: a. Because d 3 2 (x ) = 3 x dx then F (x) = x is an antiderivative of 3x . Therefore, every antiderivative of function of the form x + C is an antiderivative of 3x . 3 2 3 2 3x is of the form 3 x +C for some constant C , and every 2 b. Let f (x) = ln |x|. For x > 0, f (x) = ln(x) and 10/21/2020 3.9.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10698 d 1 (ln x) = . dx x Forx < 0, f (x) = ln(−x) and d 1 (ln(−x)) = − dx 1 = . −x x Therefore, d 1 (ln |x|) = dx Thus, F (x) = ln |x| is an antiderivative of 1 x . x . Therefore, every antiderivative of function of the form ln |x| + C is an antiderivative of 1 x 1 x is of the form ln |x| + C for some constant C and every . c. We have d (sin x) = cos x, dx so F (x) = sin x is an antiderivative of cos x. Therefore, every antiderivative of every function of the form sin x + C is an antiderivative of cos x. is of the form cos x sin x + C for some constant C and d. Since d x x (e ) = e , dx then F (x) = e is an antiderivative of e . Therefore, every antiderivative of function of the form e + C is an antiderivative of e . x x x e x is of the form e x +C for some constant C and every x Exercise 3.9.1 Find all antiderivatives of f (x) = sin x . Hint What function has a derivative of sin x ? Answer − cos x + C Indefinite Integrals We now look at the formal notation used to represent antiderivatives and examine some of their properties. These properties allow us to find antiderivatives of more complicated functions. Given a function f , we use the notation f '(x) or df dx to denote the derivative of f . Here we introduce notation for antiderivatives. If F is an antiderivative of f , we say that F (x) + C is the most general antiderivative of f and write ∫ f (x)dx = F (x) + C . (3.9.2) The symbol ∫ is called an integral sign, and ∫ f (x)dx is called the indefinite integral of f . Definition: Indefinite Integrals Given a function f , the indefinite integral of f , denoted ∫ f (x)dx, (3.9.3) is the most general antiderivative of f . If F is an antiderivative of f , then ∫ f (x)dx = F (x) + C . (3.9.4) The expression f (x) is called the integrand and the variable x is the variable of integration. Given the terminology introduced in this definition, the act of finding the antiderivatives of a function f is usually referred to as integrating f . 10/21/2020 3.9.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10698 For a function f and an antiderivative F , the functions F (x) + C , where C is any real number, is often referred to as the family of antiderivatives of f . For example, since x is an antiderivative of 2x and any antiderivative of 2x is of the form x + C , we write 2 2 2 ∫ The collection of all functions of the form x graph of this family of antiderivatives. 2 + C, 2xdx = x + C. (3.9.5) where C is any real number, is known as the family of antiderivatives of 2x. Figure shows a Figure 3.9.1 : The family of antiderivatives of 2x consists of all functions of the form x 2 +C , where C is any real number. For some functions, evaluating indefinite integrals follows directly from properties of derivatives. For example, for n ≠ −1 , n+1 x n ∫ x dx = + C, n+1 which comes directly from n+1 d dx n x ( x ) = (n + 1) n+1 n =x n+1 . This fact is known as the power rule for integrals. Power Rule for Integrals For n ≠ −1, n+1 ∫ n x x dx = + C. (3.9.6) n+1 Evaluating indefinite integrals for some other functions is also a straightforward calculation. The following table lists the indefinite integrals for several common functions. A more complete list appears in Appendix B. Table : Integration Formulas Differentiation Formula Indefinite Integral d 0 ∫ kdx = ∫ kx dx = kx + C (k) = 0 dx n+1 d n n−1 d x n (x ) = n x ∫ x dn = dx 1 ∫ x d x +C for n ≠ −1 1 (ln |x|) = dx n+1 x x dx = ln |x| + C x (e ) = e x ∫ e dx = e +C dx x d a ( dx 10/21/2020 x x x ) = a a ∫ a dx = ln(a) +C ln(a) 3.9.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10698 Differentiation Formula Indefinite Integral d (sin x) = cosx ∫ cosxdx = sin x + C (cosx) = − sin x ∫ sin xdx = − cosx + C dx d dx d 2 2 (tan x) = sec x ∫ sec xdx = tan x + C (cscx) = −cscx cot x ∫ cscx cot xdx = −cscx + C (secx) = secx tan x ∫ secx tan xdx = secx + C dx d dx d dx d 2 2 (cot x) = −csc x ∫ csc xdx = − cot x + C dx d (sin −1 1 x) = dx d (tan −1 1 ∫ − −−− − 2 √1 − x 1 x) = dx −1 (sec −1 2 dx = tan x +C −1 x +C 1 +x 1 |x|) = dx = sin 1 ∫ 2 1 +x d − −−− − √1 − x2 1 ∫ − −−− − 2 x √x − 1 − −−− − 2 x √x − 1 −1 dx = sec |x| + C From the definition of indefinite integral of f , we know ∫ f (x)dx = F (x) + C (3.9.7) if and only if F is an antiderivative of f . Therefore, when claiming that ∫ f (x)dx = F (x) + C (3.9.8) it is important to check whether this statement is correct by verifying that F '(x) = f (x). Example 3.9.2 : Verifying an Indefinite Integral Each of the following statements is of the form ∫ f (x)dx = F (x) + C . Verify that each statement is correct by showing that F '(x) = f (x). 2 a. x x ∫ (x + e )dx = b. ∫ x e +e x +C 2 x dx = x e x −e x +C Solution: a. Since 2 d x ( +e dx x + C) = x + e , x 2 the statement 2 x x ∫ (x + e )dx = +e x +C 2 is correct. 2 x Note that we are verifying an indefinite integral for a sum. Furthermore, and e are antiderivatives of x and e , respectively, and the sum x x 2 of the antiderivatives is an antiderivative of the sum. We discuss this fact again later in this section. b. Using the product rule, we see that d (x e x −e x + C) = e x + xe x −e x x = xe . dx Therefore, the statement ∫ x x e dx = x e x −e x +C is correct. 10/21/2020 3.9.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10698 Note that we are verifying an indefinite integral for a product. The antiderivative xex−ex is not a product of the antiderivatives. Furthermore, the product of antiderivatives, x e /2 is not an antiderivative of xe since 2 x x 2 d x e x ( dx 2 ) = xe x x e + 2 x ≠ xe 2 x . In general, the product of antiderivatives is not an antiderivative of a product. Exercise 3.9.2 Verify that ∫ x cos x dx = x sin x + cos x + C . Hint Calculate d (x sin x + cos x + C ). dx Answer d (x sin x + cos x + C ) = sin x + x cos x − sin x = x cos x dx In Table, we listed the indefinite integrals for many elementary functions. Let’s now turn our attention to evaluating indefinite integrals for more complicated functions. For example, consider finding an antiderivative of a sum f + g . In Example a. we showed that an antiderivative of the 2 sum x + e is given by the sum ( x x )+e 2 x —that is, an antiderivative of a sum is given by a sum of antiderivatives. This result was not specific to this example. In general, if F and G are antiderivatives of any functions f and g , respectively, then d (F (x) + G(x)) = F '(x) + G'(x) = f (x) + g(x). dx Therefore, F (x) + G(x) is an antiderivative of f (x) + g(x) and we have ∫ (f (x) + g(x))dx = F (x) + G(x) + C . Similarly, ∫ (f (x) − g(x))dx = F (x) − G(x) + C . In addition, consider the task of finding an antiderivative of kf (x), where k is any real number. Since d d (kf (x)) = k dx F (x) = kF '(x) dx for any real number k , we conclude that ∫ kf (x)dx = kF (x) + C . These properties are summarized next. Properties of Indefinite Integrals Let F and G be antiderivatives of f and g , respectively, and let k be any real number. Sums and Differences ∫ (f (x) ± g(x))dx = F (x) ± G(x) + C Constant Multiples ∫ kf (x)dx = kF (x) + C From this theorem, we can evaluate any integral involving a sum, difference, or constant multiple of functions with antiderivatives that are known. Evaluating integrals involving products, quotients, or compositions is more complicated (see Exampleb. for an example involving an antiderivative of a product.) We look at and address integrals involving these more complicated functions in Introduction to Integration. In the next example, we examine how to use this theorem to calculate the indefinite integrals of several functions. 10/21/2020 3.9.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10698 Example 3.9.3 : Evaluating Indefinite Integrals Evaluate each of the following indefinite integrals: a. 3 ∫ (5 x 2 x b. ∫ c. 2 − 7x + 3x + 4)dx − + 4 √x 3 dx x 4 ∫ dx 2 1 +x 2 t d. ∫ e. dt 2 1 +t ∫ tan x cos xdx Solution: a. Using Note, we can integrate each of the four terms in the integrand separately. We obtain 3 ∫ (5 x 2 − 7x 3 2 + 3x + 4)dx = ∫ 5 x dx − ∫ 7 x dx + ∫ 3xdx + ∫ 4dx. From the second part of Note, each coefficient can be written in front of the integral sign, which gives 3 2 3 2 ∫ 5 x dx − ∫ 7 x dx + ∫ 3xdx + ∫ 4dx = 5 ∫ x dx − 7 ∫ x dx + 3 ∫ xdx + 4 ∫ 1dx. Using the power rule for integrals, we conclude that 3 ∫ (5 x 5 2 − 7x 4 + 3x + 4)dx = x 7 − 4 3 x 3 + 3 2 x + 4x + C . 2 b. Rewrite the integrand as 2 x 3 − + 4 √x 3 − 4 √x 2 x = x + = 0. x x Then, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have 4 −2/3 ∫ (x + )dx = ∫ xdx + 4 ∫ x dx 2/3 x 1 = 2 x 1 +4 2 (−2/3)+1 x −2 ( + C ]) )+1 3 1 = 2 x 1/3 + 12 x + C. 2 c. Using Note, write the integral as 1 4∫ Then, use the fact that tan −1 (x) is an antiderivative of 1 2 1 + x2 dx. to conclude that (1 + x ) 4 ∫ −1 2 dx = 4tan (x) + C . 1 +x d. Rewrite the integrand as sin x tan x cos x = cos x = sin x. cos x Therefore, ∫ tan x cos x = ∫ sin x = − cos x + C . Exercise 3.9.3 Evaluate ∫ (4x 3 2 − 5x + x − 7)dx . Hint Integrate each term in the integrand separately, making use of the power rule. Answer 4 x 5 − 3 x 3 10/21/2020 1 + 2 x − 7x + C 2 3.9.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10698 Initial-Value Problems We look at techniques for integrating a large variety of functions involving products, quotients, and compositions later in the text. Here we turn to one common use for antiderivatives that arises often in many applications: solving differential equations. A differential equation is an equation that relates an unknown function and one or more of its derivatives. The equation dy = f (x) dx is a simple example of a differential equation. Solving this equation means finding a function y with a derivative f . Therefore, the solutions of Equation are the antiderivatives of f . If F is one antiderivative of f , every function of the form y = F (x) + C is a solution of that differential equation. For example, the solutions of dy 2 = 6x dx are given by 2 3 y = ∫ 6 x dx = 2 x . +C Sometimes we are interested in determining whether a particular solution curve passes through a certain point The problem of finding a function y that satisfies a differential equation (x0 , y0 ) —that is, y(x0 ) = y0 . dy = f (x) dx with the additional condition y(x0 ) = y0 is an example of an initial-value problem. The condition y(x satisfies the differential equation 0) = y0 is known as an initial condition. For example, looking for a function y that dy 2 = 6x dx and the initial condition y(1) = 5 is an example of an initial-value problem. Since the solutions of the differential equation are y = 2x + C , to find a function y that also satisfies the initial condition, we need to find C such that y(1) = 2(1) + C = 5 . From this equation, we see that C = 3 , and we conclude that y = 2 x + 3 is the solution of this initial-value problem as shown in the following graph. 3 3 3 Figure 3.9.2 : Some of the solution curves of the differential equation dy 2 = 6x are displayed. The function y = 2x 3 +3 satisfies the dx differential equation and the initial condition y(1) = 5. Example 3.9.4 : Solving an Initial-Value Problem Solve the initial-value problem 10/21/2020 3.9.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10698 dy = sin x, y(0) = 5. (3.9.9) dx Solution First we need to solve the differential equation. If dy = sin x dx y =∫ , then sin(x)dx = − cos x + C . (3.9.10) Next we need to look for a solution y that satisfies the initial condition. The initial condition y(0)=5 means we need a constant C such that − cos x + C = 5. Therefore, C = 5 + cos(0) = 6. (3.9.11) The solution of the initial-value problem is y = − cos x + 6. Exercise 3.9.4 Solve the initial value problem dy −2 = 3x , y(1) = 2 dx . Hint Find all antiderivatives of f (x) = 3x −2. Answer 3 y =− +5 x Initial-value problems arise in many applications. Next we consider a problem in which a driver applies the brakes in a car. We are interested in how long it takes for the car to stop. Recall that the velocity function v(t) is the derivative of a position function s(t), and the acceleration a(t) is the derivative of the velocity function. In earlier examples in the text, we could calculate the velocity from the position and then compute the acceleration from the velocity. In the next example we work the other way around. Given an acceleration function, we calculate the velocity function. We then use the velocity function to determine the position function. Example 3.9.5 : A car is traveling at the rate of 88 ft/sec (60 mph) when the brakes are applied. The car begins decelerating at a constant rate of 15 ft/sec2. a. How many seconds elapse before the car stops? b. How far does the car travel during that time? Solution a. First we introduce variables for this problem. Let t be the time (in seconds) after the brakes are first applied. Let a(t) be the acceleration of the car (in feet per seconds squared) at time t . Let v(t) be the velocity of the car (in feet per second) at time t . Let s(t) be the car’s position (in feet) beyond the point where the brakes are applied at time t . The car is traveling at a rate of 88f t/sec. Therefore, the initial velocity is v(0) = 88 ft/sec. Since the car is decelerating, the acceleration is 2 a(t) = −15f t/s . The acceleration is the derivative of the velocity, v'(t) = −15. Therefore, we have an initial-value problem to solve: v'(t) = −15, v(0) = 88. Integrating, we find that v(t) = −15t + C . Since v(0) = 88, C = 88. Thus, the velocity function is v(t) = −15t + 88. To find how long it takes for the car to stop, we need to find the time t such that the velocity is zero. Solving 88 t = −15t + 88 = 0, we obtain sec. 15 10/21/2020 3.9.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10698 b. To find how far the car travels during this time, we need to find the position of the car after 88 15 sec. We know the velocity v(t) is the derivative of the position s(t) . Consider the initial position to be s(0) = 0 . Therefore, we need to solve the initial-value problem s'(t) = −15t + 88, s(0) = 0. Integrating, we have 15 s(t) = − 2 t + 88t + C . 2 Since s(0) = 0 , the constant is C =0 . Therefore, the position function is 15 s(t) = − 2 t + 88t. 2 After t = 88 15 sec, the position is s( 88 ) ≈ 258.133 15 ft. Exercise 3.9.5 Suppose the car is traveling at the rate of 44 ft/sec. How long does it take for the car to stop? How far will the car travel? Hint v(t) = −15t + 44. Answer 2.93sec, 64.5f t Key Concepts If F is an antiderivative of f , then every antiderivative of f is of the form F (x) + C for some constant C . Solving the initial-value problem dy = f (x), y(x0 ) = y0 dx requires us first to find the set of antiderivatives of f and then to look for the particular antiderivative that also satisfies the initial condition. Glossary antiderivative a function F such that F '(x) = f (x) for all x in the domain of f is an antiderivative of f indefinite integral the most general antiderivative of f (x) is the indefinite integral of f ; we use the notation ∫ f (x)dx to denote the indefinite integral of f initial value problem a problem that requires finding a function y that satisfies the differential equation dy = f (x) dx together with the initial condition y(x 0) = y0 Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 3.9.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10698 3.9E: Exercises Exercises 3.9E . 1 For the following exercises, show that F (x) are anti derivatives of f (x). 1) F (x) = 5x 3 2 + 2x 2 + 3x + 1, f (x) = 15 x + 4x + 3 Answer 2 F '(x) = 15 x 2) F (x) = x 2 + 4x + 3 + 4x + 1, f (x) = 2x + 4 Answer F '(x) = 2x + 4 3) F (x) = x 2 x x 2 e , f (x) = e (x + 2x) Answer F '(x) = 2x e x 2 +x e x 4) F (x) = cosx, f (x) = −sinx Answer \F'(x)=-sinx 5) F (x) = e x , f (x) = e x Answer F '(x) = e x Exercises 3.9E . 2 For the following exercises, find the antiderivative of the function. 1) f (x) = 1 2 x +x Answer F (x) = − 2) f (x) = e x 1 3x3 + 2 − 3x 1 2 2 x + sinx Answer F (x) = e 3) f (x) = e x x 3 −x − cos(x) + C 2 + 3x − x Answer F (x) = e x 2 + 3x − x 4) f (x) = x − 1 + 4sin(2x) Answer 10/21/2020 3.9E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10939 2 x F (x) = − x − 2cos(2x) + C 2 Exercises 3.9E . 3 For the following exercises, find the antiderivative F (x) of each function f (x). 1) f (x) = 5x 4 5 + 4x Answer 5 F (x) = x + 2 3 6 x +C 2) f (x) = x + 12x 2 Answer 1 F (x) = 3) f (x) = 2 2 3 x + 4x +C 1 √x Answer − F (x) = 2 √x + C 4) f (x) = (√− x) 3 Answer 2 F (x) = 5 − 5 (√x ) + C 5) f (x) = x 1/3 1/3 + (2x ) Answer 3 F (x) = 6) f (x) = 4 4/3 x 4/3 + 3x +C 3 4 √2 1/3 x 2/3 x Answer F (x) = 3 2 2/3 x +C 7) f (x) = 2sin(x) + sin(2x) Answer F (x) = −2cos(x) − 8) f (x) = sec 2 1 2 cos(2x) + C (x) + 1 Answer F (x) = x + tan(x) + C 9) f (x) = sinxcosx Answer F (x) = 1 2 2 si n (x) + C 10) f (x) = si n 2 10/21/2020 (x)cos(x) 3.9E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10939 Answer 1 F (x) = 3 3 si n (x) + C 11) f (x) = 0 Answer F (x) = C 12) f (x) = 1 2 csc (x) + 2 1 2 x Answer F (x) = − 1 2 cot(x) − 1 +C x 13) f (x) = cscxcotx + 3x Answer F (x) = −csc(x) + 3 2 2 x +C 14) f (x) = 4cscxcotx − secxtanx Answer F (x) = −secx − 4cscx + C 15) f (x) = 8secx(secx − 4tanx) Answer F (x) = 8tan(x) − 32sec(x) + C 16) f (x) = 1 e 2 −4x + sinx Answer F (x) = − 1 8 e −4x − cosx + C Exercises 3.9E . 4 For the following exercises, evaluate the integral. 1) ∫ (−1)dx Answer −x + C 2) ∫ sinxdx Answer −cosx + C 3) ∫ (4x + √− x )dx Answer: 2 2x 4) ∫ + 2 3 3/2 x +C 2 3 x +2 x2 10/21/2020 dx 3.9E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10939 Answer 3x − 2 +C x 5) ∫ (secxtanx + 4x)dx Answer 2 sec(x) + 2 x +C − 6) ∫ (4√− x + √x )dx 4 Answer 8 3 3/2 x + 7) ∫ (x −1/3 4 5 5/4 x 2/3 −x +C )dx Answer 3 2 8) ∫ 2/3 x − 3 5 5/3 x +C 3 14 x +2x+1 dx x3 Answer 14x − 9) ∫ (e x 2 +e 1 − x +C 2 2x −x )dx Answer e x −e −x +C Exercises 3.9E . 5 For the following exercises, solve the initial value problem. 1) f '(x) = x −3 , f (1) = 1 Answer f (x) = − 1 3 + 2x2 2 2) f '(x) = √− x +x 2 , f (0) = 2 3) f '(x) = cosx + sec 2 (x), f ( π 4 ) =2+ √2 2 Answer f (x) = sinx + tanx + 1 4) f '(x) = x 3 5) f '(x) = 2 2 x 2 − 8x + 16x + 1, f (0) = 0 2 x − 2 , f (1) = 0 Answer f (x) = − 1 6 3 x − 2 x + 13 6 Exercises 3.9E . 6 10/21/2020 3.9E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10939 For the following exercises, find two possible functions f given the second- or third-order derivatives 1) f ′′ 2) f ′′ 2 (x) = x (x) = e +2 −x Answer Answers may vary; one possible answer is f (x) = e 3) f ′′ 4) f ′′′ −x (x) = 1 + x (x) = cosx Answer Answers may vary; one possible answer is f (x) = −sinx 5) \(f'''(x)=8e^{−2x}−sinx\ Exercise 3.9E . 7 1) A car is being driven at a rate of 40 mph when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2. How long before the car stops?. 2) Calculate how far the car travels in the time it takes to stop. Answer 1. 5.867 sec Exercise 3.9E . 8 1) You are merging onto the freeway, accelerating at a constant rate of merging speed at 60 mph? 12 ft/sec2. How long does it take you to reach 2) How far does the car travel to reach merging speed? Answer 1. 7.333 sec Exercise 3.9E . 9 A car company wants to ensure its newest model can stop in 8 sec when traveling at deceleration, find the value of deceleration that accomplishes this. 75 mph. If we assume constant Answer 2 13.75f t/sec Exercise 3.9E . 10 A car company wants to ensure its newest model can stop in less than 450 ft when traveling at constant deceleration, find the value of deceleration that accomplishes this. 60 mph. If we assume Answer Under Construction 10/21/2020 3.9E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10939 Exercises 3.9E . 11 For the following exercises, find the antiderivative of the function, assuming F (0) = 0. 1) f (x) = x 2 +2 Answer F (x) = 1 3 x 3 + 2x 2) f (x) = 4x − √− x 3) f (x) = sinx + 2x Answer 2 F (x) = x 4) f (x) = e 5) f (x) = − cosx + 1 x 1 2 (x+1) Answer F (x) = − 1 +1 (x+1) 6) \(f(x)=e^{−2x}+3x^2\ Exercises 3.9E . 12 For the following exercises, determine whether the statement is true or false. Either prove it is true or find a counterexample if it is false. 1) If f (x) is the antiderivative of v(x), then 2f (x) is the antiderivative of 2v(x). Answer True. 2) If f (x) is the antiderivative of v(x), then f (2x) is the antiderivative of v(2x). 3) If f (x) is the antiderivative of v(x), then f (x) + 1 is the antiderivative of v(x) + 1. Answer False. 4) If f (x) is the antiderivative of v(x), then (f (x)) is the antiderivative of (v(x)) 2 2 . Answer False. Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 3.9E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10939 3E: Chapter Exercises Chapter Review Exercises Exercise 3E. 1 True or False? Justify your answer with a proof or a counterexample. Assume that continuous and differentiable unless stated otherwise. 1) If f(−1) = −6 and f(1) = 2 , then there exists at least one point x ∈ [−1, 1] f(x) is such that f'(x) = 4. Answer True, by Mean Value Theorem 2) If f'(c) = 0, there is a maximum or minimum at x = c. 3) There is a function such that acceptable for this answer.) f(x) < 0, f'(x) > 0, and f ′′ (x) < 0. (A graphical “proof” is Answer True 4) There is a function such that there is both an inflection point and a critical point for some value x = a. Exercise 3E. 2 Given the graph of f' , determine where f is increasing or decreasing. Answer Increasing: (−2, 0) ∪ (4, ∞) , decreasing: (−∞, −2) ∪ (0, 4) Exercise 3E. 3 The graph of f is given below. Draw f' . 10/21/2020 3E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10940 Answer Under Construction Exercise 3E. 4 Find the linear approximation L(x) to y = x 2 + tan(πx) near x = 1 4 . Answer L(x) = 17 16 + 1 2 (1 + 4π)(x − Exercise 3E. 5 Find the differential of y = x 2 1 4 ) − 5x − 6 and evaluate for x = 2 with dx = 0.1. Answer Under Construction Exercise 3E. 6 Find the critical points and the local and absolute extrema of the following functions on the given interval. 1) f(x) = x + sin 2 (x) over [0, π] Answer Critical point: x = 2) f(x) = 3x 4 3π , 4 3 − 4x absolute minimum: x = 0, absolute maximum: x = π 2 − 12x +6 over [−3, 3] Exercise 3E. 7 Determine over which intervals the following functions are increasing, decreasing, concave up, and concave down. 1) x(t) = 3t 4 3 − 8t 2 − 18t Answer Increasing: (−1, 0) ∪ (3, ∞) Decreasing: (−∞, −1) ∪ (0, 3) 10/21/2020 3E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10940 Concave up: (−∞, Concave down: ( 1 3 1 3 − − (2 − √13)) ∪ ( − − (2 − √13), 1 3 1 3 − − (2 + √13), ∞) − − (2 + √13)) 2) y = x + sin(πx) 3) g(x) = x − √− x Answer Increasing: ( 1 4 ∞), Decreasing: (0, 1 ) 4 Concave up: (0, ∞) Concave down: nowhere 4) \(f(θ)=sin(3θ)\ Exercise 3E. 8 Evaluate the following limits. 2 3x √x +1 1) lim x→∞ √x4−1 Answer 3 2) lim x→∞ 3) lim cos( 1 x ) x−1 x→1 sin(πx) Answer − 1 π 4) lim 1/x x→∞ (3x ) Exercise 3E. 9 Use Newton’s method to find the first two iterations, given the starting point. 1) y = x 3 + 1, x0 = 0.5 Answer x1 = −1, x2 = −1 2) 1 x+1 10/21/2020 = 1 2 , x0 = 0 3E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10940 Exercise 3E. 10 Find the antiderivatives F (x) of the following functions. 1) g(x) = √− x − 1 2 x Answer 3/2 F (x) = 2x + 3 1 x +C 2) \(f(x)=2x+6cosx,F(π)=π^2+2\ Exercise 3E. 11 Graph the following functions by hand. Make sure to label the inflection points, critical points, zeros, and asymptotes. 1) y = 1 2 x(x+1) Answer: Inflection points: none Critical points: x = − 1 3 Zeros: none; vertical asymptotes: x = −1, x = 0 Horizontal asymptote: y = 0 − −−− − 2) y = x − √4 − x 2 Exercise 3E. 12 A car is being compacted into a rectangular solid. The volume is decreasing at a rate of 2m /sec . The length and width of the compactor are square, but the height is not the same length as the length and width. If the length and width walls move toward each other at a rate of 0.25 m/sec, find the rate at which the height is changing when the length and width are 2 m and the height is 1.5 m. 3 Answer The height is decreasing at a rate of 0.125 m/sec 10/21/2020 3E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10940 Exercise 3E. 13 A rocket is launched into space; its kinetic energy is given by K(t) = ( )m(t)v(t) , where K is the kinetic energy in joules, m is the mass of the rocket in kilograms, and v is the velocity of the rocket in meters/second. Assume the velocity is increasing at a rate of 15m/sec and the mass is decreasing at a rate of 10 kg/sec because the fuel is being burned. At what rate is the rocket’s kinetic energy changing when the mass is 2000 kg and the velocity is 5000 m/sec? Give your answer in mega-Joules (MJ), which is equivalent to 10 J. 1 2 2 2 6 Answer Under Construction Exercise 3E. 14 The famous Regiomontanus’ problem for angle maximization was proposed during the 15 th century. A painting hangs on a wall with the bottom of the painting a distance a feet above eye level, and the top b feet above eye level. What distance x (in feet) from the wall should the viewer stand to maximize the angle subtended by the painting, θ ? Answer − − x = √ab feet Exercise 3E. 15 An airline sells tickets from Tokyo to Detroit for $1200. There are 500 seats available and a typical flight books 350 seats. For every $10 decrease in price, the airline observes an additional five seats sold. What should the fare be to maximize profit? How many passengers would be onboard? Answer Under Construction Exercise 3E. 16 For the following functions: a. Find the interval(s) on which f (x) is increasing 10/21/2020 3E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10940 b. Find the interval(s) on which f (x) is decreasing c. Find the local (relative) maximum and local (relative) minimum (if any). 1) f (x) = x 2 e −x 2) f (x) = 3(x 2 3) f (x) = x + 2/3 − 4) 1 x Answer Under Construction Exercise 3E. 17 For the following functions: a. Find the interval(s) on which g(x) is concave up b. Find the interval(s) on which g(x) is concave down c. Find the inflection points (if any) 1) g(x) = 10x 3 2) g(x) = e 5 + 3x 2 −x /2 3) g(x) = 5x 3 5 + 2x 4) g(x) = ln(1 + x 2 5) g(x) = (3 − x 2 ) 2 ) Answer Under Construction Exercise 3E. 18 Find the horizontal asymptote −−−−−−−− − 2 y = f (x) = x − √x + 2x − 6 Answer Under Construction Exercise 3E. 19 1 f (t) = 2 1−t a. Find vertical and horizontal asymptotes (if any) b. Find the intervals when f (t) is concave up and concave down Answer Under Construction Exercise 3E. 20 2 f (t) = e 10/21/2020 −t 3E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10940 a. Find the local maxima and minima for the function f (t) b. Find the intervals when f (t) is increasing and decreasing c. Find the intervals when f (t) is concave up and concave down d. Find the inflection points for f (t) Answer Under Construction Exercise 3E. 21 Find all vertical and horizontal asymptotes for the function and justify your work using limits. 2 f (x) = x −5x+6 2 x −4x+3 Answer Under Construction Exercise 3E. 22 Using the following information: y = x 2 x −1 , y ′ 2 2 =− x +1 2 2 ( x −1) , and y ′′ 2x(x +3) = 2 3 ( x −1) a) Identify any asymptotes for the function y using limits. b) Find where the function y is increasing/decreasing and any relative extrema. c) Find where the function y is concave up/down and any points of inflection. d) Graph the function y Answer Under Construction Exercise 3E. 23 Let f (x) = x − x 5/4 1/4 Determine the intervals on which f is increasing and decreasing and find any local extrema. Answer Under Construction Exercise 3E. 24 Let g(x) = ln(x + 1) 2 Determine the intervals of concavity of g(x) and locate any inflection points. Answer Under Construction Exercise 3E. 25 Determine the intervals of concavity of g(x) and locate any inflection points: 10/21/2020 3E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10940 1) g(x) = (ln x) 2 2) g(x) = x − 2 tan −1 (x) Answer Under Construction Exercise 3E. 26 Find interval(s) on which the function is concave up and the interval(s) on which the function is concave down, and inflection point(s): −1 y = f (x) = x − 2tan (x) Answer Under Construction Exercise 3E. 27 Let f (x) = x 2 2 e −16x Determine the intervals on which f is increasing and decreasing and find any local extrema. Answer Under Construction Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 3E.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10940 CHAPTER OVERVIEW 1 10/21/2020 4: INTEGRAL CALCULUS Iceboats are a common sight on the lakes of Wisconsin and Minnesota on winter weekends. Iceboats are similar to sailboats, but they are fitted with runners, or “skates,” and are designed to run over the ice, rather than on water. Iceboats can move very quickly, and many ice boating enthusiasts are drawn to the sport because of the speed. Top iceboat racers can attain speeds up to five times the wind speed. If we know how fast an iceboat is moving, we can use integration to determine how far it travels. We revisit this question later in the chapter (see [link]). Figure 4.0.15.0.1: Iceboating is a popular winter sport in parts of the northern United States and Europe. (credit: modification of work by Carter Brown, Flickr) Determining distance from velocity is just one of many applications of integration. In fact, integrals are used in a wide variety of mechanical and physical applications. In this chapter, we first introduce the theory behind integration and use integrals to calculate areas. From there, we develop the Fundamental Theorem of Calculus, which relates differentiation and integration. We then study some basic integration techniques and briefly examine some applications. CONTRIBUTORS Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY 3/0 license. Download for free at http://cnx.org. 2 10/21/2020 Topic hierarchy 4.0: ANTIDERVATIVES AND INDEFINITE INTEGRATION (REVISITED) 4.0E: EXERCISES 4.1: INTEGRATION BY SUBSTITUTION 4.1E: EXERCISES 4.2: DEFINITE INTEGRAL- AN INTRODUCTION 4.2E: EXERCISES 4.3: APPROXIMATING AREAS 4.3E: EXERCISES 4.4: THE DEFINITE INTEGRAL 4.4E: EXERCISES 4.5: THE FUNDAMENTAL THEOREM OF CALCULUS 4.5E: EXERCISES 4.6: INTEGRATION FORMULAS AND THE NET CHANGE THEOREM 4.6E: EXERCISES 4.7: DEFINITE INTEGRALS BY SUBSTITUTION. 4.7E: EXERCISES 4.8: AREA BETWEEN TWO CURVES 4.8E: EXERCISES 4.9: APPLICATIONS OF DEFINITE INTEGRALS 4.9E: EXERCISES 4E: EXERCISES 3 10/21/2020 4.0: Antidervatives and Indefinite Integration (Revisited) We have spent considerable time considering the derivatives of a function and their applications. In the following chapters, we are going to starting thinking in "the other direction." That is, given a function f (x), we are going to consider functions F (x) such that F (x) = f (x). There are numerous reasons this will prove to be useful: these functions will help us compute areas, volumes, mass, force, pressure, work, and much more. ′ Given a function y = f (x), a differential equation is one that incorporates y , x, and the derivatives of y . For instance, a simple differential equation is: y ′ Solving a differential equation amounts to finding a function that equation; can you find a function y such that y = 2x ? = 2x. y (4.0.1) that satisfies the given equation. Take a moment and consider ′ Can you find another? And yet another? Hopefully one was able to come up with at least one solution: y = x . "Finding another" may have seemed impossible until one realizes that a function like y = x + 1 also has a derivative of 2x. Once that discovery is made, finding "yet another" is not difficult; the function y = x + 123, 456, 789 also has a derivative of 2x. The differential equation y = 2x has many solutions. This leads us to some definitions. 2 2 2 ′ Definition 4.0.1 : Antiderivatives and Indefinite Integrals Let a function f (x) be given. An antiderivative of f (x) is a function F (x) such that F ′ . (x) = f (x) The set of all antiderivatives of f (x) is the indefinite integral of f , denoted by ∫ f (x) dx. (4.0.2) Make a note about our definition: we refer to an antiderivative of f , as opposed to the antiderivative of f , since there is always an infinite number of them. We often use upper-case letters to denote antiderivatives. Knowing one antiderivative of f allows us to find infinitely more, simply by adding a constant. Not only does this give us more antiderivatives, it gives us all of them. Theorem 4.0.1 : Antiderivative Forms Let F (x) and G(x) be antiderivatives of f (x). Then there exists a constant C such that G(x) = F (x) + C . Given a function f and one of its antiderivatives F , we know all antiderivatives of constant C . Using Definition 4.0.1, we can say that ∫ f (x) dx = F (x) + C . (4.0.3) f have the form F (x) + C for some (4.0.4) Let's analyze this indefinite integral notation. 10/21/2020 4.0.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/13743 Figure 4.0.1 : Understanding the indefinite integral notation. Figure 4.0.1 shows the typical notation of the indefinite integral. The integration symbol, representing "take the sum." We will later see how sums and antiderivatives are related. ∫ , is, in reality, an "elongated S," The function we want to find an antiderivative of is called the integrand. It contains the differential of the variable we are integrating with respect to. The ∫ symbol and the differential dx are not "bookends" with a function sandwiched in between; rather, the symbol ∫ means "find all antiderivatives of what follows," and the function f (x) and dx are multiplied together; the dx does not "just sit there." Let's practice using this notation. Example 4.0.1 : Evaluating indefinite integrals Evaluate ∫ sin x dx. Solution We are asked to find all functions F (x) such that F (x) = − cos x , because (− cos x) = sin x . ′ F (x) = sin x . Some thought will lead us to one solution: d dx The indefinite integral of sin x is thus − cos x, plus a constant of integration. So: ∫ sin x dx = − cos x + C . (4.0.5) A commonly asked question is "What happened to the dx?" The unenlightened response is "Don't worry about it. It just goes away." A full understanding includes the following. This process of antidifferentiation is really solving a differential question. The integral ∫ presents us with a differential, y = − cos x + C . dy = sin x dx sin x dx (4.0.6) . It is asking: "What is y ?" We found lots of solutions, all of the form Letting dy = sin x dx , rewrite ∫ sin x dx as ∫ dy. (4.0.7) This is asking: "What functions have a differential of the form dy ?" The answer is "Functions of the form y + C , where C is a constant." What is y ? We have lots of choices, all differing by a constant; the simplest choice is y = − cos x . Understanding all of this is more important later as we try to find antiderivatives of more complicated functions. In this section, we will simply explore the rules of indefinite integration, and one can succeed for now with answering "What happened to the dx?" with "It went away." Let's practice once more before stating integration rules. Example 4.0.2 : Evaluating indefinite integrals 10/21/2020 4.0.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/13743 Evaluate ∫ (3x 2 + 4x + 5) dx . Solution We seek a function F (x) whose derivative is 3x term, so we can likely do that here. 2 What functions have a derivative of constant. 2 3x . When taking derivatives, we can consider functions term--by-- + 4x + 5 ? Some thought will lead us to a cubic, specifically 3 x + C1 , where C1 is a What functions have a derivative of 4x? Here the x term is raised to the first power, so we likely seek a quadratic. Some thought should lead us to 2x + C , where C is a constant. 2 2 2 Finally, what functions have a derivative of 5? Functions of the form 5x + C , where C is a constant. 3 3 Our answer appears to be 2 ∫ (3 x 3 + 4x + 5) dx = x 2 + C1 + 2 x + C2 + 5x + C3 . (4.0.8) We do not need three separate constants of integration; combine them as one constant, giving the final answer of 2 ∫ (3 x 3 + 4x + 5) dx = x It is easy to verify our answer; take the derivative of x 3 3 + 2x 2 + 2x + 5x + C + 5x + C . (4.0.9) and see we indeed get 3x 2 + 4x + 5 . This final step of "verifying our answer" is important both practically and theoretically. In general, taking derivatives is easier than finding antiderivatives so checking our work is easy and vital as we learn. We also see that taking the derivative of our answer returns the function in the integrand. Thus we can say that: d (∫ f (x) dx) = f (x). (4.0.10) dx Differentiation "undoes" the work done by antidifferentiation. Theorem 27 gave a list of the derivatives of common functions we had learned at that point. We restate part of that list here to stress the relationship between derivatives and antiderivatives. This list will also be useful as a glossary of common antiderivatives as we learn. Theorem 4.0.2 : Derivatives and Antiderivatives Common Differentiation Rules 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. d dx d dx d dx d dx d dx d dx d dx d dx d dx d dx d dx d dx d dx d dx ′ (cf (x)) = c ⋅ f (x) ′ ′ (f (x) ± g(x)) = f (x) ± g (x) (C) = 0 (x) = 1 n n−1 (x ) = n ⋅ x ( sin x) = cosx ( cosx) = − sin x ( tan x) = sec 2 x ( csc x) = − csc x cot x ( sec x) = sec x tan x ( cot x) = − csc x 2 x x (e ) = e x x (a ) = ln a ⋅ a ( ln x) = 1 Common Indefinite Integration Rules 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. ∫ c ⋅ f (x) dx = c ⋅ ∫ f (x) dx ∫ (f (x) ± g(x)) dx = ∫ f (x) dx ± ∫ g(x) dx ∫ 0 dx = C ∫ 1 dx = ∫ dx = x + C n ∫x dx = 1 n+1 n+1 x +C ∫ cosx dx = sin x + C ∫ sin x dx = − cosx + C ∫ sec 2 x dx = tan x + C ∫ csc x cot x dx = − csc x + C ∫ sec x tan x dx = sec x + C ∫ csc x ∫e x ∫a ∫ 1 x 2 x dx = − cot x + C x dx = e dx = +C 1 ln a x ⋅a +C dx = ln |x| + C x We highlight a few important points from Theorem 4.0.2: 10/21/2020 4.0.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/13743 Rule #1 states ∫ c ⋅ f (x) dx = c ⋅ ∫ f (x) dx . This is the Constant Multiple Rule: we can temporarily ignore constants when finding antiderivatives, just as we did when computing derivatives (i.e., (3 x ) is just as easy to compute as (x )). An example: d 2 dx d 2 dx ∫ 5 cos x dx = 5 ⋅ ∫ cos x dx = 5 ⋅ (sin x + C ) = 5 sin x + C . (4.0.11) In the last step we can consider the constant as also being multiplied by 5, but "5 times a constant" is still a constant, so we just write "C ,". Rule #2 is the Sum/Difference Rule: we can split integrals apart when the integrand contains terms that are added/subtracted, as we did in Example 4.0.2. So: 2 ∫ (3 x + 4x + 5) dx 2 =∫ 3x =3∫ dx + ∫ 2 x 1 =3⋅ dx + 4 ∫ 1 3 x +4 ⋅ 3 3 =x 4x dx + ∫ x dx + ∫ 2 x 5 dx (4.0.12) 5 dx (4.0.13) + 5x + C (4.0.14) 2 2 + 2x + 5x + C (4.0.15) In practice we generally do not write out all these steps, but we demonstrate them here for completeness. Rule #5 is the Power Rule of indefinite integration. There are two important things to keep in mind: 1. Notice the restriction that n ≠ −1 . This is important: ∫ dx ≠ " x + C "; rather, see Rule #14. 2. We are presenting antidifferentiation as the "inverse operation" of differentiation. Here is a useful quote to remember: "Inverse operations do the opposite things in the opposite order." When taking a derivative using the Power Rule, we first multiply by the power, then second subtract 1 from the power. To find the antiderivative, do the opposite things in the opposite order: first add one to the power, then second divide by the power. 1 1 x 0 0 Note that Rule #14 incorporates the absolute value of x. The exercises will work the reader through why this is the case; for now, know the absolute value is important and cannot be ignored. Initial Value Problems In Section 2.3 we saw that the derivative of a position function gave a velocity function, and the derivative of a velocity function describes acceleration. We can now go "the other way:" the antiderivative of an acceleration function gives a velocity function, etc. While there is just one derivative of a given function, there are infinite antiderivatives. Therefore we cannot ask "What is the velocity of an object whose acceleration is −32ft/s ?", since there is more than one answer. 2 We can find the answer if we provide more information with the question, as done in the following example. Often the additional information comes in the form of an initial value, a value of the function that one knows beforehand. Example 4.0.3 : Solving initial value problems The acceleration due to gravity of a falling object is Find the equation of the object's velocity. −32 ft/s . At time 2 t =3 , a falling object had a velocity of −10 ft/s. Solution We want to know a velocity function, v(t) . We know two things: 1. The acceleration, i.e., v (t) = −32 , and 2. the velocity at a specific time, i.e., v(3) = −10 . ′ Using the first piece of information, we know that indefinite integral of −32: 10/21/2020 v(t) is an antiderivative of 4.0.4 CC-BY-NC-SA ′ v (t) = −32 . So we begin by finding the https://math.libretexts.org/@go/page/13743 ∫ (−32) dt = −32t + C = v(t). (4.0.16) Now we use the fact that v(3) = −10 to find C : v(t) = −32t + C (4.0.17) v(3) = −10 (4.0.18) −32(3) + C = −10 (4.0.19) C = 86 (4.0.20) Thus v(t) = −32t + 86 . We can use this equation to understand the motion of the object: when velocity of $v(0) = 86$ ft/s. Since the velocity is positive, the object was moving upward. t =0 , the object had a When did the object begin moving down? Immediately after v(t) = 0 : 43 −32t + 86 = 0 ⇒ t = ≈ 2.69s. (4.0.21) 16 Recognize that we are able to determine quite a bit about the path of the object knowing just its acceleration and its velocity at a single point in time. Example 4.0.4 : Solving initial value problems Find f (t), given that f ′′ (t) = cos t ,f ′ (0) = 3 and f (0) = 5 . Solution We start by finding f ′ (t) , which is an antiderivative of f ∫ So f ′ (t) = sin t + C f ′′ (t) dt = ∫ ′′ (t) : ′ cos t dt = sin t + C = f (t). for the correct value of C . We are given that f ′ f (0) = 3 Using the initial value, we have found f ′ ⇒ ′ (0) = 3 sin 0 + C = 3 (4.0.22) , so: ⇒ C = 3. (4.0.23) (t) = sin t + 3. We now find f (t) by integrating again. f (t) = ∫ ′ f (t) dt = ∫ (sin t + 3) dt = − cos t + 3t + C . (4.0.24) We are given that f (0) = 5 , so − cos 0 + 3(0) + C =5 (4.0.25) −1 + C = 5 (4.0.26) C =6 (4.0.27) Thus f (t) = − cos t + 3t + 6 . This section introduced antiderivatives and the indefinite integral. We found they are needed when finding a function given information about its derivative(s). For instance, we found a position function given a velocity function. In the next section, we will see how position and velocity are unexpectedly related by the areas of certain regions on a graph of the velocity function. Then, in Section 5.4, we will see how areas and antiderivatives are closely tied together. Contributors Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/ Integrated by Justin Marshall. 10/21/2020 4.0.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/13743 4.0E: Exercises Exercise 4.0.1 1. Define the term “antiderivative” in your own words 2. Is it more accurate to refer to “the” antiderivative of f (x) or “an” antiderivative of f (x)? 3. Use your own words to define the indefinite integral of f (x). 4. Fill in the blanks: “Inverse operations do the ____ things in the _____ order.” 5. What is an “initial value problem”? 6. The derivative of a position function is a _____ function. 7. The antiderivative of an acceleration function is a ______ function. Answer Under Construction Exercise 4.0.2 Evaluate the indefinite integrals: 1. ∫ 3x 3 2. ∫ x 8 dx dx 3. ∫ (10x 2 4. ∫ − 2) dx dt 5. ∫ 1 ds 6. ∫ 7. ∫ 8. ∫ 1 dt 2 3t 1 dt 2 t 1 dx √x 9. ∫ sec 2 θ dθ 10. ∫ sin θ dθ 11. ∫ (sec x tan x + csc x cot x) dx 12. ∫ 5e θ 13. ∫ 3 dt t 14. ∫ dθ t 5 2 dt 15. ∫ (2t + 3) dt 16. ∫ (t 3 2 2 + 3)(t 17. ∫ x x 18. ∫ e dx 2 π 3 − 2t) dt dx 19. ∫ a dx Answer Under Construction 10/21/2020 4.0E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/13746 Exercise 4.0.3 This problem investigates why Theorem 35 states that ∫ dx = ln |x| + C . (a) What is the domain of y = ln x ? (b) Find (ln x). (c) What is the domain of y = ln(−x) ? (d) Find ((ln(−x)). (e) You should find that 1/x has two types of antiderivatives, depending on whether x > 0 or x < 0 . In one expression, give a formula for ∫ dx that takes these different domains into account, and explain your answer. 1 x d dx d dx 1 x Answer Under Construction Exercise 4.0.4 Find f (x) described by the given initial value problem. 1. f ′ 2. f ′ 3. f ′ 4. f ′ 5. f ′ 6. f ′′ 7. f ′′ 8. f ′′ 9. f ′′ (x) = sin x and f (0) = 2 (x) = 5 e x 3 (x) = 4 x (x) = sec x (x) = 7 and f (0) = 10 2 − 3x 2 and f (−1) = 9 x and f (π/4) = 5 and f (2) = 1 ′ (x) = 5 and f (0) = 7, f (0) = 3 ′ (x) = 7x and f (1) = −1, f (1) = 10 (x) = 5 e x ′ and f (0) = 3, f (0) = 5 ′ (θ) = sin θ and f (π) = 2, f (π) = 4 10. f ′′ 11. f ′′ 2 (x) = 24 x x +2 ′ − cos x and f (0) = 5, f (0) = 0 ′ (x) = 0 and f (1) = 3, f (1) = 1 Answer Under Construction Exercise 4.0.5 Use information gained from the first and second derivative to sketch f (x) = 1 x e +1 . Answer Under Construction Exercise 4.0.6 Given y = x 2 e x cos x , find dy . Answer Under Construction 10/21/2020 4.0E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/13746 4.1: Integration by Substitution We motivate this section with an example. Let f (x) = (x 2 ′ 2 9 f (x) = 10(x + 3x − 5 ) Now consider this: What is ∫ (20x + 30)(x + 3x − 5 ) 2 2 dx ′ (x) using the Chain Rule. It is: 9 + 3x − 5 ) . (4.1.1) ? We have the answer in front of us; 9 ∫ (20x + 30)(x . We can compute f ⋅ (2x + 3) = (20x + 30)(x 9 2 10 + 3x − 5 ) 2 + 3x − 5 ) 10 dx = (x + 3x − 5 ) + C. (4.1.2) How would we have evaluated this indefinite integral without starting with f (x) as we did? This section explores integration by substitution. It allows us to "undo the Chain Rule." Substitution allows us to evaluate the above integral without knowing the original function first. The underlying principle is to rewrite a "complicated" integral of the form ∫ f (x) dx as a not--so--complicated integral ∫ h(u) du . We'll formally establish later how this is done. First, consider again our introductory indefinite integral, ∫ (20x + 30)(x + 3x − 5 ) dx . Arguably the most "complicated" part of the integrand is (x + 3x − 5 ) . We wish to make this simpler; we do so through a substitution. Let u = x + 3x − 5 . Thus 2 9 2 9 2 2 (x 9 + 3x − 5 ) 9 =u . (4.1.3) We have established u as a function of x, so now consider the differential of u: du = (2x + 3)dx. (4.1.4) Keep in mind that (2x + 3) and dx are multiplied; the dx is not "just sitting there." Return to the original integral and do some substitutions through algebra: 2 ∫ (20x + 30)(x 9 + 3x − 5 ) dx 2 =∫ 10(2x + 3)(x 9 + 3x − 5 ) 2 =∫ dx (4.1.5) 9 10( x + 3x − 5 ) (2x + 3) dx   u 9 =∫ 10 u 10 =u 2 = (x (4.1.6) du du +C (4.1.7) 2 (replace u with x 10 + 3x − 5 ) + 3x − 5) (4.1.8) +C (4.1.9) One might well look at this and think "I (sort of) followed how that worked, but I could never come up with that on my own," but the process is learnable. This section contains numerous examples through which the reader will gain understanding and mathematical maturity enabling them to regard substitution as a natural tool when evaluating integrals. We stated before that integration by substitution "undoes" the Chain Rule. Specifically, let functions and consider the derivative of their composition: d ′ ′ (F (g(x))) = F (g(x))g (x). F (x) and g(x) be differentiable (4.1.10) dx Thus ∫ ′ ′ F (g(x))g (x) dx = F (g(x)) + C . Integration by substitution works by recognizing the "inside" function u = g(x) , we can rewrite the derivative as d ′ g(x) ′ (F (u)) = F (u)u . (4.1.11) and replacing it with a variable. By setting (4.1.12) dx 10/21/2020 4.1.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 Since du = g ′ , we can rewrite the above integral as (x)dx ∫ ′ ′ F (g(x))g (x) dx = ∫ ′ F (u)du = F (u) + C = F (g(x)) + C . (4.1.13) This concept is important so we restate it in the context of a theorem. Theorem 4.1.1 : Integration by Substitution Let F and g be differentiable functions, where the range of g is an interval I contained in the domain of F . Then ∫ If u = g(x) , then du = g ′ ′ (4.1.14) and (x)dx ′ ∫ ′ F (g(x))g (x) dx = F (g(x)) + C . ′ ′ F (g(x))g (x) dx = ∫ F (u) du = F (u) + C = F (g(x)) + C . (4.1.15) The point of substitution is to make the integration step easy. Indeed, the step ∫ F (u) du = F (u) + C looks easy, as the antiderivative of the derivative of F is just F , plus a constant. The "work" involved is making the proper substitution. There is not a step-by-step process that one can memorize; rather, experience will be one's guide. To gain experience, we now embark on many examples. ′ Example 4.1.1 : Integrating by substitution Evaluate ∫ x sin(x 2 + 5) dx . Solution Knowing that substitution is related to the Chain Rule, we choose to let u be the "inside" function of sin(x not always a good choice, but it is often the best place to start.) 2 + 5) . (This is Let u = x + 5 , hence du = 2x dx. The integrand has an x dx term, but not a 2x dx term. (Recall that multiplication is commutative, so the x does not physically have to be next to dx for there to be an x dx term.) We can divide both sides of the du expression by 2: 2 1 du = 2x dx ⇒ du = x dx. (4.1.16) 2 We can now substitute. ∫ 2 x sin(x + 5) dx = ∫ 2 sin( x + 5 )x dx   u 1 2 (4.1.17) du 1 =∫ sin u du (4.1.18) 2 1 =− cos u + C 2 (now replace u with x + 5) (4.1.19) 2 1 =− 2 cos(x + 5) + C . (4.1.20) 2 Thus side. 2 ∫ x sin(x + 5) dx = − 1 2 2 cos(x + 5) + C . We can check our work by evaluating the derivative of the right hand Exercise 4.1.1 Evaluate ∫ x cos(x 2 + 5) dx . Answer \frac12\sin(x^2+5)+C. 10/21/2020 4.1.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 Example 4.1.2 : Integrating by substitution Evaluate ∫ cos(5x) dx. Solution Again let u replace the "inside" function. Letting u = 5x, we have du = 5dx . Since our integrand does not have a term, we can divide the previous equation by 5 to obtain du = dx . We can now substitute. 5dx 1 5 ∫ cos(5x) dx = ∫ cos( 5x ) dx   u 1 5 (4.1.21) du 1 =∫ cos u du (4.1.22) 5 1 = sin u + C (4.1.23) sin(5x) + C . (4.1.24) 5 1 = 5 We can again check our work through differentiation. Exercise 4.1.2 Evaluate ∫ sin(5x) dx . Answer -\frac{1}{5}\cos(5x)+C. . The previous example exhibited a common, and simple, type of substitution. The "inside" function was a linear function (in this case, y = 5x). When the inside function is linear, the resulting integration is very predictable, outlined here. Key Idea 10: Substitution With A Linear Function Consider ∫ F ′ (ax + b) dx , where a ≠ 0 and b are constants. Letting u = ax + b gives du = a ⋅ dx , leading to the result ∫ 1 ′ F (ax + b) dx = F (ax + b) + C . (4.1.25) a Thus ∫ sin(7x − 4) dx = − cos(7x − 4) + C . Our next example can use Key Idea 10, but we will only employ it after going through all of the steps. 1 7 Example 4.1.3 : Integrating by substituting a linear function Evaluate ∫ 7 −3x+1 dx . Solution View this a composition of functions f (g(x)), where f (x) = 7/x and g(x) = −3x + 1 . Employing our understanding of substitution, we let u = −3x + 1 , the inside function. Thus du = −3dx . The integrand lacks a −3; hence divide the previous equation by −3 to obtain −du/3 = dx. We can now evaluate the integral through substitution. 7 ∫ 7 du dx = ∫ (4.1.26) −3x + 1 u −3 −7 = du ∫ 3 (4.1.27) u −7 = ln |u| + C (4.1.28) ln | − 3x + 1| + C . (4.1.29) 3 7 =− 3 10/21/2020 4.1.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 Using Key Idea 10 is faster, recognizing that u is linear and until they are comfortable with this particular shortcut. a = −3 . One may want to continue writing out all the steps Exercise 4.1.3 Evaluate ∫ 7 3x+1 dx . Answer \frac{7}{3|\ln|3x+1| + C. Not all integrals that benefit from substitution have a clear "inside" function. Several of the following examples will demonstrate ways in which this occurs. Example 4.1.4 : Integrating by substitution Evaluate ∫ sin x cos x dx . Solution There is not a composition of function here to exploit; rather, just a product of functions. Do not be afraid to experiment; when given an integral to evaluate, it is often beneficial to think "If I let u be this, then du must be that ..." and see if this helps simplify the integral at all. In this example, let's set u = sin x . Then du = cos x dx , which we have as part of the integrand! The substitution becomes very straightforward: ∫ sin x cos x dx = ∫ 1 = u du 2 u (4.1.30) +C (4.1.31) 2 1 = 2 sin x + C. (4.1.32) 2 One would do well to ask "What would happen if we let u = cos x?" The result is just as easy to find, yet looks very different. The challenge to the reader is to evaluate the integral letting u = cos x and discover why the answer is the same, yet looks different. Exercise 4.1.4 Evaluate ∫ sin 3 x cos x dx .. Answer \frac{1}{3}\sin^4 x + C. Our examples so far have required "basic substitution." The next example demonstrates how substitutions can be made that often strike the new learner as being "nonstandard." Example 4.1.5 : Integrating by substitution − − − − − Evaluate ∫ x √x + 3 dx . Solution Recognizing the composition of functions, set substitution. But at this stage, we have: ∫ u = x +3 . Then − − − − − x √ x + 3 dx = ∫ du = dx , giving what seems initially to be a simple − x √u du. (4.1.33) We cannot evaluate an integral that has both an x and an u in it. We need to convert the x to an expression involving just u. 10/21/2020 4.1.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 Since we set u = x + 3 , we can also state that u −3 = x . Thus we can replace x in the integrand with u −3 . It will also 1 be helpful to rewrite √− u as u . 2 ∫ − − − − − x √ x + 3 dx = ∫ (u − 3)u 1 du 2 3 =∫ (u 2 = 2 1 − 3u 5 u 2 (4.1.34) 2 ) du (4.1.35) 3 − 2u 2 +C (4.1.36) 5 2 = 5 (x + 3 ) 2 3 − 2(x + 3 ) 2 + C. (4.1.37) 5 Checking your work is always a good idea. In this particular case, some algebra will be needed to make one's answer match the integrand in the original problem. Example 4.1.6 : Integrating by substitution Evaluate ∫ 1 x ln x dx . Solution This is another example where there does not seem to be an obvious composition of functions. The line of thinking used in Example 4.1.5 is useful here: choose something for u and consider what this implies du must be. If u can be chosen such that du also appears in the integrand, then we have chosen well. Choosing u = 1/x makes du = −1/x which is part of the integrand. Thus: 2 dx ; that does not seem helpful. However, setting 1 ∫ dx = ∫ x ln x 1 1 dx ln x x   1/u u = ln x makes , du = 1/x dx (4.1.38) du 1 =∫ du (4.1.39) u = ln |u| + C (4.1.40) = ln | ln x| + C . (4.1.41) The final answer is interesting; the natural log of the natural log. Take the derivative to confirm this answer is indeed correct. Integrals Involving Trigonometric Functions Section 6.3 delves deeper into integrals of a variety of trigonometric functions; here we use substitution to establish a foundation that we will build upon. The next three examples will help fill in some missing pieces of our antiderivative knowledge. We know the antiderivatives of the sine and cosine functions; what about the other standard functions tangent, cotangent, secant and cosecant? We discover these next. Example 4.1.7 : Integration by substitution: antiderivatives of tan x Evaluate ∫ tan x dx. Solution The previous paragraph established that we did not know the antiderivatives of tangent, hence we must assume that we have learned something in this section that can help us evaluate this indefinite integral. Rewrite tan x as sin x/ cos x. While the presence of a composition of functions may not be immediately obvious, recognize that cos x is "inside" the 1/x function. Therefore, we see if setting u = cos x returns usable results. We have that du = − sin x dx , hence −du = sin x dx . We can integrate: 10/21/2020 4.1.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 sin x ∫ tan x dx = ∫ dx (4.1.42) cos x =∫ 1 sin x dx cos x   −du (4.1.43) u −1 =∫ du (4.1.44) u = − ln |u| + C (4.1.45) = − ln | cos x| + C . (4.1.46) Some texts prefer to bring the −1 inside the logarithm as a power of cos x, as in: − ln | cos x| + C −1 = ln |(cos x ) | +C (4.1.47) 1 ∣ ∣ = ln∣ ∣+C ∣ cos x ∣ (4.1.48) = ln | sec x| + C . (4.1.49) Thus the result they give is ∫ tan x dx = ln | sec x| + C . These two answers are equivalent. Example 4.1.8 : Integrating by substitution: antiderivatives of sec x Evaluate ∫ sec x dx . Solution This example employs a wonderful trick: multiply the integrand by "1" so that we see how to integrate more clearly. In this case, we write "1" as sec x + tan x 1 = . (4.1.50) sec x + tan x This may seem like it came out of left field, but it works beautifully. Consider: sec x + tan x ∫ sec x dx = ∫ sec x ⋅ dx (4.1.51) dx. (4.1.52) sec x + tan x sec 2 x + sec x tan x =∫ sec x + tan x Now let u = sec x + tan x ; this means du = (sec x tan x + sec 2 x) dx , which is our numerator. Thus: du =∫ (4.1.53) u = ln |u| + C (4.1.54) = ln | sec x + tan x| + C . (4.1.55) We can use similar techniques to those used in Examples 4.1.6 and 4.1.7 to find antiderivatives of cot x and csc x (which the reader can explore in the exercises.) We summarize our results here. Theorem 4.1.1 : Antiderivatives of Trigonometric Functions 1. ∫ sin x dx = − cos x + C 2. ∫ cos x dx = sin x + C 3. ∫ tan x dx = − ln | cos x| + C 4. ∫ csc x dx = − ln | csc x + cot x| + C 5. ∫ sec x dx = ln | sec x + tan x| + C 6. ∫ cot x dx = ln | sin x| + C We explore one more common trigonometric integral. 10/21/2020 4.1.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 Example 4.1.9 : Integration by substitution: powers of cos x and sin x Evaluate ∫ cos 2 x dx . Solution 2 We have a composition of functions as cos 2 x = ( cos x ) . However, setting u = cos x means du = − sin x dx , which we do not have in the integral. Another technique is needed. The process we'll employ is to use a Power Reducing formula for formula), which states 2 cos 2 cos x (perhaps consult the back of this text for this 1 + cos(2x) x = . (4.1.56) dx (4.1.57) 2 The right hand side of this equation is not difficult to integrate. We have: ∫ 1 + cos(2x) 2 cos x dx = ∫ 2 1 =∫ ( 1 + 2 cos(2x)) dx. (4.1.58) 2 Now use Key Idea 10: 1 sin(2x) 1 = x+ 2 (4.1.59) 2 sin(2x) 1 = +C 2 x+ 2 + C. (4.1.60) 4 We'll make significant use of this power--reducing technique in future sections. Simplifying the Integrand It is common to be reluctant to manipulate the integrand of an integral; at first, our grasp of integration is tenuous and one may think that working with the integrand will improperly change the results. Integration by substitution works using a different logic: as long as equality is maintained, the integrand can be manipulated so that its form is easier to deal with. The next two examples demonstrate common ways in which using algebra first makes the integration easier to perform. Example 4.1.10 : Integration by substitution: simplifying first Evaluate ∫ 3 2 x +4 x +8x+5 x2 +2x+1 dx . Solution One may try to start by setting u equal to either the numerator or denominator; in each instance, the result is not workable. When dealing with rational functions (i.e., quotients made up of polynomial functions), it is an almost universal rule that everything works better when the degree of the numerator is less than the degree of the denominator. Hence we use polynomial division. We skip the specifics of the steps, but note that when times with a remainder of 3x + 3 . Thus 3 x 2 + 4x 2 x 2 x + 2x + 1 is divided into + 8x + 5 2 + 4x + 8x + 5 , it goes in x +2 3x + 3 = x +2 + + 2x + 1 3 x 2 x . (4.1.61) + 2x + 1 Integrating x + 2 is simple. The fraction can be integrated by setting u = x + 2x + 1 , giving du = (2x + 2) dx . This is very similar to the numerator. Note that du/2 = (x + 1) dx and then consider the following: 2 10/21/2020 4.1.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 3 x 2 + 4x ∫ 2 x + 8x + 5 3x + 3 dx = ∫ (x + 2 + 2 + 2x + 1 x ) dx (4.1.62) + 2x + 1 3(x + 1) = ∫ (x + 2) dx + ∫ dx 2 x 1 = 3 du 2 x + 2x + C1 + ∫ 2 1 = 1 (4.1.64) u 2 3 2 x + 2x + C1 + 2 = (4.1.63) + 2x + 1 2 3 2 x + 2x + ln |u| + C2 2 ln | x 2 (4.1.65) + 2x + 1| + C . (4.1.66) 2 In some ways, we "lucked out" in that after dividing, substitution was able to be done. In later sections we'll develop techniques for handling rational functions where substitution is not directly feasible. Example 4.1.11 : Integration by alternate methods Evaluate ∫ 2 x +2x+3 √x dx with, and without, substitution. Solution 1 We already know how to integrate this particular example. Rewrite √− x as x and simplify the fraction: 2 2 x + 2x + 3 3 =x 1 + 2x 2 + 3x 2 − 1 . 2 (4.1.67) 1/2 x We can now integrate using the Power Rule: 2 x + 2x + 3 ∫ 3 dx = ∫ (x 1 + 2x 2 + 3x 2 − 1 2 ) dx (4.1.68) x1/2 2 = 4 5 x + 2 5 3 x 2 1 + 6x 2 +C (4.1.69) 3 This is a perfectly fine approach. We demonstrate how this can also be solved using substitution as its implementation is rather clever. 1 Let u = √− x =x ; therefore 2 1 du = x − 1 1 2 dx = 2 This gives us ∫ 2 x +2x+3 √x 2 dx = ∫ (x 1 − 2 √x dx ⇒ 2du = − √x dx. (4.1.70) 1 + 2x + 3) ⋅ 2 du . What are we to do with the other x terms? Since u = x , 2 2 u =x , etc. We can then replace x and x with appropriate powers of u. We thus have 2 2 x ∫ + 2x + 3 − √x dx 2 = ∫ (x =∫ 2 = + 2x + 3) ⋅ 2 du 4 2(u 5 u 4 + 5 4 5 x 5 3 u + 3) du + 6u + C 3 2 = 2 + 2u 2 + 3 x 2 1 + 6x 2 + C, 3 which is obviously the same answer we obtained before. In this situation, substitution is arguably more work than our other method. The fantastic thing is that it works. It demonstrates how flexible integration is. Substitution and Inverse Trigonometric Functions 10/21/2020 4.1.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 When studying derivatives of inverse functions, we learned that d −1 ( tan 1 x) = . 2 dx (4.1.71) 1 +x Applying the Chain Rule to this is not difficult; for instance, d −1 ( tan 5 5x) = . 2 dx (4.1.72) 1 + 25x We now explore how Substitution can be used to "undo" certain derivatives that are the result of the Chain Rule applied to Inverse Trigonometric functions. We begin with an example. Example 4.1.12 : Integrating by substitution: inverse trigonometric functions Evaluate ∫ 1 2 25+x dx . Solution The integrand looks similar to the derivative of the arctangent function. Note: 1 1 2 = 25 + x (4.1.73) 2 x 25(1 + 25 ) 1 = 25(1 + ( 1 x 5 (4.1.74) 2 ) ) 1 = 25 1 +( x 5 . 2 (4.1.75) ) Thus 1 ∫ 1 2 1 dx = ∫ 25 25 + x 1 +( 2 x dx. (4.1.76) ) 5 This can be integrated using Substitution. Set u = x/5, hence du = dx/5 or dx = 5du . Thus 1 ∫ 1 2 1 dx = ∫ 25 25 + x 1 +( 1 = 1 = 5 2 dx (4.1.77) ) 1 ∫ 5 x 2 du (4.1.78) 1 +u −1 tan u +C (4.1.79) 5 1 = −1 tan 5 x ( )+C (4.1.80) 5 Example 4.1.12 demonstrates a general technique that can be applied to other integrands that result in inverse trigonometric functions. The results are summarized here. Theorem 4.1.2 : Integrals Involving Inverse Trigonomentric Functions Let a > 0 . 1. ∫ 2. ∫ 3. ∫ 1 2 2 a +x dx = 1 √a2 −x2 1 2 2 x√x −a 1 a −1 tan −1 dx = sin dx = 1 a ( ( sec x a x a −1 )+C )+C |x| ( a )+C Let's practice using Theorem 4.1.2. Example 4.1.13 : Integrating by substitution: inverse trigonometric functions 10/21/2020 4.1.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 Evaluate the given indefinite integrals. 1 ∫ 2 dx, 1 − − − − − − − dx ∫ 9 +x 2 x√ x and ∫ 1 − 1 − −−− − dx. √ 5 − x2 (4.1.81) 100 Solution Each can be answered using a straightforward application of Theorem 4.1.2. ∫ 1 dx = 2 9+x 1 ∫ 1 2 x√x − ∫ 1 −1 x tan 3 3 +C dx = 10 sec −1 , as a = 3 . 10x + C , as a = 1 10 . 100 1 √5−x2 x −1 = sin √5 +C – , as a = √5 . Most applications of Theorem 4.1.2 are not as straightforward. The next examples show some common integrals that can still be approached with this theorem. Example 4.1.14 : Integrating by substitution: completing the square Evaluate ∫ 1 2 x −4x+13 dx . Solution Initially, this integral seems to have nothing in common with the integrals in Theorem 4.1.2. As it lacks a square root, it almost certainly is not related to arcsine or arcsecant. It is, however, related to the arctangent function. We see this by completing the square in the denominator. We give a brief reminder of the process here. Start with a quadratic with a leading coefficient of 1. It will have the form of add/subtract it back into the expression. I.e., 2 2 x + bx + c 2 = x + bx + c . Take 1/2 of b , square it, and 2 b + bx + 2 x b − 4  +c (4.1.82) 4 2 (x+b/2) 2 b = (x + ) 2 b +c − 2 (4.1.83) 4 In our example, we take half of −4 and square it, getting 4. We add/subtract it into the denominator as follows: \] 1 2 x 1 = − 4x + 13 (4.1.84) 2 x − 4x + 4 − 4 + 13  2 (x−2) 1 = 2 (x − 2 ) (4.1.85) +9 \] We can now integrate this using the arctangent rule. Technically, we need to substitute first with employ Key Idea 10 instead. Thus we have 1 ∫ 2 x 1 dx = ∫ − 4x + 13 1 2 (x − 2 ) dx = +9 −1 , but we can x −2 tan 3 u = x −2 + C. (4.1.86) 3 Example 4.1.15 : Integrals requiring multiple methods Evaluate ∫ 10/21/2020 4−x √16−x2 dx . 4.1.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 Solution This integral requires two different methods to evaluate it. We get to those methods by splitting up the integral: 4 −x ∫ 4 − −−−− − √ 16 − x2 dx = ∫ x dx − ∫ − −−−− − √ 16 − x2 − −−−− − √ 16 − x2 The first integral is handled using a straightforward application of Theorem substitution, with u = 16 − x . We handle each separately. dx. (4.1.87) ; the second integral is handled by 4.1.2 2 ∫ ∫ 4 √16−x2 x √16−x2 −1 dx = 4 sin dx x 4 + C. : Set u = 16 − x , so du = −2xdx and xdx = −du/2. We have 2 −du/2 x − −−−− − dx = ∫ √ 16 − x2 ∫ (4.1.88) − √u 1 1 =− ∫ 2 − √u du (4.1.89) − = −√u + C (4.1.90) − −−−− − 2 = −√ 16 − x + C . (4.1.91) Combining these together, we have 4 −x ∫ −1 x dx = 4 sin − −−−− − √ 16 − x2 − −−−− − 2 + √ 16 − x + C . (4.1.92) 4 Exercise 4.1.1 Use substitution to find the antiderivative of ∫ 3x 2 3 (x 2 − 3 ) dx. Hint Let u = x 3 − 3. Answer ∫ 2 3 3 x (x 1 2 − 3 ) dx = 3 (x 3 − 3) +C (4.1.93) 3 Sometimes we need to adjust the constants in our integral if they don’t match up exactly with the expressions we are substituting. Example 4.1.2 : Using Substitution with Alteration Use substitution to find the antiderivative of ∫ − − − − − 2 z√ z − 5 dz. (4.1.94) Solution Rewrite the integral as ∫ z(z − 5) dz. Let u = z − 5 and du = 2z dz . Now we have a problem because du = 2z dz and the original expression has only z dz . We have to alter our expression for du or the integral in u will be twice as large 2 1/2 2 as it should be. If we multiply both sides of the du equation by u =z . we can solve this problem. Thus, (4.1.95) du = 2z dz (4.1.96) 1 du = 10/21/2020 2 −5 1 2 2 1 (2z) dz = z dz. (4.1.97) 2 4.1.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 1 Write the integral in terms of u, but pull the outside the integration symbol: 2 ∫ z(z 2 1 1/2 − 5) dz = 1/2 ∫ u du. (4.1.98) 2 Integrate the expression in u: 1 1/2 ∫ u 3/2 1 du = ( 2 u ) 2 +C 3 2 1 =( 2 )( +C 3 1 3/2 = 3/2 )u 2 u +C 3 1 = (z 2 3/2 − 5) +C 3 Exercise 4.1.2 Use substitution to find the antiderivative of ∫ x 2 3 (x 9 + 5) dx. Hint Multiply the du equation by 1 3 . Answer 3 (x 10 + 5) +C (4.1.99) 30 Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that is in terms of the original variable. When we are done, u should be the only variable in the integrand. In some cases, this means solving for the original variable in terms of u. This technique should become clear in the next example. Example 4.1.4 : Finding an Antiderivative Using u-Substitution Use substitution to find the antiderivative of x − − − − − dx. √x − 1 ∫ (4.1.100) Solution If we let u = x − 1, then du = dx . But this does not account for the x in the numerator of the integrand. We need to express x in terms of u. If u = x − 1 , then x = u + 1. Now we can rewrite the integral in terms of u: ∫ x − − − − − dx = ∫ √x − 1 u +1 − √u + du = ∫ − √u 1 1/2 − du = ∫ (u √u −1/2 +u ) du. (4.1.101) Then we integrate in the usual way, replace u with the original expression, and factor and simplify the result. Thus, 1/2 ∫ (u −1/2 +u 2 ) du = 3/2 u 1/2 + 2u +C 3 2 = 3/2 (x − 1 ) 1/2 + 2(x − 1 ) +C 3 1/2 = (x − 1 ) 2 [ (x − 1) + 2] + C 3 10/21/2020 4.1.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 1/2 = (x − 1 ) 2 ( 2 x− 3 1/2 = (x − 1 ) 3 2 ( 2 1/2 (x − 1 ) ) 3 4 x+ 3 = 6 + ) 3 (x + 2) + C . 3 Using Substitution with Integrals of Trigonometric Functions Example 4.1.4 : Using Substitution with Integrals of Trigonometric Functions sin t Use substitution to evaluate the integral ∫ dt. 3 cos t Solution We know the derivative of cos t is − sin t , so we set u = cos t . Then du = − sin t dt. Substituting into the integral, we have sin t ∫ 3 cos du dt = − ∫ u3 t . (4.1.102) Evaluating the integral, we get du −∫ −3 = −∫ u3 u 1 du = −(− −2 )u + C. (4.1.103) 2 Putting the answer back in terms of t, we get sin t ∫ 3 cos 1 dt = t 1 2 +C = 2u 2 2 cos + C. (4.1.104) t Exercise 4.1.4 Use substitution to evaluate the integral cos t ∫ 2 sin dt. (4.1.105) t Hint Use the process from Example to solve the problem. Answer 1 − +C sin t Exercise 4.1.4 Use substitution to evaluate the indefinite integral ∫ 3 cos t sin t dt. (4.1.106) Hint Use the process from Example to solve the problem. Answer 4 cos t − +C (4.1.107) 4 10/21/2020 4.1.13 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 Example 4.1.5 : ∫ sec(t) dt. (4.1.108) ∫ csc(t) dt. (4.1.109) Exercise 4.1.5 Answer Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page. Integrals Involving Logarithmic Functions and involving Exponential Function A common mistake when dealing with exponential expressions is treating the exponent one the same way we treat exponents in polynomial expressions. We cannot use the power rule for the exponent one. This can be especially confusing when we have both exponentials and polynomials in the same expression, as in the previous checkpoint. In these cases, we should always double-check to make sure we’re using the right rules for the functions we’re integrating. Example 4.1.5 : Square Root of an Exponential Function Find the antiderivative of the exponential function e x − − − − − √1 + ex . Solution First rewrite the problem using a rational exponent: − − − − − x x x 1/2 x ∫ e √1 + e dx = ∫ e (1 + e ) dx. Using substitution, choose u = 1 + e x . u = 1 +e x x Then, du = e x 1/2 ∫ e (1 + e ) x dx . We have (Figure) 1/2 dx = ∫ u du. Then 3/2 1/2 ∫ u u du = 2 +C = 3/2 3/2 u 3 2 +C = x 3/2 (1 + e ) +C 3 Figure 4.1.1 : The graph shows an exponential function times the square root of an exponential function. Exercise 4.1.6 Find the antiderivative of e x (3 e x 2 − 2) . Hint Let u = 3e x − 2u = 3 e x − 2. Answer x ∫ e (3 e x 2 1 − 2 ) dx = (3 e x 3 − 2) 9 10/21/2020 4.1.14 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 Example 4.1.7 : Using Substitution with an Exponential Function Use substitution to evaluate the indefinite integral ∫ 3x 2 3 e 2x dx. Solution Here we choose to let u equal the expression in the exponent on e. Let 3 u = 2x and 2 du = 6 x dx .. Again, du is off by a constant multiplier; the original function contains a factor of 3x , not 6x . Multiply both sides of the equation by 2 2 the integrand in u equals the integrand in x. Thus, 2 ∫ 3x e 1 3 2x dx = u ∫ e du 2 1 2 so that . Integrate the expression in u and then substitute the original expression in x back into the u integral: 1 u 1 ∫ e du = 2 e u 1 +C = 2 2 3 e x + C. 2 Exercise 4.1.8 Evaluate the indefinite integral ∫ 2x 3 4 ex dx . Hint Let u = x 4 . Answer 3 1 4 ∫ 2 x ex dx = 4 e x 2 Example 4.1.9 : Finding an Antiderivative Involving ln x Find the antiderivative of the function 3 . (4.1.110) x − 10 Solution First factor the 3 outside the integral symbol. Then use the u −1 3 ∫ 1 dx = 3 ∫ x − 10 du dx = 3 ∫ x − 10 rule. Thus, = 3 ln |u| + C = 3 ln |x − 10| + C , x ≠ 10. (4.1.111) u See Figure. Figure 4.1.3 : The domain of this function is x ≠ 10. Exercise 4.1.9 Find the antiderivative of 10/21/2020 4.1.15 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 1 . (4.1.112) x +2 Hint Follow the pattern from Example to solve the problem. Answer ln |x + 2| + C (4.1.113) Example 4.1.10 : Finding an Antiderivative of a Rational Function Find the antiderivative of 3 2x + 3x x4 + 3 x2 . (4.1.114) Solution This can be rewritten as ∫ (2x by factoring out the 2. Thus, 3 4 + 3x)(x 2 −1 + 3x ) dx. 3 du = (4 x Use substitution. Let u = x 4 3 + 6x)dx = 2(2 x 1 3 du = (2 x 2 + 3x , then du = 4x + 3x)dx 3 + 6x. Alter du (4.1.115) + 3x)dx. (4.1.116) 2 Rewrite the integrand in u: 3 ∫ (2 x 4 + 3x)(x 2 −1 + 3x ) 1 dx = ∫ −1 u du. (4.1.117) 2 Then we have 1 ∫ −1 u 2 1 du = 1 ln |u| + C = 2 4 ln ∣ x 2 + 3x ∣ +C . (4.1.118) 2 Example 4.1.11 : Finding an Antiderivative of a Logarithmic Function Find the antiderivative of the log function \[\drac{\ln(2x)}{x}.\) Solution Let u = ln(2x). Then du = 2 dx . Now, x 1 ∫ \dracln(2x)x. dx = ∫ 1 udu = 2 2 u 1 +C = \2 2 (ln x ) + C. (4.1.119) 2 Key Concepts Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term ‘substitution’ refers to changing variables or substituting the variable u and du for appropriate expressions in the integrand. Formulas for derivatives of inverse trigonometric functions developed in Derivatives of Exponential and Logarithmic Functions lead directly to integration formulas involving inverse trigonometric functions. Use the formulas listed in the rule on integration formulas resulting in inverse trigonometric functions to match up the correct format and make alterations as necessary to solve the problem. Substitution is often required to put the integrand in the correct form. Key Equations Substitution with Indefinite Integrals 10/21/2020 4.1.16 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 ∫ f [g(x)]g'(x)dx = ∫ f (u)du = F (u) + C = F (g(x)) + C Integrals That Produce Inverse Trigonometric Functions ∫ du u −1 ( )+C − − − − − − = si n 2 2 a √a − u du ∫ 2 a ∫ 1 2 = +u −1 tan a u ( )+C a du 1 u −1 sec ( )+C − − − − − − = 2 2 a a u √u − a Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution Noncommercial (BY-NC) License. http://www.apexcalculus.com/ Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 4.1.17 CC-BY-NC-SA https://math.libretexts.org/@go/page/10312 4.1E: Exercises Exercise 4.1E . 1 1. Why is u-substitution referred to as change of variable? 2. If f = g∘ h d , when reversing the chain rule, x(g ∘ h)(x) = g'(h(x))h'(x) d , should you take u = g(x) or u= h(x)? Answer . u = h(x) Exercise 4.1E . 2 In the following exercises, verify each identity using differentiation. Then, using the indicated u-substitution, identify f such that the integral takes the form ∫ 1. ∫ 2 − − − − − x √x + 1 x = f (u)du. 3/2 (x + 1 ) (3x − 2) + C ; u = x + 1 15 2 x 2 − − − − − 2 √x − 1 (3 x + 4x + 8) + C ; u = x − 1 − − − − − dx(x > 1) = 15 √x − 1 2. ∫ Answer 2 (u + 1) f (u) = 3. ∫ − √u − −−−− − 2 x √4 x 1 2 + 9 dx = (4 x 3/2 + 9) 2 + C ; u = 4x +9 12 x 4. ∫ 1 − −−−− − √4 x2 + 9 dx = − −−−− − 2 2 √4 x + 9 + C ; u = 4 x + 9 4 Answer 1 du = 8xdx; f (u) = − 8 √u x 5. ∫ 2 (4 x 1 2 dx = − + 9) 2 8(4 x 2 ; u = 4x +9 + 9) Answer 1 du = 8xdx; f (u) = 2 8u Exercise 4.1E . 3 In the following exercises, find the antiderivative using the indicated substitution. 1. ∫ 4 (x + 1 ) dx; u = x + 1 Answer 1 5 (x + 1 ) +C 5 10/21/2020 4.1E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/13730 2. ∫ (x − 1 ) dx; u = x − 1 3. ∫ (2x − 3 ) 5 −7 dx; u = 2x − 3 Answer 1 − +C 6 12(3 − 2x) 4. ∫ −11 (3x − 2 ) dx; u = 3x − 2 x 2 − −−− − dx; u = x + 1 2 √x + 1 5. ∫ Answer − −−− − √x2 + 1 + C x 2 − −−− − dx; u = 1 − x √1 − x2 6. ∫ 7. ∫ 2 3 (x − 1)(x 2 − 2x ) dx; u = x − 2x Answer 1 2 (x 4 − 2x ) +C 8 8. ∫ 2 (x 9. ∫ 3 2 − 2x)(x 2 3 − 3 x ) dx; u = x 3 2 2 = 3x 2 cos θdθ; u = sinθ(H int : cos θ = 1 − si n θ) Answer 3 \displaystylesinθ − 10. ∫ sin θ 3 +C 3 2 2 si n θdθ; u = cosθ(H int : si n θ = 1 − cos θ) Exercise 4.1E . 4 In the following exercises, use a suitable change of variables to determine the indefinite integral. 1. ∫ 99 x(1 − x ) dx Answer 101 100 (1 − x) (1 − x) − +C 101 100 2. ∫ t(1 − t ) 3. ∫ (11x − 7 ) 2 10 dt −3 dx Answer 10/21/2020 4.1E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/13730 1 − +C 2 22(7 − 11 x ) 4. \displaystyle∫(7x−11)^4dx\) 5. ∫ 3 cos θsinθdθ Answer 4 cos θ − +C 4 6. ∫ si n θcosθdθ 7. ∫ cos (πt)sin(πt)dt 7 2 Answer 3 cos (πt) − +C 3π 8. ∫ si n xcos xdx(H int : si n x + cos x = 1) 9. ∫ tsin(t )cos(t )dt 2 3 2 2 2 2 Answer 1 − 2 2 cos (t ) + C 4 10. ∫ 2 2 3 3 t cos (t )sin(t )dt 2 x 11. ∫ 3 2 (x dx − 3) Answer 1 − 3 3(x +C − 3) 3 12. ∫ 13. ∫ x − −−− − dx √1 − x2 y 5 dy (1 − y 3 )3/2 Answer 2(y − 14. ∫ 15. ∫ 3 − 2) − −−− − 3 3 √1 − y 99 cosθ(1 − cosθ) 3 10 (1 − cos θ) sinθdθ 2 cos θsinθdθ Answer 10/21/2020 4.1E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/13730 1 3 11 (1 − cos θ) +C 33 16. ∫ (cosθ − 1)(cos θ − 2cosθ) sinθdθ 17. ∫ (si n θ − 2sinθ)(si n θ − 3si n θ 3cosθdθ 2 3 2 3 2 ) Answer 1 3 2 4 (si n θ − 3si n θ) +C 12 Exercise 4.1E . 5 3 1) ∫ 2) ∫ 3)∫ 4) ∫ 5) ∫ 6) ∫ x dx − −−− − √1 − x2 (csc x)(cot x)(e csc x ) dx 1 e x +e −x dx 1 − −−− − √4 − x2 dx − −−− − √x2 − 4 dx x 1 2 x dx + 6x + 13 Answer Under Construction Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada) 10/21/2020 4.1E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/13730 4.2: Definite Integral- An Introduction We start with an easy problem. An object travels in a straight line at a constant velocity of 5 ft/s for 10 seconds. How far away from its starting point is the object? We approach this problem with the familiar "Distance feet. = Rate × Time" equation. In this case, Distance = 5ft/s × 10s = 50 It is interesting to note that this solution of 50 feet can be represented graphically. Consider Figure 4.2.1, where the constant velocity of 5ft/s is graphed on the axes. Shading the area under the line from t = 0 to t = 10 gives a rectangle with an area of 50 square units; when one considers the units of the axes, we can say this area represents 50 ft. Figure 4.2.1 : The area under a constant velocity function corresponds to distance traveled. Now consider a slightly harder situation (and not particularly realistic): an object travels in a straight line with a constant velocity of 5ft/s for 10 seconds, then instantly reverses course at a rate of 2ft/s for 4 seconds. (Since the object is traveling in the opposite direction when reversing course, we say the velocity is a constant −2ft/s.) How far away from the starting point is the object -- what is its displacement? Here we use "Distance = Rate 1 × Time + Rate 1 2 × Time ," which is Distance 2 = 5 ⋅ 10 + (−2) ⋅ 4 = 42 ft. (4.2.1) Hence the object is 42 feet from its starting location. We can again depict this situation graphically. In Figure [10, 14], respectively. The displacement of the object is 4.2.3 "Area above the t--axis we have the velocities graphed as straight lines on [0, 10] and − Area below the t-axis, (4.2.2) which is easy to calculate as 50 − 8 = 42 feet. 10/21/2020 4.2.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/13741 Figure 4.2.2 : The total displacement is the area above the t --axis minus the area below the t --axis. Now consider a more difficult problem. Example 4.2.1 : Finding position using velocity The velocity of an object moving straight up/down under the acceleration of gravity is given as v(t) = −32t + 48 , where time t is given in seconds and velocity is in ft/s. When t = 0 , the object had a height of 0 ft. 1. What was the initial velocity of the object? 2. What was the maximum height of the object? 3. What was the height of the object at time t = 2 ? Solution It is straightforward to find the initial velocity; at time t = 0 , v(0) = −32 ⋅ 0 + 48 = 48 ft/s. To answer questions about the height of the object, we need to find the object's position function s(t) . This is an initial value problem, which we studied in the previous section. We are told the initial height is 0, i.e., s(0) = 0 . We know s (t) = v(t) = −32t + 48 . To find s , we find the indefinite integral of v(t) : ′ ∫ 2 v(t) dt = ∫ (−32t + 48) dt = −16 t Since s(0) = 0 , we conclude that C =0 and s(t) = −16t 2 + 48t + 48t + C = s(t). (4.2.3) . To find the maximum height of the object, we need to find the maximum of s . Recalling our work finding extreme values, we find the critical points of s by setting its derivative equal to 0 and solving for t : ′ s (t) = −32t + 48 = 0 ⇒ t = 48/32 = 1.5s. (4.2.4) (Notice how we ended up just finding when the velocity was 0ft/s!) The first derivative test shows this is a maximum, so the maximum height of the object is found at 2 s(1.5) = −16(1.5 ) + 48(1.5) = 36ft. (4.2.5) The height at time t = 2 is now straightforward to compute: it is s(2) = 32 ft. While we have answered all three questions, let's look at them again graphically, using the concepts of area that we explored earlier. Figure 4.2.3 shows a graph of v(t) on axes from t = 0 to t = 3 . It is again straightforward to find the graph to find the maximum height of the object? 10/21/2020 4.2.2 CC-BY-NC-SA v(0) . How can we use https://math.libretexts.org/@go/page/13741 Figure 4.2.3 : A graph of v(t) = −32t + 48 ; the shaded areas help determine displacement. Recall how in our previous work that the displacement of the object (in this case, its height) was found as the area under the velocity curve, as shaded in the figure. Moreover, the area between the curve and the t --axis that is below the t --axis counted as "negative" area. That is, it represents the object coming back toward its starting position. So to find the maximum distance from the starting point -- the maximum height -- we find the area under the velocity line that is above the t --axis, i.e., from t = 0 to t = 1.5 . This region is a triangle; its area is 1 Area = 1 Base × Height = 2 × 1.5s × 48ft/s = 36ft, (4.2.6) 2 which matches our previous calculation of the maximum height. Finally, we find the total signed area under the velocity function from t = 0 to t = 2 to find the s(2) , the height at which is a displacement, the distance from the current position to the starting position. That is, Displacement = Area above the t−axis − Area below t-axis. t =2 , (4.2.7) The regions are triangles, and we find 1 Displacement = 1 (1.5s)(48ft/s) − 2 (.5s)(16ft/s) = 32ft. (4.2.8) 2 This also matches our previous calculation of the height at t = 2 . Notice how we answered each question in this example in two ways. Our first method was to manipulate equations using our understanding of antiderivatives and derivatives. Our second method was geometric: we answered questions looking at a graph and finding the areas of certain regions of this graph. The above example does not prove a relationship between area under a velocity function and displacement, but it does imply a relationship exists. Section 5.4 will fully establish fact that the area under a velocity function is displacement. Given a graph of a function y = f (x), we will find that there is great use in computing the area between the curve y = f (x) and the x-axis. Because of this, we need to define some terms. Definition 4.2.1 : The Definite Integral, Total Signed Area Let y = f (x) be defined on a closed interval [a, b]. The total signed area from x = a to x = b under f is: (area under f and above the x-axis on [a, b]). −(area above f and under the x-axis on [a, b]). (4.2.9) The definite integral of f on [a, b] is the total signed area of f on [a, b], denoted b ∫ f (x) dx, (4.2.10) a There a and b are the bounds of integration. 10/21/2020 4.2.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/13741 By our definition, the definite integral gives the "signed area under f ." We usually drop the word "signed" when talking about the definite integral, and simply say the definite integral gives "the area under f \," or, more commonly, "the area under the curve." The previous section introduced the indefinite integral, which related to antiderivatives. We have now defined the definite integral, which relates to areas under a function. The two are very much related, as we'll see when we learn the Fundamental Theorem of Calculus in Section 4.5. Recall that earlier we said that the "∫ " symbol was an "elongated S" that represented finding a "sum." In the context of the definite integral, this notation makes a bit more sense, as we are adding up areas under the function f . We practice using this notation. Example 4.2.2 : Evaluating definite integrals Consider the function f given in Figure 4.2.4. Figure 4.2.4 : A graph of f (x) in Example 2 Find: 1. ∫ 2. ∫ 3 3. ∫ 4. ∫ 5 5. ∫ 1 0 5 3 0 3 0 1 f (x) dx f (x) dx f (x) dx 5f (x) dx f (x) dx Solution 1. ∫ 3 0 ∫ 3 0 2. ∫ 5 3 f (x)dx is the area under (3)(1) = 1.5 . f (x)dx = f (x)dx f on the interval . This region is a triangle, so the area is [0, 3] 1 2 represents the area of the triangle found under the x--axis on found under the x--axis, this is "negative area." Therefore ∫ 5 3 3. ∫ 4. ∫ 5 0 3 0 . The area is [3, 5] f (x) dx = −1 1 2 (2)(1) = 1 ; since it is . is the total signed area under f on [0, 5]. This is 1.5 + (−1) = 0.5 . 5f (x)dx is the area under 5f on [0, 3]. This is sketched in Figure 4.2.5. Again, the region is a triangle, with height 5 f (x)dx times that of the height of the original triangle. Thus the area is ∫ 3 0 10/21/2020 5f (x) dx = 15/2 = 7.5. 4.2.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/13741 Figure 4.2.5 : A graph of 5f in Example 4.2.2. (Yes, it looks just like the graph of f in Figure 4.2.4, just with a different y -scale.) 1 5. ∫ f (x)dx is the area under Therefore the area is 0. 1 f on the "interval" . This describes a line segment, not a region; it has no width. [1, 1] The area definition of the definite integral allows us to use geometry compute the definite integral of some simple functions. Example 4.2.3 : Evaluating definite integrals using geometry Evaluate the following definite integrals: 5 1. ∫ 3 (2x − 4) dx 2. ∫ −2 − −−− − 2 √ 9 − x dx. (4.2.11) −3 Solution − −−− − Figure 4.2.7 : A graph of f (x) = 2x − 4 in (a) and f (x) = √9 − x in (b), from Example 4.2.4 2 1. It is useful to sketch the function in the integrand, as shown in Figure 4.2.7a. We see we need to compute the areas of two regions, which we have labeled R and R . Both are triangles, so the area computation is straightforward: 1 2 1 R1 : 1 (4)(8) = 16 2 R2 : (3)6 = 9. (4.2.12) 2 Region R lies under the x--axis, hence it is counted as negative area (we can think of the triangle's height as being " −8"), so 1 10/21/2020 4.2.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/13741 5 ∫ (2x − 4) dx = −16 + 9 = −7. (4.2.13) −2 2. Recognize that the integrand of this definite integral describes a half circle, as sketched in Figure 4.2.7b, with radius 3. Thus the area is: 3 ∫ −3 − −−− − 1 9 2 2 √ 9 − x dx = πr = π. 2 (4.2.14) 2 Example 4.2.4 : Understanding motion given velocity Consider the graph of a velocity function of an object moving in a straight line, given in Figure 4.2.8, where the numbers in the given regions gives the area of that region. Assume that the definite integral of a velocity function gives displacement. Find the maximum speed of the object and its maximum displacement from its starting position. Solution Figure 4.2.8 : A graph of a velocity in Example 4.2.5. Since the graph gives velocity, finding the maximum speed is simple: it looks to be 15ft/s. At time t = 0 , the displacement is 0; the object is at its starting position. At time t = a , the object has moved backward 11 feet. Between times t = a and t = b , the object moves forward 38 feet, bringing it into a position 27 feet forward of its starting position. From t = b to t = c the object is moving backwards again, hence its maximum displacement is 27 feet from its starting position. In our examples, we have either found the areas of regions that have nice geometric shapes (such as rectangles, triangles and circles) or the areas were given to us. Consider Figure 4.2.9, where a region below y = x is shaded. What is its area? The function y = x is relatively simple, yet the shape it defines has an area that is not simple to find geometrically. 2 2 10/21/2020 4.2.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/13741 Figure 4.2.9 : What is the area below y = x on [0, 3]? The region is not a usual geometric shape. 2 In the next section, we will explore how to find the areas of such regions. Contributors Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/ Integrated by Justin Marshall. 10/21/2020 4.2.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/13741 4.2E: Exercises Exercise 4.2E . 1 1. What is "total signed area"? 2. What is "displacement"? 3. What is ∫ 3 3 sin x dx 4. Give a single definite integral that has the same value as ∫ 1 0 (2x + 3) dx + ∫ 2 1 (2x + 3) dx . Answer Under Construction Exercise 4.2E . 2 A graph of a function f (x) is given. Using the geometry of the graph, evaluate the definite integrals. 1. (a) ∫ (b) ∫ 1 (c) ∫ (d) ∫ 3 (e) ∫ (f) ∫ 4 0 2 0 0 3 1 2 1 0 (−2x + 4) dx (−2x + 4) dx (−2x + 4) dx (−2x + 4) dx (−2x + 4) dx (−6x + 12) dx 2. (a) ∫ 2 (b) ∫ (c) ∫ 3 0 0 5 0 (d) ∫ (e) ∫ 5 2 3 5 (f) ∫ 3 0 f (x) dx f (x) dx f (x) dx f (x) dx f (x) dx f (x) dx 10/21/2020 4.2E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/13747 3. (a) ∫ 2 (b) ∫ (c) ∫ 4 0 2 4 2 (d) ∫ (e) ∫ 1 0 3 2 3 (f) ∫ 2 f (x) dx f (x) dx 2f (x) dx 4x dx (2x − 4) dx (4x − 8) dx 4. (a) ∫ (b) ∫ 1 (c) ∫ (d) ∫ 3 (e) ∫ (f) ∫ 4 0 2 0 0 3 2 1 4 1 (x − 1) dx (x − 1) dx (x − 1) dx (x − 1) dx (x − 1) dx ((x − 1) + 1) dx 5. (a) ∫ 2 (b) ∫ (c) ∫ 4 0 2 4 0 (d) ∫ 4 0 f (x) dx f (x) dx f (x) dx 5f (x) dx Answer Under Construction Exercise 4.2E . 3 A graph of a function f (x) is given; the numbers inside the shaded regions give the area of that region. Evaluate the definite integrals using this area information. 10/21/2020 4.2E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/13747 1. (a) ∫ (b) ∫ 1 (c) ∫ (d) ∫ 3 0 2 0 0 2 1 f (x) dx f (x) dx f (x) dx −3f (x) dx 2. (a) ∫ 2 (b) ∫ (c) ∫ 4 0 2 4 0 (d) ∫ 1 0 f (x) dx f (x) dx f (x) dx f (x) dx 3. (a) ∫ −1 (b) ∫ (c) ∫ 2 (d) ∫ 1 f (x) dx −2 1 1 f (x) dx −1 0 f (x) dx f (x) dx 4. 10/21/2020 4.2E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/13747 (a) ∫ 2 (b) ∫ (c) ∫ 2 0 0 3 1 (d) ∫ 4 2 2 5x 2 (x dx + 1) dx 2 (x − 1 ) dx ((x − 2) + 5) dx Answer Under Construction Exercise 4.2E . 4 A graph of the velocity function of an object moving in a straight line is given. Answer the questions based on that graph. 1. (a) What is the object's maximum velocity? (b) What is the object's maximum displacement? (c) What is the object's total displacement on [0,3]? 2. (a) What is the object's maximum velocity? (b) What is the object's maximum displacement? (c) What is the object's total displacement on [0,5]? Answer Under Construction Exercise 4.2E . 5 An object is thrown straight up with a velocity, in ft/s, given by v(t) = −32t + 64 , where t is in seconds, from a height of 48 feet. (a) What is the object's maximum velocity? (b) What is the object's maximum displacement? (c) When does the maximum displacement occur? (d) When will the object reach a height of 0? (Hint: find when the displacement is -48ft.) 10/21/2020 4.2E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/13747 Answer Under Construction Exercise 4.2E . 6 An object is thrown straight up with a velocity, in ft/s, given by v(t) = −32t + 96 , where t is seconds, from a height of 64 feet. (a) What is the object's initial velocity? (b) What is the object's displacement 0? (c) How long does it take for the object to return to its initial height? (d) When will the object reach a height of 210ft? Answer Under Construction Exercise 4.2E . 7 Use these values to evaluate the given definite integrals. ∫ 2 0 ∫ 3 0 2 ∫ 3 2 1. ∫ 2 2. ∫ 3 3. ∫ 3 , f (x) dx = 7 , g(x) dx = −3 0 ∫ f (x) dx = 5 g(x) dx = 5 , and . (f (x) + g(x)) dx 0 (f (x) − g(x)) dx 0 (3f (x) + 2g(x)) dx 2 4. Find values for a and b such that ∫ 3 0 (af (x) + bg(x)) dx = 0 Answer Under Construction Exercise 4.2E . 8 Use these values to evaluate the given definite integrals. ∫ 3 0 ∫ 5 3 ∫ 5 3 ∫ 5 0 1. ∫ 3 2. ∫ 0 3. ∫ 3 0 5 3 s(t) dt = 10 s(t) dt = 8 , , r(t) dt = −1 r(t) dt = 11 , and . (s(t) + r(t)) dt (s(t) − r(t)) dt (πs(t) − 7r(t)) dt 4. Find values for a and b such that ∫ 5 0 (ar(t) + bs(t)) dt = 0 Answer 10/21/2020 4.2E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/13747 Under Construction Exercise 4.2E . 9 Evaluate the given indefinite integral: 1. ∫ (x 3 2 − 2x + 7x − 9) dx 2. ∫ (sin x − cos x + sec 3. ∫ (√t + 3 4. ∫ ( 1 x 1 2 t 2 x) dx t + 2 ) dt − csc x cot x) dx Answer Under Construction 10/21/2020 4.2E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/13747 4.3: Approximating Areas Archimedes was fascinated with calculating the areas of various shapes—in other words, the amount of space enclosed by the shape. He used a process that has come to be known as the method of exhaustion, which used smaller and smaller shapes, the areas of which could be calculated exactly, to fill an irregular region and thereby obtain closer and closer approximations to the total area. In this process, an area bounded by curves is filled with rectangles, triangles, and shapes with exact area formulas. These areas are then summed to approximate the area of the curved region. In this section, we develop techniques to approximate the area between a curve, defined by a function f (x), and the x-axis on a closed interval [a, b]. Like Archimedes, we first approximate the area under the curve using shapes of known area (namely, rectangles). By using smaller and smaller rectangles, we get closer and closer approximations to the area. Taking a limit allows us to calculate the exact area under the curve. Let’s start by introducing some notation to make the calculations easier. We then consider the case when f (x) is continuous and nonnegative. Later in the chapter, we relax some of these restrictions and develop techniques that apply in more general cases. Sigma (Summation) Notation As mentioned, we will use shapes of known area to approximate the area of an irregular region bounded by curves. This process often requires adding up long strings of numbers. To make it easier to write down these lengthy sums, we look at some new notation here, called sigma notation (also known as summation notation). The Greek capital letter Σ, sigma, is used to express long sums of values in a compact form. For example, if we want to add all the integers from 1 to 20 without sigma notation, we have to write 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20. (4.3.1) We could probably skip writing a couple of terms and write 1 + 2 + 3 + 4 + ⋯ + 19 + 20, (4.3.2) which is better, but still cumbersome. With sigma notation, we write this sum as 20 ∑i (4.3.3) i=1 which is much more compact. Typically, sigma notation is presented in the form n ∑ ai (4.3.4) i=1 where a describes the terms to be added, and the i is called the index. Each term is evaluated, then we sum all the values, beginning with the value when i = 1 and ending with the value when i = n. For example, an expression like ∑ s is interpreted as s + s + s + s + s + s . Note that the index is used only to keep track of the terms to be added; it does not factor into the calculation of the sum itself. The index is therefore called a dummy variable. We can use any letter we like for the index. Typically, mathematicians use i, j, k, m, and n for indices. i 7 i=2 i 2 3 4 5 6 7 Let’s try a couple of examples of using sigma notation. Example 4.3.1 : Using Sigma Notation a. Write in sigma notation and evaluate the sum of terms 3 for i = 1, 2, 3, 4, 5. b. Write the sum in sigma notation: i 1 1 1+ 1 + 4 + 9 1 + 16 . (4.3.5) 25 Solution a. Write 5 i ∑3 2 = 3 +3 3 +3 4 +3 5 +3 = 363. (4.3.6) i=1 5 b. The denominator of each term is a perfect square. Using sigma notation, this sum can be written as ∑ i=1 1 2 . i Exercise 4.3.1 Write in sigma notation and evaluate the sum of terms 2 for i = 3, 4, 5, 6. i Hint Use the solving steps in Example as a guide. Answer 10/21/2020 4.3.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10977 6 i ∑i=3 2 3 4 =2 +2 5 6 +2 +2 = 120 The properties associated with the summation process are given in the following rule. Rule: Properties of Sigma Notation Let a , a , … , a and b , b , … , b represent two sequences of terms and let c be a constant. The following properties hold for all positive integers n and for integers m, with 1 ≤ m ≤ n. 1 2 1. ∑ 2. ∑ 3. ∑ 4. ∑ 5. ∑ n i=1 n 1 i=1 n i=1 n i=1 n c = nc n c i=1 n 2 n ai = c ∑ i=1 n ai n (ai + bi ) = ∑ ai + ∑ (ai − bi ) = ∑ ai − ∑ i=1 n i=1 m ai = ∑ i=1 i=1 n i=1 n ai + ∑ i=m+1 bi bi ai Proof We prove properties 2. and 3. here, and leave proof of the other properties to the Exercises. 2. We have n n ∑ c ai = c a1 + c a2 + c a3 + ⋯ + c an = c(a1 + a2 + a3 + ⋯ + an ) = c ∑ i=1 i=1 ai . 3. We have n ∑(ai + bi ) = (a1 + b1 ) + (a2 + b2 ) + (a3 + b3 ) + ⋯ + (an + bn ) (4.3.7) = (a1 + a2 + a3 + ⋯ + an ) + (b1 + b2 + b3 + ⋯ + bn ) (4.3.8) i=1 n n = ∑ ai + ∑ bi . i=1 (4.3.9) i=1 □ A few more formulas for frequently found functions simplify the summation process further. These are shown in the next rule, for sums and powers of integers, and we use them in the next set of examples. Rule: Sums and Powers of Integers 1. The sum of n integers is given by n n(n + 1) ∑i = 1 +2 +⋯ +n = . (4.3.10) 2 i=1 2. The sum of consecutive integers squared is given by n 2 ∑i 2 =1 2 +2 2 +⋯ +n n(n + 1)(2n + 1) = . (4.3.11) 6 i=1 3. The sum of consecutive integers cubed is given by n 2 3 ∑i 3 =1 3 +2 3 +⋯ +n 2 n (n + 1 ) = . (4.3.12) 4 i=1 Example 4.3.2 : Evaluation Using Sigma Notation Write using sigma notation and evaluate: a. The sum of the terms (i − 3) for i = 1, 2, … , 200. b. The sum of the terms (i − i ) for i = 1, 2, 3, 4, 5, 6 2 3 2 Solution a. Multiplying out (i − 3) , we can break the expression into three terms. 2 200 ∑ i=1 2 (i − 3 ) 200 =∑ i=1 200 =∑ i=1 2 i 2 (i 200 −∑ i=1 = ∑i = 1 − 6i + 9) 200 6i + ∑ 200 i=1 2 i 9 200 −6 ∑ i=1 200 i +∑ i=1 200(200 + 1)(400 + 1) = 200(200 + 1) − 6[ 6 10/21/2020 9 ] + 9(200) 2 4.3.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10977 = 2, 686, 700 − 120, 600 + 1800 = 2, 567, 900 b. Use sigma notation property iv. and the rules for the sum of squared terms and the sum of cubed terms. 6 3 ∑i=1 (i 2 −i ) = ∑i = 1 2 6 3 i 2 6 (6 + 1 ) 6 2 − ∑i=1 i 6(6 + 1)(2(6) + 1) = − 4 1764 = 6 546 − 4 6 = 350 Exercise 4.3.2 Find the sum of the values of 4 + 3i for i = 1, 2, … , 100. Hint Use the properties of sigma notation to solve the problem. Answer 15,550 Example 4.3.3 : Finding the Sum of the Function Values Find the sum of the values of f (x) = x over the integers 1, 2, 3, … , 10. 3 Solution Using the formula, we have 2 10 3 ∑i=0 i 2 (10 ) (10 + 1 ) = 100(121) = 4 = 3025 4 Exercise 4.3.3 20 Evaluate the sum indicated by the notation ∑ k=1 (2k + 1) . Hint Use the rule on sum and powers of integers. Answer 440 Approximating Area Now that we have the necessary notation, we return to the problem at hand: approximating the area under a curve. Let f (x) be a continuous, nonnegative function defined on the closed interval [a, b]. We want to approximate the area A bounded by f (x) above, the x-axis below, the line x = a on the left, and the line x = b on the right (Figure 4.3.1). Figure 4.3.1 : An area (shaded region) bounded by the curve f (x) at top, the x-axis at bottom, the line x = a to the left, and the line x = b at right. 10/21/2020 4.3.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10977 How do we approximate the area under this curve? The approach is a geometric one. By dividing a region into many small shapes that have known area formulas, we can sum these areas and obtain a reasonable estimate of the true area. We begin by dividing the interval [a, b] into n subintervals of equal width, b −a n . We do this by selecting equally spaced points x 0, x1 , x2 , … , xn with x 0 = a, xn = b, and b −a xi − xi−1 = (4.3.13) n for i = 1, 2, 3, … , n. We denote the width of each subinterval with the notation Δx, so Δx = b−a n and xi = x0 + iΔx (4.3.14) for i = 1, 2, 3, … , n. This notion of dividing an interval [a, b] into subintervals by selecting points from within the interval is used quite often in approximating the area under a curve, so let’s define some relevant terminology. Definition: Partitions A set of points P = xi for i = 0, 1, 2, … , n with a = x < x < x < ⋯ < xn = b, whichdividestheinterval\([a, b] into subintervals of the form [x_0,x_1],[x_1,x_2],…,[x_{n−1},x_n] is called a partition of [a, b]. If the subintervals all have the same width, the set of points forms a regular partition of the interval [a,b]. 0 1 2 We can use this regular partition as the basis of a method for estimating the area under the curve. We next examine two methods: the leftendpoint approximation and the right-endpoint approximation. Rule: Left-Endpoint Approximation On each subinterval [x , x ] (for i = 1, 2, 3, … , n), construct a rectangle with width Δx and height equal to f (x ), which is the function value at the left endpoint of the subinterval. Then the area of this rectangle is f (x )Δx. Adding the areas of all these rectangles, we get an approximate value for A (Figure 4.3.2). We use the notation Ln to denote that this is a left-endpoint approximation of A using n subintervals. i−1 i i−1 i−1 n A ≈ Ln = f (x0 )Δx + f (x1 )Δx + ⋯ + f (xn − 1)Δx = ∑ f (xi−1 )Δx (4.3.15) i=1 Figure 4.3.2 : In the left-endpoint approximation of area under a curve, the height of each rectangle is determined by the function value at the left of each subinterval. The second method for approximating area under a curve is the right-endpoint approximation. It is almost the same as the left-endpoint approximation, but now the heights of the rectangles are determined by the function values at the right of each subinterval. Rule: Right-Endpoint Approximation Construct a rectangle on each subinterval [x , x ], only this time the height of the rectangle is determined by the function value f(xi) at the right endpoint of the subinterval. Then, the area of each rectangle is f (x )Δx and the approximation for A is given by i−1 i i n A ≈ Rn = f (x1 )Δx + f (x2 )Δx + ⋯ + f (xn )Δx = ∑ f (xi )Δx. (4.3.16) i=1 The notation R indicates this is a right-endpoint approximation for A (Figure 4.3.3). n 10/21/2020 4.3.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10977 Figure 4.3.3 :In the right-endpoint approximation of area under a curve, the height of each rectangle is determined by the function value at the right of each subinterval. Note that the right-endpoint approximation differs from the left-endpoint approximation in Figure 4.3.2. 2 The graphs in Figure 4.3.4 represent the curve x f (x) = 2 . In Figure 4.3.4b we divide the region represented by the interval [0, 3] into six subintervals, each of width 0.5. Thus, Δx = 0.5. We then form six rectangles by drawing vertical lines perpendicular to x , the left endpoint of each subinterval. We determine the height of each rectangle by calculating f (x ) for i = 1, 2, 3, 4, 5, 6. The intervals are [0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2], [2, 2.5], [2.5,. 3] We find the area of each rectangle by multiplying the height by the width. Then, the sum of the rectangular areas approximates the area between f (x) and the x-axis. When the left endpoints are used to calculate height, we have a left-endpoint approximation. Thus, i−1 i−1 6 A ≈ L6 = ∑ i=1 f (xi−1 )Δx = f (x0 )Δx + f (x1 )Δx + f (x2 )Δx + f (x3 )Δx + f (x4 )Δx + f (x5 )Δx = f (0)0.5 + f (0.5)0.5 + f (1)0.5 + f (1.5)0.5 + f (2)0.5 + f (2.5)0.5 = (0)0.5 + (0.125)0.5 + (0.5)0.5 + (1.125)0.5 + (2)0.5 + (3.125)0.5 = 0 + 0.0625 + 0.25 + 0.5625 + 1 + 1.5625 = 3.4375 Figure 4.3.4 : Methods of approximating the area under a curve by using (a) the left endpoints and (b) the right endpoints. In Figure , we draw vertical lines perpendicular to x such that x is the right endpoint of each subinterval, and calculate f (x ) for . We multiply each f (x ) by Δx to find the rectangular areas, and then add them. This is a right-endpoint approximation of the area under f (x). Thus, 4.3.4b i i = 1, 2, 3, 4, 5, 6 6 A ≈ R6 = ∑ i=1 i i i f (xi )Δx = f (x1 )Δx + f (x2 )Δx + f (x3 )Δx + f (x4 )Δx + f (x5 )Δx + f (x6 )Δx = f (0.5)0.5 + f (1)0.5 + f (1.5)0.5 + f (2)0.5 + f (2.5)0.5 + f (3)0.5 = (0.125)0.5 + (0.5)0.5 + (1.125)0.5 + (2)0.5 + (3.125)0.5 + (4.5)0.5 = 0.0625 + 0.25 + 0.5625 + 1 + 1.5625 + 2.25 = 5.6875. Example 4.3.4 : Approximating the Area Under a Curve Use both left-endpoint and right-endpoint approximations to approximate the area under the curve of f (x) = x on the interval [0, 2]; use n = 4 . 2 Solution First, divide the interval [0, 2] into n equal subintervals. Using n = 4, Δx = [0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2] are shown in f (0) = 0, f (0.5) = 0.25, f (1) = 1, f (1.5) = 2.25. Figure Then, . 4.3.5 (2 − 0) Using = 0.5 4 a . This is the width of each rectangle. The intervals left-endpoint approximation, the heights L4 = f (x0 )Δx + f (x1 )Δx + f (x2 )Δx + f (x3 )Δx = 0(0.5) + 0.25(0.5) + 1(0.5) + 2.25(0.5) = 1.75 are (4.3.17) Figure 4.3.5 : The graph shows the left-endpoint approximation of the area under f (x) = x from 0 to 2. 2 10/21/2020 4.3.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10977 The right-endpoint approximation is shown in Figure 4.3.6. The intervals are the same, Δx = 0.5, but now use the right endpoint to calculate the height of the rectangles. We have R4 = f (x1 )Δx + f (x2 )Δx + f (x3 )Δx + f (x4 )Δx = 0.25(0.5) + 1(0.5) + 2.25(0.5) + 4(0.5) = 3.75 (4.3.18) Figure 4.3.6 : The graph shows the right-endpoint approximation of the area under f (x) = x from 0 to 2. 2 The left-endpoint approximation is 1.75; the right-endpoint approximation is 3.75. Exercise 4.3.4 Sketch left-endpoint and right-endpoint approximations for f (x) = 1 on [1, 2]; use n = 4 . Approximate the area using both methods. x Hint Follow the solving strategy in Example step-by-step. Answer The left-endpoint approximation is 0.7595. The right-endpoint approximation is 0.6345. See the below Media. Looking at Figure 4.3.4 and the graphs in Example 4.3.4, we can see that when we use a small number of intervals, neither the left-endpoint approximation nor the right-endpoint approximation is a particularly accurate estimate of the area under the curve. However, it seems logical that if we increase the number of points in our partition, our estimate of A will improve. We will have more rectangles, but each rectangle will be thinner, so we will be able to fit the rectangles to the curve more precisely. We can demonstrate the improved approximation obtained through smaller intervals with an example. Let’s explore the idea of increasing n , first in a left-endpoint approximation with four rectangles, then eight rectangles, and finally 32 rectangles. Then, let’s do the same thing in a right-endpoint approximation, using the same sets of intervals, of the same curved region. Figure 4.3.7 shows the area of the region under the curve f (x) = (x − 1 ) + 4 on the interval [0, 2] using a left-endpoint approximation where n = 4. he width of each rectangle is 3 2 −0 Δx = 1 = 4 . (4.3.19) 2 The area is approximated by the summed areas of the rectangles, or L4 = f (0)(0.5) + f (0.5)(0.5) + f (1)(0.5) + f (1.5)0.5 = 7.5 10/21/2020 4.3.6 CC-BY-NC-SA (4.3.20) https://math.libretexts.org/@go/page/10977 Figure 4.3.7 : With a left-endpoint approximation and dividing the region from a to b into four equal intervals, the area under the curve is approximately equal to the sum of the areas of the rectangles. Figure 4.3.8 shows the same curve divided into eight subintervals. Comparing the graph with four rectangles in Figure with this graph with eight rectangles, we can see there appears to be less white space under the curve when n = 8. This white space is area under the curve we are unable to include using our approximation. The area of the rectangles is L8 = f (0)(0.25) + f (0.25)(0.25) + f (0.5)(0.25) + f (0.75)(0.25) + f (1)(0.25) + f (1.25)(0.25) + f (1.5)(0.25) (4.3.21) + f (1.75)(0.25) = 7.75 Figure 4.3.8 : The region under the curve is divided into n = 8 rectangular areas of equal width for a left-endpoint approximation. The graph in Figure 4.3.9 shows the same function with 32 rectangles inscribed under the curve. There appears to be little white space left. The area occupied by the rectangles is L32 = f (0)(0.0625) + f (0.0625)(0.0625) + f (0.125)(0.0625) + ⋯ + f (1.9375)(0.0625) = 7.9375. (4.3.22) Figure 4.3.9 : Here, 32 rectangles are inscribed under the curve for a left-endpoint approximation. We can carry out a similar process for the right-endpoint approximation method. A right-endpoint approximation of the same curve, using four rectangles (Figure 4.3.10), yields an area R4 = f (0.5)(0.5) + f (1)(0.5) + f (1.5)(0.5) + f (2)(0.5) = 8.5 (4.3.23) Figure 4.3.10 : Now we divide the area under the curve into four equal subintervals for a right-endpoint approximation. 10/21/2020 4.3.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10977 Dividing the region over the interval [0, 2] into eight rectangles results in Δx = 2 −0 = 0.25. 8 The graph is shown in Figure 4.3.11. The area is R8 = f (0.25)(0.25) + f (0.5)(0.25) + f (0.75)(0.25) + f (1)(0.25) + f (1.25)(0.25) + f (1.5)(0.25) + f (1.75)(0.25) (4.3.24) + f (2)(0.25) = 8.25 Figure 4.3.11 : Here we use right-endpoint approximation for a region divided into eight equal subintervals. Last, the right-endpoint approximation with n = 32 is close to the actual area (Figure 4.3.12). The area is approximately R32 = f (0.0625)(0.0625) + f (0.125)(0.0625) + f (0.1875)(0.0625) + ⋯ + f (2)(0.0625) = 8.0625 (4.3.25) Figure 4.3.12 : The region is divided into 32 equal subintervals for a right-endpoint approximation. Based on these figures and calculations, it appears we are on the right track; the rectangles appear to approximate the area under the curve better as n gets larger. Furthermore, as n increases, both the left-endpoint and right-endpoint approximations appear to approach an area of 8 square units. Table shows a numerical comparison of the left- and right-endpoint methods. The idea that the approximations of the area under the curve get better and better as n gets larger and larger is very important, and we now explore this idea in more detail. Table 4.3.15: Converging Values of Left- and Right-Endpoint Approximations as n Increases Value of n Approximate Area L Approximate Area R n = 4 7.5 8.5 n = 8 7.75 8.25 n = 32 7.94 8.06 n n Forming Riemann Sums So far we have been using rectangles to approximate the area under a curve. The heights of these rectangles have been determined by evaluating the function at either the right or left endpoints of the subinterval [x , x ]. In reality, there is no reason to restrict evaluation of the function to one of these two points only. We could evaluate the function at any point ci in the subinterval [x , x ], and use f (x ) as the height of our rectangle. This gives us an estimate for the area of the form i−1 i i−1 i ∗ i n ∗ A ≈ ∑ f (x )Δx. (4.3.26) i i=1 A sum of this form is called a Riemann sum, named for the 19th-century mathematician Bernhard Riemann, who developed the idea. Note Let f (x) be defined on a closed interval [a, b] and let P be a regular partition of [a, b]. Let Δx be the width of each subinterval [x each i, let x be any point in [x , x ]. A Riemann sum is defined for f (x) as i−1 ∗ i i−1 , xi ] and for i n ∗ ∑ f (x )Δx. i (4.3.27) i=1 10/21/2020 4.3.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10977 Recall that with the left- and right-endpoint approximations, the estimates seem to get better and better as n get larger and larger. The same thing happens with Riemann sums. Riemann sums give better approximations for larger values of n. We are now ready to define the area under a curve in terms of Riemann sums. Definition n Let f (x) be a continuous, nonnegative function on an interval [a, b], and let ∑ the curve y = f (x) on [a, b] is given by i=1 ∗ f (x )Δx i be a Riemann sum for f (x). Then, the area under n ∗ A = lim ∑ f (x )Δx. (4.3.28) i n→∞ i=1 See a graphical demonstration of the construction of a Riemann sum. Some subtleties here are worth discussing. First, note that taking the limit of a sum is a little different from taking the limit of a function f (x) as x goes to infinity. Limits of sums are discussed in detail in the chapter on Sequences and Series; however, for now we can assume that the computational techniques we used to compute limits of functions can also be used to calculate limits of sums. Second, we must consider what to do if the expression converges to different limits for different choices of x . Fortunately, this does not happen. Although the proof is beyond the scope of this text, it can be shown that if f (x) is continuous on the closed interval [a, b], then li m ∑ f (x )Δx exists and is unique (in other words, it does not depend on the choice of x ). ∗ i n→∞ n ∗ ∗ i=1 i i We look at some examples shortly. But, before we do, let’s take a moment and talk about some specific choices for x . Although any choice for x gives us an estimate of the area under the curve, we don’t necessarily know whether that estimate is too high (overestimate) or too low (underestimate). If it is important to know whether our estimate is high or low, we can select our value for x to guarantee one result or the other. ∗ ∗ i i ∗ i If we want an overestimate, for example, we can choose x such that for i = 1, 2, 3, … , n, f (x ) ≥ f (x) for all x ∈ [x − 1, x ] . In other words, we choose x so that for i = 1, 2, 3, … , n, f (x ) is the maximum function value on the interval [x , x ]. If we select x in this way, then the Riemann sum ∑ f (x )Δx is called an upper sum. Similarly, if we want an underestimate, we can choose {x∗i} so that for i=1,2,3,…,n,f(x∗i) is the minimum function value on the interval [x , x ]. In this case, the associated Riemann sum is called a lower sum. Note that if f(x) is either increasing or decreasing throughout the interval [a, b], then the maximum and minimum values of the function occur at the endpoints of the subintervals, so the upper and lower sums are just the same as the left- and right-endpoint approximations. ∗ ∗ i i n ∗ i=1 i i−1 ∗ ∗ i i i i ∗ i−1 i i i Example 4.3.5 : Finding Lower and Upper Sums Find a lower sum for f (x) = 10 − x on [1, 2]; let n = 4 subintervals. 2 Solution With n =4 over the interval 1 [1, 2], Δx = 4 . We can list the intervals as . Because the function is [1, 1.25], [1.25, 1.5], [1.5, 1.75], [1.75, 2] decreasing over the interval [1, 2], Figure shows that a lower sum is obtained by using the right endpoints. Figure 4.3.13 : The graph of f (x) = 10 − x is set up for a right-endpoint approximation of the area bounded by the curve and the x-axis on [1, 2], and it shows a lower sum. 2 The Riemann sum is 4 2 2 ∑k=1 (10 − x )(0.25) = 0.25[10 − (1.25 ) 2 + 10 − (1.5 ) 2 + 10 − (1.75 ) 2 + 10 − (2 ) ] = 0.25[8.4375 + 7.75 + 6.9375 + 6] = 7.28. The area of 7.28 is a lower sum and an underestimate. Exercise 4.3.5 a. Find an upper sum for f (x) = 10 − x on [1, 2]; let n = 4. b. Sketch the approximation. 2 10/21/2020 4.3.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10977 Hint f (x) is decreasing on [1, 2], so the maximum function values occur at the left endpoints of the subintervals. Answer a. Upper sum=8.0313. b. Example 4.3.6 : Finding Lower and Upper Sums for f (x) = sin x Find a lower sum for f (x) = sin x over the interval [a, b] = [0, π ; let n = 6. ] 2 Solution Let’s first look at the graph in Figure to get a better idea of the area of interest. Figure 4.3.14 : The graph of y = sin x is divided into six regions: Δx = The intervals are π [0, π ], [ 12 π , 12 π ], [ 6 π , 6 π ], [ π , 4 4 π ], [ 3 5π , 3 12 ,] and 5π [ π , 12 . Note that ] 2 f (x) = sin x π/2 endpoint approximation gives us the lower sum. A left-endpoint approximation is the Riemann sum ∑ i=0 π π ) + sin( 12 π )( 12 π ) + sin( 12 π )( 6 π ) + sin( 12 π )( 4 π ) + sin( 12 sin xi ( ) π [0, , so a left- ] 2 .We have 12 5π ) + sin( 12 . π π )( 3 12 is increasing on the interval 5 A ≈ sin(0)( π = 6 π )( 12 ) = 0.863. (4.3.29) 12 Exercise 4.3.6 Using the function f (x) = sin x over the interval [0, π ], 2 find an upper sum; let n = 6. Hint Follow the steps from Example. Answer A ≈ 1.125 Key Concepts The use of sigma (summation) notation of the form ∑ a is useful for expressing long sums of values in compact form. For a continuous function defined over an interval [a, b], the process of dividing the interval into n equal parts, extending a rectangle to the graph of the function, calculating the areas of the series of rectangles, and then summing the areas yields an approximation of the area of that region. n i=1 The width of each rectangle is Δx = 10/21/2020 b −a n i . 4.3.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/10977 Riemann sums are expressions of the form ∑ f (x )Δx, and can be used to estimate the area under the curve y = f (x). Left- and right-endpoint approximations are special kinds of Riemann sums where the values of x are chosen to be the left or right endpoints of the subintervals, respectively. Riemann sums allow for much flexibility in choosing the set of points x at which the function is evaluated, often with an eye to obtaining a lower sum or an upper sum. n ∗ i=1 i ∗ i ∗ i Key Equations Properties of Sigma Notation n ∑ i=1 c = nc n n ∑i=1 c ai = c ∑i=1 ai n ∑ i=1 n ∑ i=1 n ∑ i=1 n (ai + bi ) = ∑ i=1 n (ai − bi ) = ∑ i=1 m n ai + ∑ i=1 n ai − ∑ i=1 n ai = ∑ ai + ∑ i=1 i=m+1 bi bi ai Sums and Powers of Integers n ∑ i=1 n ∑ i=1 n(n + 1) i = 1 +2 +⋯ +n = 2 2 i 2 =1 2 +2 2 +⋯ +n n(n + 1)(2n + 1) = 6 2 n ∑ i=0 3 i 3 =1 3 +2 3 +⋯ +n 2 n (n + 1 ) = 4 Left-Endpoint Approximation n A ≈ Ln = f (x0 )Δx + f (x1 )Δx + ⋯ + f (xn−1 )Δx = ∑i=1 f (xi−1 )Δx Right-Endpoint Approximation n A ≈ Rn = f (x1 )Δx + f (x2 )Δx + ⋯ + f (xn )Δx = ∑ i=1 f (xi )Δx Glossary left-endpoint approximation an approximation of the area under a curve computed by using the left endpoint of each subinterval to calculate the height of the vertical sides of each rectangle lower sum a sum obtained by using the minimum value of $f(x)$ on each subinterval partition a set of points that divides an interval into subintervals regular partition a partition in which the subintervals all have the same width riemann sum an estimate of the area under the curve of the form A ≈ ∑ n i=1 ∗ f (x )Δx i right-endpoint approximation the right-endpoint approximation is an approximation of the area of the rectangles under a curve using the right endpoint of each subinterval to construct the vertical sides of each rectangle sigma notation (also, summation notation) the Greek letter sigma (Σ ) indicates addition of the values; the values of the index above and below the sigma indicate where to begin the summation and where to end it upper sum a sum obtained by using the maximum value of $f(x)$ on each subinterval Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 4.3.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/10977 4.3E: Exercises Exercise 4.3E . 1 1. A fundamental calculus technique is to use ________ to refine approximations to get an exact answer. 2. What is the upper bound in the summation ∑ 14 i=7 ? (48i − 201) 3. This section approximates definite integrals using what geometric shape? 4. T/F: A sum using the Right Hand Rule is an example of a Riemann Sum. Answer Under Construction Exercise 4.3E . 2 Write out each term of the summation and compute the sum. 1. ∑ 4 2 i i=2 2. ∑ 3 i=−1 (4i − 2) 3. ∑ 2 i=−2 4. ∑ 5 1 i=1 i 6 5. ∑ sin(πi/2) i i=1 (−1 ) i 6. ∑ 4 ( i=1 7. ∑ 5 1 i − 1 i+1 ) i i=0 (−1 ) cos(πi) Answer Under Construction Exercise 4.3E . 3 Write each sum in summation notation. 1. 3 + 6 + 9 + 12 + 15 2. −1 + 0 + 3 + 8 + 15 + 24 + 35 + 48 + 63 3. 1 2 + 2 3 + 4. 1 − e + e 3 + 4 2 4 5 −e 3 +e 4 Answer Under Construction Exercise 4.3E . 4 Evaluate the summation using Theorem 37. 1. ∑ 2 i=1 1 2. ∑ i=1 10/21/2020 5i 2 0(3 i − 2i) 4.3E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/13732 3. ∑ 5(2 i 4. ∑ 0(−4 i 1 3 i=1 1 − 10) 3 i=1 1 5. ∑ 3 2 0(i i=1 2 + 10 i − 3t − 7i + 11) + 2i + 7) 6. 1 + 2 + 3+. . . +99 + 100 7. 1 + 4 + 9+. . . +361 + 400 Answer Under Construction Exercise 4.3E . 5 Theorem 37 states: n ∑ ai = ∑ +i = 1 i=1 n n ∑ n ai + ∑ i=k+1 k ai = ∑ i=k+1 k ai − ∑ i=1 i=1 ai ai , so . Use this fact, along with other parts of Theorem 37, to evaluate the summations given 2 1. ∑ 0i 2. ∑ 5i i=11 2 3 i=16 3. ∑ 1 24 i=7 1 4. ∑ 3 04 i i=5 Answer Under Construction Exercise 4.3E . 6 b A definite integral ∫ f (x) dx is given. (a) Graph f (x) on [a,b]. (b) Add to the sketch rectangles using the provided rule. (c) Approximate ∫ f (x) dx by summing the areas of the rectangles. a b a 1. ∫ 3 2. ∫ 2 3. ∫ π 4. ∫ 3 5. ∫ 2 6. ∫ 9 2 −3 0 0 0 1 1 x dx , with 6 rectangles using the Left Hand Rule. 2 (5 − x ) dx sin x dx x 2 dx , with 4 rectangles using the Midpoint Rule. , with 6 rectangles using the Right Hand Rule. , with 5 rectangles using the Left Hand Rule. , with 3 rectangles using the Midpoint Rule. ln x dx 1 x dx , with 4 rectangles using the Right Hand Rule. Answer Under Construction Exercise 4.3E . 7 A definite integral ∫ b a f (x) dx is given. (a) Find a formula to approximate ∫ b a 10/21/2020 f (x) dx using n subintervals and the provided rule. 4.3E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/13732 (b) Evaluate the formula using n = 10, 100 and 1000. (c) Find the limit of the formula, as n → ∞ , to find the exact value of ∫ b a 1. ∫ 1 2. ∫ 1 3. ∫ 3 4. ∫ 4 5. ∫ 10 6. ∫ 1 3 x 0 2 −1 −1 1 3x , using the Right Hand Rule. , using the Left Hand Rule. dx (3x − 1) dx 2 (2 x −10 0 dx − 3) dx 3 , using the Midpoint Rule. , using the Left Hand Rule. , using the Right Hand Rule. (5 − x) dx (x . f (x) dx 2 − x ) dx , using the Right Hand Rule. Answer Under Construction Exercise 4.3E . 8 Find an antiderivative of the given function. 1. f (x) = 5 sec 2. f (x) = 2 x 7 x 3. g(t) = 4t 5 3 − 5t +8 4. g(t) = 5 ⋅ 8 t 5. g(t) = cos t + sin t 6. f (x) = 1 √x Answer Under Construction Exercise 4.3E . 9 State whether the given sums are equal or unequal. 10 10 1. ∑ i and ∑ k i=1 k=1 Answer They are equal; both represent the sum of the first 10 whole numbers 10 15 2. ∑ i and ∑(i − 5) i=1 i=6 Answer They are equal; both represent the sum of the first 10 whole numbers 10 9 3. ∑ i(i − 1) and ∑(j + 1)j i=1 j=0 Answer 10/21/2020 4.3E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/13732 They are equal by substituting j = i − 1. 10 10 4. ∑ i(i − 1) and ∑(k 2 i=1 − k) k=1 Answer They are equal; the first sum factors the terms of the second. Exercise 4.3E . 10 In the following exercises, use the rules for sums of powers of integers to compute the sums. 10 1. ∑ i i=5 10 2. ∑ i 2 i=5 Answer 385 − 30 = 355 Exercise 4.3E . 11 100 100 Suppose that ∑ a i = 15 i=1 and ∑ b i = −12. i=1 In the following exercises, compute the sums. 100 1. ∑(a i + bi ) i=1 100 2. ∑(a i − bi ) i=1 Answer 15 − (−12) = 27 100 3. ∑(3a i − 4 bi ) i=1 100 4. ∑(5a i + 4 bi ) i=1 Answer 5(15) + 4(−12) = 27 Exercise 4.3E . 12 In the following exercises, use summation properties and formulas to rewrite and evaluate the sums. 20 1. ∑ 100(k 2 − 5k + 1) k=1 10/21/2020 4.3E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/13732 50 2. ∑(j 2 − 2j) j=1 Answer 50 ∑j j=1 20 3. ∑ (j 2 − 10j) j=11 25 4. ∑[(2k) 2 − 100k] k=1 Answer 25 25 2 4∑k 4(25)(26)(51) − 100 ∑ k = k=1 − 50(25)(26) = −10, 400 9 k=1 Exercise 4.3E . 13 Let L denote the left-endpoint sum using n subintervals and let R denote the corresponding right-endpoint sum. n n In the following exercises, compute the indicated left and right sums for the given functions on the indicated interval. 1. L for f (x) = 4 1 x −1 on [2, 3] 2. R for g(x) = cos(πx) on [0, 1] 4 Answer R4 = 0.25 1 3. L for f (x) = 6 on [2, 5] x(x − 1) 4. R for f (x) = 6 1 x(x − 1) on [2, 5] Answer R6 = 0.372 1 5. R for 4 2 x 6. L for 4 on [−2, 2] +1 1 x2 + 1 on [−2, 2] Answer L4 = 2.20 7. R for x − 2x + 1 on [0, 2] 8. L for x − 2x + 1 on [0, 2] 2 4 2 8 Answer L8 = 0.6875 10/21/2020 4.3E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/13732 Exercise 4.3E . 14 1. Compute the left and right Riemann sums— L and R , respectively—for f (x) = (2 − |x|) on average value and compare it with the area under the graph of f. 4 4 2. Compute the left and right Riemann sums— L and R , respectively—for their average value and compare it with the area under the graph of f. 6 [−2, 2]. f (x) = (3 − |3 − x|) 6 on Compute their [0, 6]. Compute Answer L6 = 9.000 = R6 . The graph of f is a triangle with area 9. 3. Compute the left and right Riemann sums— their values. L4 4. Compute the left and right Riemann sums— and L6 R4 and , respectively—for R6 − −−− − 2 f (x) = √4 − x , respectively—for on [−2, 2] − − − − − − − − − − 2 f (x) = √9 − (x − 3) and compare on [0, 6] and compare their values. Answer . They are equal. L6 = 13.12899 = R6 Exercise 4.3E . 15 Express the following endpoint sums in sigma notation but do not evaluate them. 1. L for f (x) = x on [1, 2] 2. L for f (x) = √4 − x on [−2, 2] 2 30 − −−− − 2 10 Answer 10 4 10 20 (i − 1) ∑ √4 − (−2 + 4 L10 = 3. R − −−−−−−−−−−−−−− − ) 10 i=1 for f (x) = sinx on [0, π] 4. R for lnx on [1, e] 100 Answer e−1 R100 = 100 i ∑ ln(1 + (e − 1) 100 ) 100 i=1 Exercise 4.3E . 16 In the following exercises, graph the function then use a calculator or a computer program to evaluate the following left and right endpoint sums. Is the area under the curve between the left and right endpoint sums? 1. L and R for y = x 2. L and R for y = x on the interval [0, 1] 100 100 100 100 2 − 3x + 1 on the interval [−1, 1] 2 Answer 10/21/2020 4.3E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/13732 The plot shows that the left Riemann sum is an underestimate because the function is increasing. Similarly, the right Riemann sum is an overestimate. The area lies between the left and right Riemann sums. Ten rectangles are shown for visual clarity. This behavior persists for more rectangles. R100 = 0.33835, L100 = 0.32835. 3. L 50 and R 50 for y = x +1 2 x 4. L 100 and R on the interval [2, 4] −1 for y = x on the interval [−1, 1] 3 100 Answer . The left endpoint sum is an underestimate because the function is increasing. Similarly, a right endpoint approximation is an overestimate. The area lies between the left and right endpoint estimates. L100 = −0.02, R100 = 0.02 5. L 50 6. L and R 100 50 and R for y = tan(x) on the interval [0, 100 for y = e 2x π ] 4 on the interval [−1, 1] Answer . The plot shows that the left Riemann sum is an underestimate because the function is increasing. Ten rectangles are shown for visual clarity. This behavior persists for more rectangles. L100 = 3.555, R100 = 3.670 10/21/2020 4.3E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/13732 Exercise 4.3E . 17 1. Let t denote the time that it took Tejay van Garteren to ride the jth stage of the Tour de France in 2014. If there were a j 21 total of 21 stages, interpret ∑ t . j j=1 31 2. Let r denote the total rainfall in Portland on the jth day of the year in 2009. Interpret ∑ r . j j j=1 Answer The sum represents the cumulative rainfall in January 2009. 3. Let d denote the hours of daylight and δ denote the increase in the hours of daylight from day j − 1 to day j in Fargo, j j 365 North Dakota, on thejth day of the year. Interpret d1 + ∑ δ . j j=2 4. To help get in shape, Joe gets a new pair of running shoes. If Joe runs 1 mi each day in week 1 and adds daily routine each week, what is the total mileage on Joe’s shoes after 25 weeks? 1 10 mi to his Answer 25 The total mileage is 7 × ∑(1 + (i − 1) i=1 7 ) = 7 × 25 + 10 × 12 × 25 = 385mi 10 . Exercise 4.3E . 18 1. The following table gives approximate values of the average annual atmospheric rate of increase in carbon dioxide (CO2) each decade since 1960, in parts per million (ppm). Estimate the total increase in atmospheric CO2 between 1964 and 2013. Decade Ppm/y 1964-1973 1.07 1976-1983 1.34 1984-1993 1.40 1994-2003 1.87 2004-2013 2.07 Average Annual Atmospheric CO2 Increase, 1964–2013 Source: http://www.esrl.noaa.gov/gmd/ccgg/trends/. 2. The following table gives the approximate increase in sea level in inches over 20 years starting in the given year. Estimate the net change in mean sea level from 1870 to 2010. 10/21/2020 Starting Year 20- Year Change 1870 0.3 1890 1.5 4.3E.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/13732 Starting Year 20- Year Change 1910 0.2 1930 2.8 1950 0.7 1970 1.1 1990 1.5 Approximate 20-Year Sea Level Increases, 1870–1990 Source: http://link.springer.com/article/10....712-011-9119-1 Answer Add the numbers to get 8.1-in. net increase. 3. The following table gives the approximate increase in dollars in the average price of a gallon of gas per decade since 1950. If the average price of a gallon of gas in 2010 was $2.60, what was the average price of a gallon of gas in 1950? Starting Year 10- Year Change 1950 0.03 1960 0.05 1970 0.86 1980 −0.03 1990 0.29 2000 1.12 Approximate 10-Year Gas Price Increases, 1950–2000 Source: http://epb.lbl.gov/homepages/Rick_Di...011-trends.pdf. 4. The following table gives the percent growth of the U.S. population beginning in July of the year indicated. If the U.S. population was 281,421,906 in July 2000, estimate the U.S. population in July 2010. 10/21/2020 Year % Change/Year 2000 1.12 2001 0.99 2002 0.93 2003 0.86 4.3E.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/13732 Year % Change/Year 2004 0.93 2005 0.93 2006 0.97 2007 0.96 2008 0.95 2009 0.88 Annual Percentage Growth of U.S. Population, 2000–2009 Source: http://www.census.gov/popest/data. (Hint: To obtain the population in July 2001, multiply the population in July 2000 by 1.0112 to get 284,573,831.) Answer 309,389,957 Exercise 4.3E . 19 In the following exercises, estimate the areas under the curves by computing the left Riemann sums, L 8. 1. 2. Answer: L8 = 3 + 2 + 1 + 2 + 3 + 4 + 5 + 4 = 24 10/21/2020 4.3E.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/13732 3. 4. Answer L8 = 3 + 5 + 7 + 6 + 8 + 6 + 5 + 4 = 44 Exercise 4.3E . 20 Use a computer algebra system to compute the Riemann sum: − −−− − 1. L , for N = 10, 30, 50 for f (x) = √1 − x on [−1, 1]. 2. L , for N = 10, 30, 50 for f (x) = N N 2 1 − −−− − √1 + x2 on [−1, 1]. Answer L10 ≈ 1.7604, L30 ≈ 1.7625, L50 ≈ 1.76265 3. L , for N N = 10, 30, 50 for f (x) = si n 2 x on [0, 2π]. Compare these estimates with π. Exercise 4.3E . 21 In the following exercises, use a calculator or a computer program to evaluate the endpoint sums N = 1, 10, 100. RN and LN for How do these estimates compare with the exact answers, which you can find via geometry? 1. y = cos(πx) on the interval [0, 1] 10/21/2020 4.3E.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/13732 Answer and R R1 = −1, L1 = 1, R10 = −0.1, L10 = 0.1, L100 = 0.01, 100 = −0.1. By symmetry of the graph, the exact area is zero. 2. y = 3x + 2 on the interval [3, 5] Exercise 4.3E . 22 In the following exercises, use a calculator or a computer program to evaluate the endpoint sums RN and LN for N = 1, 10, 100. 1. y = x 4 2 − 5x +4 32 on the interval [−2, 2], which has an exact area of 15 Answer R1 = 0, L1 = 0, R10 = 2.4499, L10 = 2.4499, R100 = 2.1365, L100 = 2.1365 2. y = lnx on the interval [1, 2], which has an exact area of \(\displaystyle 2ln(2)−1\ Exercise 4.3E . 23 1. Explain why, if f (a) ≥ 0 and f is increasing on [a, b], that the left endpoint estimate is a lower bound for the area below the graph of f on [a, b]. Answer If [c, d]is a subinterval of [a, b] under one of the left-endpoint sum rectangles, then the area of the rectangle contributing to the left-endpoint estimate is f (c)(d − c) . But, f (c) ≤ f (x) for c ≤ x ≤ d , so the area under the graph of f between c and d is f (c)(d − c) plus the area below the graph of f but above the horizontal line segment at height f (c), which is positive. As this is true for each left-endpoint sum interval, it follows that the left Riemann sum is less than or equal to the area below the graph of f on [a, b]. 2. Explain why, if f (b) ≥ 0 and f is decreasing on below the graph of f on [a, b]. 3. Show that, in general, R N [a, b], that the left endpoint estimate is an upper bound for the area f (b) − f (a) − LN = (b − a) × . N Answer b −a LN = N i −1 ∑ f (a + (b − a) N b −a ) = N i=1 N −1 i ∑ f (a + (b − a) N and R ) N b −a = N i=0 N i ∑ f (a + (b − a) N i=1 ) . N The left sum has a term corresponding to i = 0 and the right sum has a term corresponding to i = N . In R so N , any term corresponding to i = 1, 2, … , N − 1 occurs once with a plus sign and once with a minus sign, each such term cancels and one is left with − LN b −a RN − LN = N (f (a + (b − a)) N 0 ) − (f (a) + (b − a) N b −a ) = N 4. Explain why, if f is increasing on [ a, b], the error between either most \(\displaystyle (b−a)\frac{f(b)−f(a)}{N]|). (f (b) − f (a)). N LN or RN and the area A below the graph of f is at Exercise 4.3E . 24 For each of the three graphs: 10/21/2020 4.3E.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/13732 a. Obtain a lower bound completely by the curve. L(A) for the area enclosed by the curve by adding the areas of the squares enclosed b. Obtain an upper bound U (A) for the area by adding to L(A) the areas B(A) of the squares enclosed partially by the curve. Answer Graph 1: a. L(A) = 0, B(A) = 20 b. U (A) = 20. Graph 2: a. L(A) = 9 10/21/2020 4.3E.13 CC-BY-NC-SA https://math.libretexts.org/@go/page/13732 b. B(A) = 11, U (A) = 20. Graph 3: a. L(A) = 11.0 b. B(A) = 4.5, U (A) = 15.5. In the previous exercise, explain why four boxes of equal area. L(A) gets no smaller while U (A) gets no larger as the squares are subdivided into Answer Under Construction Exercise 4.3E . 25 A unit circle is made up of n wedges equivalent to the inner wedge in the figure. The base of the inner triangle is 1 unit and its height is sin( π ). n The base of the outer triangle is B = cos( π π ) + sin( n π )tan( n Use this information to argue that the area of a unit circle is equal to π.\) and the height is H = Bsin( ) n 2π ) n . Answer Let A be the area of the unit circle. The circle encloses n congruent triangles each of area n 2π sin( 2 Similarly, ) ≤ A. n BH 1 = 2 π (cos( 2 ) + sin( n n n → ∞, )tan( πsin( ) = n ( circle π n 2π sin( 2 the π 2π n ))sin( n 2π n is 2π ) → π ) n contained , so , so we conclude inside n A ≤ sin( 2 π ≤A n congruent 2π )(cos( n . Also, as triangles π )) + sin( n π )tan( n π n → ∞, cos( ) ) n of . π ) + sin( n n 2 each π 2π sin( , so area As π )tan( n ) ) → 1 n , so we also have A ≤ π . By the squeeze theorem for limits, we conclude that A = π. 10/21/2020 4.3E.14 CC-BY-NC-SA https://math.libretexts.org/@go/page/13732 4.4: The Definite Integral In the preceding section we defined the area under a curve in terms of Riemann sums: n ∗ A = lim ∑ f (x )Δx. (4.4.1) i n→∞ i=1 However, this definition came with restrictions. We required f (x) to be continuous and nonnegative. Unfortunately, real-world problems don’t always meet these restrictions. In this section, we look at how to apply the concept of the area under the curve to a broader set of functions through the use of the definite integral. Definition and Notation The definite integral generalizes the concept of the area under a curve. We lift the requirements that nonnegative, and define the definite integral as follows. Definition f (x) be continuous and Definition: definite integral If f (x) is a function defined on an interval [a, b], the definite integral of f from a to b is given by n b ∫ ∗ f (x)dx = lim ∑ f (x )Δx, n→∞ a (4.4.2) i i=1 provided the limit exists. If this limit exists, the function f (x) is said to be integrable on [a,b], or is an integrable function. The integral symbol in the previous definition should look familiar. We have seen similar notation in the chapter on Applications of Derivatives, where we used the indefinite integral symbol (without the a and b above and below) to represent an antiderivative. Although the notation for indefinite integrals may look similar to the notation for a definite integral, they are not the same. A definite integral is a number. An indefinite integral is a family of functions. Later in this chapter we examine how these concepts are related. However, close attention should always be paid to notation so we know whether we’re working with a definite integral or an indefinite integral. Integral notation goes back to the late seventeenth century and is one of the contributions of Gottfried Wilhelm Leibniz, who is often considered to be the codiscoverer of calculus, along with Isaac Newton. The integration symbol ∫ is an elongated S, suggesting sigma or summation. On a definite integral, above and below the summation symbol are the boundaries of the interval, [a, b]. The numbers a and b are x-values and are called the limits of integration; specifically, a is the lower limit and b is the upper limit. To clarify, we are using the word limit in two different ways in the context of the definite integral. First, we talk about the limit of a sum as n → ∞. Second, the boundaries of the region are called the limits of integration. We call the function f (x) the integrand, and the dx indicates that f (x) is a function with respect to x, called the variable of integration. Note that, like the index in a sum, the variable of integration is a dummy variable, and has no impact on the computation of the integral. We could use any variable we like as the variable of integration: b ∫ b f (x)dx = ∫ a b f (t)dt = ∫ a f (u)du (4.4.3) a Previously, we discussed the fact that if f (x) is continuous on [a, b], then the limit unique. This leads to the following theorem, which we state without proof. n limn→∞ ∑ i=1 ∗ f (x )Δx i exists and is Continuous Functions Are Integrable If f (x) is continuous on [a, b], then f is integrable on [a, b]. Functions that are not continuous on [a, b] may still be integrable, depending on the nature of the discontinuities. For example, functions with a finite number of jump discontinuities on a closed interval are integrable. It is also worth noting here that we have retained the use of a regular partition in the Riemann sums. This restriction is not strictly necessary. Any partition can be used to form a Riemann sum. However, if a nonregular partition is used to define the 10/21/2020 4.4.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/13794 definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must take the limit as the width of the largest subinterval goes to zero. This introduces a little more complex notation in our limits and makes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions for the Riemann sums. Example 4.4.1 : Evaluating an Integral Using the Definition 2 Use the definition of the definite integral to evaluate ∫ sum. Use a right-endpoint approximation to generate the Riemann 2 x dx. 0 Solution We first want to set up a Riemann sum. Based on the limits of integration, we have a = 0 and b = 2 . For i = 0, 1, 2, … , n, let P = x be a regular partition of [0, 2]. Then i b −a 2 Δx = = . n (4.4.4) n Since we are using a right-endpoint approximation to generate Riemann sums, for each i, we need to calculate the function value at the right endpoint of the interval [x , x ]. The right endpoint of the interval is x , and since P is a regular partition, i−1 i i 2 xi = x0 + iΔx = 0 + i[ 2i ] = . n (4.4.5) n Thus, the function value at the right endpoint of the interval is 2 2i 2 f (xi ) = x =( i 2 ) 4i = n . n2 (4.4.6) Then the Riemann sum takes the form 2 n ∑ i=1 i=1 Using the summation formula for ∑ n 2 i i=1 i=1 f (xi )Δx = 3 ( 2 2 2 =∑ i=1 n n 8i n ) 3 8 = n 3 n ∑ i=1 n 2 i . , we have 8 n ∑ 4i n f (xi )Δx = ∑ n ∑ 2 i=1 n i n(n + 1)(2n + 1) 8 = 3 [ ] n 6 3 8 = 2n 3 2 + 3n +n [ ] n 6 3 16 n 2 + 24 n = +n 3 6n 8 = 4 + 1 + 3 n 2 . 6n Now, to calculate the definite integral, we need to take the limit as n → ∞ . We get 2 ∫ 0 2 8 n x dx = limn→∞ ∑i=1 f (xi )Δx = limn→∞ ( 4 + 3 n 8 = limn→∞ ( 3 1 + 6n2 ) 4 ) + limn→∞ ( n 1 ) + limn→∞ ( 8 2 6n ) = 8 +0 +0 = 3 3 . Exercise 4.4.1 Use the definition of the definite integral to evaluate ∫ 3 0 (2x − 1)dx . Use a right-endpoint approximation to generate the Riemann sum. Hint 10/21/2020 4.4.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/13794 Use the solving strategy from Example. Answer 6 Evaluating Definite Integrals Evaluating definite integrals this way can be quite tedious because of the complexity of the calculations. Later in this chapter we develop techniques for evaluating definite integrals without taking limits of Riemann sums. However, for now, we can rely on the fact that definite integrals represent the area under the curve, and we can evaluate definite integrals by using geometric formulas to calculate that area. We do this to confirm that definite integrals do, indeed, represent areas, so we can then discuss what to do in the case of a curve of a function dropping below the x-axis. Example 4.4.2 : Using Geometric Formulas to Calculate Definite Integrals Use the formula for the area of a circle to evaluate ∫ 6 3 − − − − − − − − − − 2 √9 − (x − 3) dx . Solution The function describes a semicircle with radius 3. To find 6 ∫ − − − − − − − − − − 2 √ 9 − (x − 3) dx (4.4.7) 3 we want to find the area under the curve over the interval [3, 6]. The formula for the area of a circle is A = πr . The area 2 1 of a semicircle is just one-half the area of a circle, or semicircle, or A = ( 1 2 )π r A =( 2 )π r 2 . The shaded area in Figure covers one-half of the . Thus, 4 6 ∫ − − − − − − − − − − 2 √ 9 − (x − 3) 1 = 9 2 π(3 ) = 4 3 π ≈ 7.069. (4.4.8) 4 Figure 4.4.1 : The value of the integral of the function f (x) over the interval [3, 6] is the area of the shaded region. Exercise 4.4.2 Use the formula for the area of a trapezoid to evaluate ∫ 4 2 (2x + 3)dx . Hint Graph the function f (x) and calculate the area under the function on the interval [2, 4]. Answer 18 square units Area and the Definite Integral When we defined the definite integral, we lifted the requirement that f (x) be nonnegative. But how do we interpret “the area under the curve” when f (x) is negative? 10/21/2020 4.4.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/13794 Net Signed Area Let us return to the Riemann sum. Consider, for example, the function f (x) = 2 − 2x (shown in Figure) on the interval [0, 2]. Use n = 8 and choose {x } as the left endpoint of each interval. Construct a rectangle on each subinterval of height f (x ) and width Δx. When f (x ) is positive, the product f (x )Δx represents the area of the rectangle, as before. When f (x ) is negative, however, the product f (x )Δx represents the negative of the area of the rectangle. The Riemann sum then becomes 2 ∗ i ∗ ∗ i i ∗ i ∗ ∗ i i 8 ∗ ∑ f (x )Δx = (4.4.9) i i=1 (Area of rectangles above thex-axis)−(Area of rectangles below thex-axis) Figure 4.4.2 : For a function that is partly negative, the Riemann sum is the area of the rectangles above the x-axis less the area of the rectangles below the x-axis. Taking the limit as n → ∞, the Riemann sum approaches the area between the curve above the x-axis and the x-axis, less the area between the curve below the x-axis and the x-axis, as shown in Figure. Then, n 2 ∫ f (x)dx = lim ∑ f (ci )Δx = A1 − A2 . 0 The quantity A 1 − A2 (4.4.10) n→∞ i=1 is called the net signed area. Figure 4.4.3 : In the limit, the definite integral equals area A1 less area A2, or the net signed area. Notice that net signed area can be positive, negative, or zero. If the area above the x-axis is larger, the net signed area is positive. If the area below the x-axis is larger, the net signed area is negative. If the areas above and below the x-axis are equal, the net signed area is zero. 10/21/2020 4.4.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/13794 Example 4.4.3 : Finding the Net Signed Area Find the net signed area between the curve of the function f (x) = 2x and the x-axis over the interval [−3, 3]. Solution The function produces a straight line that forms two triangles: one from x =3 x = −3 (Figure). Using the geometric formula for the area of a triangle, A = 1 A1 = 3(6) = 9 2 1 bh 2 to and the other from x =0 x =0 to , the area of triangle A1, above the axis, is , where 3 is the base and 2(3) = 6 is the height. The area of triangle A2, below the axis, is 1 A2 = (3)(6) = 9, 2 where 3 is the base and 6 is the height. Thus, the net area is 3 ∫ −3 2xdx = A1 − A2 = 9 − 9 = 0. Figure 4.4.4 : The area above the curve and below the x-axis equals the area below the curve and above the x-axis. Analysis If A1 is the area above the x-axis and A2 is the area below the x-axis, then the net area is A two triangles are equal, the net area is zero. 1 − A2 . Since the areas of the Exercise 4.4.3 Find the net signed area of f (x) = x − 2 over the interval [0, 6], illustrated in the following image. Hint Use the solving method described in Example. Answer 6 Total Area 10/21/2020 4.4.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/13794 One application of the definite integral is finding displacement when given a velocity function. If v(t) represents the velocity of an object as a function of time, then the area under the curve tells us how far the object is from its original position. This is a very important application of the definite integral, and we examine it in more detail later in the chapter. For now, we’re just going to look at some basics to get a feel for how this works by studying constant velocities. When velocity is a constant, the area under the curve is just velocity times time. This idea is already very familiar. If a car travels away from its starting position in a straight line at a speed of 75 mph for 2 hours, then it is 150 mi away from its original position (Figure). Using integral notation, we have ∫ 2 0 75dt = 150 . Figure 4.4.5 : The area under the curve v(t) = 75 tells us how far the car is from its starting point at a given time. In the context of displacement, net signed area allows us to take direction into account. If a car travels straight north at a speed of 60 mph for 2 hours, it is 120 mi north of its starting position. If the car then turns around and travels south at a speed of 40 mph for 3 hours, it will be back at it starting position (Figure). Again, using integral notation, we have 2 ∫ 0 5 60 dt + ∫ −40 dt = 120 − 120 = 0. (4.4.11) 2 In this case the displacement is zero. Figure 4.4.6 : The area above the axis and the area below the axis are equal, so the net signed area is zero. 10/21/2020 4.4.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/13794 Suppose we want to know how far the car travels overall, regardless of direction. In this case, we want to know the area between the curve and the x-axis, regardless of whether that area is above or below the axis. This is called the total area. Graphically, it is easiest to think of calculating total area by adding the areas above the axis and the areas below the axis (rather than subtracting the areas below the axis, as we did with net signed area). To accomplish this mathematically, we use the absolute value function. Thus, the total distance traveled by the car is 2 ∫ 5 |60|dt + ∫ 0 2 | − 40|dt = ∫ 2 5 60dt + ∫ 0 40dt = 120 + 120 = 240. (4.4.12) 2 Bringing these ideas together formally, we state the following definitions. Definition: net signed area Let f (x) be an integrable function defined on an interval [a, b]. Let A represent the area between f (x) and the x-axis that lies above the axis and let A represent the area between f (x) and the x-axis that lies below the axis. Then, the net signed area between f (x) and the x-axis is given by 1 2 b ∫ f (x)dx = A1 − A2 . (4.4.13) a The total area between f (x) and the x-axis is given by b ∫ |f (x)|dx = A1 + A2 . (4.4.14) a Example 4.4.4 : Finding the Total Area Find the total area between f (x) = x − 2 and the x-axis over the interval [0, 6]. Solution Calculate the x-intercept as (2, 0) (set y = 0, solve for x). To find the total area, take the area below the x-axis over the subinterval [0, 2] and add it to the area above the x-axis on the subinterval [2, 6] (Figure). Figure 4.4.7 : The total area between the line and the x-axis over [0, 6] is A plus A . 2 1 We have 6 ∫ 0 |(x − 2)|dx = A2 + A1 . Then, using the formula for the area of a triangle, we obtain 1 A2 = 1 bh = 2 1 A1 = ⋅2⋅2 =2 2 1 bh = 2 ⋅4⋅4 =8 2 . The total area, then, is A1 + A2 = 8 + 2 = 10 . Exercise 4.4.4 10/21/2020 4.4.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/13794 Find the total area between the function f (x) = 2x and the x-axis over the interval [−3, 3]. Hint Review the solving strategy in Example. Answer 18 Properties of the Definite Integral The properties of indefinite integrals apply to definite integrals as well. Definite integrals also have properties that relate to the limits of integration. These properties, along with the rules of integration that we examine later in this chapter, help us manipulate expressions to evaluate definite integrals. Rule: Properties of the Definite Integral 1. a ∫ f (x) dx = 0 (4.4.15) a If the limits of integration are the same, the integral is just a line and contains no area. 2. a ∫ b f (x) dx = − ∫ b f (x)dx (4.4.16) a If the limits are reversed, then place a negative sign in front of the integral. 3. b ∫ b [f (x) + g(x)] dx = ∫ a b f (x)dx + ∫ a g(x)dx (4.4.17) g(x)dx (4.4.18) a The integral of a sum is the sum of the integrals. 4. b ∫ b [f (x) − g(x)] dx = ∫ a b f (x)dx − ∫ a a The integral of a difference is the difference of the integrals 5. b ∫ b cf (x) dx = c ∫ a f (x) (4.4.19) a for constant c. The integral of the product of a constant and a function is equal to the constant multiplied by the integral of the function. 6. b ∫ a c f (x) dx = ∫ a b f (x) dx + ∫ f (x) dx (4.4.20) c Although this formula normally applies when c is between a and b, the formula holds for all values of a , b , and c , provided f (x) is integrable on the largest interval. Example 4.4.5 : Using the Properties of the Definite Integral 10/21/2020 4.4.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/13794 Use the properties of the definite integral to express the definite integral of [−2, 1] as the sum of three definite integrals. 3 f (x) = −3 x + 2x + 2 over the interval Solution 1 Using integral notation, we have ∫ 3 −2 ∫ 1 −2 3 (−3 x + 2x + 2)dx = ∫ (−3 x 1 3 −2 We apply properties 3. and 5. to get + 2x + 2)dx. −3 x dx + ∫ 1 −2 2xdx + ∫ 1 −2 = −3 ∫ 1 3 2dx x dx + 2 ∫ −2 1 xdx + ∫ −2 1 −2 2dx. Exercise 4.4.5 Use the properties of the definite integral to express the definite integral of f (x) = 6x [1, 3] as the sum of four definite integrals. 3 2 − 4x + 2x − 3 over the interval Hint Use the solving strategy from Example and the properties of definite integrals. Answer 3 3 3 6∫ 1 3 2 x dx − 4 ∫ x dx + 2 ∫ 1 3 xdx − ∫ 1 3dx (4.4.21) 1 Example 4.4.6 : Using the Properties of the Definite Integral If it is known that ∫ 8 0 f (x)dx = 10 and ∫ 5 f (x)dx = 5 0 , find the value of (∫^8_5f(x)dx\). Solution By property 6., b ∫ a c f (x)dx = ∫ b f (x)dx + ∫ a c f (x)dx. Thus, ∫ 8 0 f (x)dx = ∫ 5 f (x)dx + ∫ 0 8 5 10 = 5 + ∫ 8 5 5 =∫ 8 5 f (x)dx f (x)dx f (x)dx . Exercise 4.4.6 If it is known that ∫ 5 1 f (x)dx = −3 and ∫ 5 2 f (x)dx = 4 , find the value of ∫ 2 1 f (x)dx. Hint Use the solving strategy from Example and the rule on properties of definite integrals. Answer −7 Comparison Properties of Integrals A picture can sometimes tell us more about a function than the results of computations. Comparing functions by their graphs as well as by their algebraic expressions can often give new insight into the process of integration. Intuitively, we might say that if a function f (x) is above another function g(x), then the area between f (x) and the x-axis is greater than the area between g(x) and the x-axis. This is true depending on the interval over which the comparison is made. The properties of 10/21/2020 4.4.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/13794 definite integrals are valid whether a < b, a = b , or a > b . The following properties, however, concern only the case and are used when we want to compare the sizes of integrals. a ≤b , Comparison Theorem i. If f (x) ≥ 0 for a ≤ x ≤ b , then b ∫ f (x)dx ≥ 0. (4.4.22) a ii. If f (x) ≥ g(x) for a ≤ x ≤ b , then b ∫ b f (x)dx ≥ ∫ a g(x)dx. (4.4.23) a iii. If m and M are constants such that m ≤ f (x) ≤ M for a ≤ x ≤ b , then b m(b − a) ≤ ∫ f (x)dx ≤ M (b − a). (4.4.24) a Example 4.4.7 : Comparing Two Functions over a Given Interval − −−− − − − − − − Compare f (x) = √1 + x and g(x) = √1 + x over the interval [0, 1]. 2 Solution Graphing these functions is necessary to understand how they compare over the interval [0, 1]. Initially, when graphed on a graphing calculator, f (x) appears to be above g(x) everywhere. However, on the interval [0, 1], the graphs appear to be on top of each other. We need to zoom in to see that, on the interval [0, 1], g(x) is above f (x). The two functions intersect at x = 0 and x = 1 (Figure). Figure 4.4.8 : (a) The function f (x) appears above the function g(x) except over the interval [0, 1] (b) Viewing the same graph with a greater zoom shows this more clearly. We can see from the graph that over the interval [0, 1], g(x) ≥ f (x). Comparing the integrals over the specified interval [0, 1], we also see that ∫ g(x)dx ≥ ∫ f (x)dx (Figure). The thin, red-shaded area shows just how much difference there is between these two integrals over the interval [0, 1]. 1 0 1 0 Figure 4.4.9 : (a) The graph shows that over the interval [0, 1], g(x) ≥ f (x), where equality holds only at the endpoints of the interval. (b) Viewing the same graph with a greater zoom shows this more clearly. 10/21/2020 4.4.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/13794 Average Value of a Function We often need to find the average of a set of numbers, such as an average test grade. Suppose you received the following test scores in your algebra class: 89, 90, 56, 78, 100, and 69. Your semester grade is your average of test scores and you want to know what grade to expect. We can find the average by adding all the scores and dividing by the number of scores. In this case, there are six test scores. Thus, 89 + 90 + 56 + 78 + 100 + 69 482 = ≈ 80.33. 6 (4.4.25) 6 Therefore, your average test grade is approximately 80.33, which translates to a B− at most schools. Suppose, however, that we have a function v(t) that gives us the speed of an object at any time t, and we want to find the object’s average speed. The function v(t) takes on an infinite number of values, so we can’t use the process just described. Fortunately, we can use a definite integral to find the average value of a function such as this. Let f (x) be continuous over the interval [a, b] and let [a, b] be divided into n subintervals of width Δx = (b − a)/n . Choose a representative x in each subinterval and calculate f (x ) for i = 1, 2, … , n. In other words, consider each f (x ) as a sampling of the function over each subinterval. The average value of the function may then be approximated as ∗ ∗ ∗ i i i ∗ ∗ 1 2 ∗ f (x ) + f (x ) + ⋯ + f (xn ) , (4.4.26) n which is basically the same expression used to calculate the average of discrete values. But we know Δx = b −a , n so n = b −a , and we get Δx ∗ ∗ 1 2 ∗ f (x ) + f (x ) + ⋯ + f (xn ) ∗ ∗ 1 2 ∗ f (x ) + f (x ) + ⋯ + f (xn ) = n . (4.4.27) (b − a) Δx Following through with the algebra, the numerator is a sum that is represented as ∑ f (x ∗ i), and we are dividing by a fraction. To divide by a fraction, invert the denominator and multiply. Thus, an approximate value for the average value of the function is given by n i=1 n ∑ ∗ i=1 f (x ) i (b − a) Δx =( n 1 ∗ )∑ i=1 b −a f (x ) = ( i n )∑ i=1 b −a ∗ f (x )Δx. i Δx This is a Riemann sum. Then, to get the exact average value, take the limit as n goes to infinity. Thus, the average value of a function is given by 1 b −a 1 n limn→∞ ∑ i=1 f (xi )Δx = ∫ b −a b a f (x)dx. Definition: average value of the function Let f (x) be continuous over the interval [a, b]. Then, the average value of the function f (x) (or f ave b 1 fave = ) on [a, b] is given by ∫ b −a f (x)dx. (4.4.28) a Example 4.4.8 : Finding the Average Value of a Linear Function Find the average value of f (x) = x + 1 over the interval [0, 5]. Solution First, graph the function on the stated interval, as shown in Figure. 10/21/2020 4.4.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/13794 Figure 4.4.10 :The graph shows the area under the function (x) = x + 1 over [0, 5]. The region is a trapezoid lying on its side, so we can use the area formula for a trapezoid represents height, and a and b represent the two parallel sides. Then, ∫ 5 0 1 x + 1dx = 1 h(a + b) = 2 35 ⋅ 5 ⋅ (1 + 6) = 2 2 1 A = h(a + b), 2 where h . Thus the average value of the function is 1 ∫ 5 −0 5 0 1 x + 1dx = 35 ⋅ 5 7 = 2 2 . Exercise 4.4.7 Find the average value of f (x) = 6 − 2x over the interval [0, 3]. Hint Use the average value formula, and use geometry to evaluate the integral. Answer 3 Key Concepts The definite integral can be used to calculate net signed area, which is the area above the x-axis less the area below the x-axis. Net signed area can be positive, negative, or zero. The component parts of the definite integral are the integrand, the variable of integration, and the limits of integration. Continuous functions on a closed interval are integrable. Functions that are not continuous may still be integrable, depending on the nature of the discontinuities. The properties of definite integrals can be used to evaluate integrals. The area under the curve of many functions can be calculated using geometric formulas. The average value of a function can be calculated using definite integrals. Key Equations Definite Integral n b ∗ ∫ f (x)dx = lim ∑ f (x )Δx i n→∞ a i=1 Properties of the Definite Integral a ∫ f (x)dx = 0 a 10/21/2020 4.4.12 CC-BY-NC-SA https://math.libretexts.org/@go/page/13794 a ∫ b f (x)dx = − ∫ b f (x)dx a b ∫ b [f (x) + g(x)]dx = ∫ a b f (x)dx + ∫ a b ∫ b [f (x) − g(x)]dx = ∫ a b f (x)dx − ∫ a b ∫ g(x)dx a g(x)dx a b cf (x)dx = c ∫ a f (x) for constant c a b ∫ c f (x)dx = ∫ a a b f (x)dx + ∫ f (x)dx c Glossary average value of a function (or f ) the average value of a function on an interval can be found by calculating the definite integral of the function and dividing that value by the length of the interval ave definite integral a primary operation of calculus; the area between the curve and the x-axis over a given interval is a definite integral integrable function a function is integrable if the limit defining the integral exists; in other words, if the limit of the Riemann sums as n goes to infinity exists integrand the function to the right of the integration symbol; the integrand includes the function being integrated limits of integration these values appear near the top and bottom of the integral sign and define the interval over which the function should be integrated net signed area the area between a function and the x-axis such that the area below the x-axis is subtracted from the area above the x-axis; the result is the same as the definite integral of the function total area total area between a function and the x-axis is calculated by adding the area above the x-axis and the area below the x-axis; the result is the same as the definite integral of the absolute value of the function variable of integration indicates which variable you are integrating with respect to; if it is x, then the function in the integrand is followed by dx Contributors Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org. 10/21/2020 4.4.13 CC-BY-NC-SA https://math.libretexts.org/@go/page/13794 4.4E: Exercises Exercise 4.4E . 1 In the following exercises, express the limits as integrals. n 1. ∗ lim ∑(x )Δx i n→∞ over [1, 3] i=1 n 2. ∗ 2 lim ∑(5(x ) i n→∞ ∗ 3 − 3(x ) )Δx i over [0, 2] i=1 Answer 2 2 ∫ (5 x 3 − 3 x )dx 0 n 3. 2 ∗ lim ∑ si n (2π x )Δx i n→∞ over [0, 1] i=1 n 4. 2 ∗ lim ∑ cos (2π x )Δx i n→∞ over [0, 1] i=1 Answer 1 2 ∫ cos (2πx)dx 0 Exercise 4.4E . 2 In the following exercises, given L or R as indicated, express their limits as n → ∞ as definite integrals, identifying the correct intervals. n n 1 1. L n = n i −1 ∑ n = n i=1 n 1 2. R n i ∑ n i=1 n Answer 1 ∫ xdx 0 3. Ln = n 2 n n = ) n i=1 n 3 4. R i −1 ∑(1 + 2 i ∑(3 + 3 n i=1 ) n Answer 6 ∫ xdx 3 10/21/2020 4.4E.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/13795 2π 5. L n = n 1 6. R n n i −1 ∑ 2π = i=1 n i ∑(1 + n i −1 cos(2π n i )log((1 + n i=1 ) n 2 ) ) n Answer 2 ∫ 2 xlog(x )dx 1 Exercise 4.4E . 3 In the following exercises, evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the x-axis. 1. 2. Answer 1 + 2 ⋅ 2 + 3 ⋅ 3 = 14 3. 10/21/2020 4.4E.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/13795 4. Answer 1 −4 +9 = 6 5. 6. Answer 1 − 2π + 9 = 10 − 2π Exercise 4.4E . 4 10/21/2020 4.4E.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/13795 In the following exercises, evaluate the integral using area formulas. 3 1. ∫ (3 − x)dx 0 3 2. ∫ (3 − x)dx 2 Answer The integral is the area of the triangle, 1 2 3 3. ∫ (3 − |x|)dx −3 6 4. ∫ (3 − |x − 3|)dx 0 Answer The integral is the area of the triangle, 9. 2 5. ∫ − −−− − √4 − x2 dx −2 5 6. ∫ − − − − − − − − − − 2 √4 − (x − 3) dx 1 Answer The integral is the area 12 7. ∫ 1 2 πr = 2π. 2 − − − − − − − − − − − 2 √36 − (x − 6) dx 0 3 8. ∫ (3 − |x|)dx −2 Answer The integral is the area of the “big” triangle less the “missing” triangle, 9 − 1 . 2 Exercise 4.4E . 5 In the following exercises, use averages of values at the left (L) and right (R) endpoints to compute the integrals of the piecewise linear functions with graphs that pass through the given list of points over the indicated intervals. 1. (0, 0), (2, 1), (4, 3), (5, 0), (6, 0), (8, 3)over [0, 8] 2. (0, 2), (1, 0), (3, 5), (5, 5), (6, 2), (8, 0)over [0, 8] Answer L+R L = 2 + 0 + 10 + 5 + 4 = 21, R = 0 + 10 + 10 + 2 + 0 = 22, = 21.5 2 3. (−4, −4), (−2, 0), (0, −2), (3, 3), (4, 3)over [−4, 4] 4. (−4, 0), (−2, 2), (0, 0), (1, 2), (3, 2), (4, 0)over [−4, 4] Answer 10/21/2020 4.4E.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/13795 L+R L = 0 + 4 + 0 + 4 + 2 = 10, R = 4 + 0 + 2 + 4 + 0 = 10, = 10 2 Exercise 4.4E . 6 4 Suppose that ∫ 2 f (x)dx = 5 0 exercises, compute the integrals. and ∫ 4 f (x)dx = −3 , and 0 ∫ 2 g(x)dx = −1 and 0 ∫ g(x)dx = 2 . In the following 0 4 1. ∫ (f (x) + g(x))dx 0 4 2. ∫ (f (x) + g(x))dx 2 Answer 4 4 ∫ f (x)dx + ∫ 2 g(x)dx = 8 − 3 = 5 2 2 3. ∫ (f (x) − g(x))dx 0 4 4. ∫ (f (x) − g(x))dx 2 Answer 4 4 ∫ f (x)dx − ∫ 2 g(x)dx = 8 + 3 = 11 2 2 5. ∫ (3f (x) − 4g(x))dx 0 4 6. ∫ (4f (x) − 3g(x))dx 2 Answer 4 4∫ 4 f (x)dx − 3 ∫ 2 g(x)dx = 32 + 9 = 41 2 Exercise 4.4E . 7 A In the following exercises, use the identity ∫ −A π sint 1. ∫ −π 2 dt 0 f (x)dx = ∫ −A A f (x)dx + ∫ f (x)dx to compute the integrals. 0 (Hint: sin(−t) = −sin(t)) 1 +t √π 2. ∫ t dt √−π 1 + cost Answer The integrand is odd; the integral is zero. 3 3. ∫ (2 − x)dx (Hint: Look at the graph of f.) 1 10/21/2020 4.4E.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/13795 4 4. ∫ 3 (x − 3 ) dx (Hint: Look at the graph of f.) 2 Answer The integrand is antisymmetric with respect to x = 3. The integral is zero. Exercise 4.4E . 8 1 In the following exercises, given that ∫ 1 1 xdx = 2 0 2 ,∫ , 3 0 1 1 x dx = and ∫ 3 1 x dx = 0 4 , compute the integrals. 1 1. ∫ 2 (1 + x + x 3 + x )dx 0 1 2. ∫ 2 (1 − x + x 3 − x )dx 0 Answer 1 1− 1 1 + 2 7 − = 3 4 12 1 3. ∫ 2 (1 − x ) dx 0 1 4. ∫ 3 (1 − 2x ) dx 0 Answer 1 2 ∫ (1 − 2x + 4 x 4 3 − 8 x )dx = 1 − 1 + 3 0 1 5. ∫ 2 −2 = − 4 (6x − 3 2 x )dx 3 0 1 6. ∫ 3 (7 − 5 x )dx 0 Answer 5 7− 23 = 4 4 Exercise 4.4E . 9 In the following exercises, use the comparison theorem. 3 1. Show that ∫ 2 (x − 6x + 9)dx ≥ 0. 0 3 2. Show that ∫ (x − 3)(x + 2)dx ≤ 0. −2 Answer The integrand is negative over [−2, 3]. 1 3. Show that ∫ 0 10/21/2020 − −−− − 3 √1 + x dx ≤ ∫ 1 − −−− − 2 √1 + x dx . 0 4.4E.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/13795 2 2 − − − − − √1 + x dx ≤ ∫ 4. Show that ∫ 1 − −−− − 2 √1 + x dx. 1 Answer 2 x ≤x − −−− − − − − − over [1, 2], so √− 1 + x ≤ √1 + x 2 π/2 5. Show that ∫ π sintdt ≥ ) (Hint: sint ≥ 4 0 over [1, 2]. 2t over [0, π π ]) 2 π/4 – costdt ≥ π √2/4 6. Show that ∫ . −π/4 Answer – √2 cos(t) ≥ 2 . Multiply by the length of the interval to get the inequality. Exercise 4.4E . 10 In the following exercises, find the average value f ave 1. f (x) = x , a = −1, b = 1 2. f (x) = x , a = −1, b = 1 2 5 of f between a and b, and find a point c, where f (c) = f ave Answer fave = 0; c = 0 − −−− − 2 , a = 0, b = 2 3. f (x) = √4 − x 4. f (x) = (3 − |x|), a = −3, b = 3 Answer 3 2 when c = ± 3 2 5. f (x) = sinx, a = 0, b = 2π 6. f (x) = cosx, a = 0, b = 2π Answer π fave = 0; c = 3π , 2 2 Exercise 4.4E . 11 In the following exercises, approximate the average value using Riemann sums compare with the exact given answer? 1. y = ln(x) over the interval [1, 4]; the exact solution is L100 and R100 . How does your answer ln(256) − 1. 3 2. y = e x/2 over the interval [0, 1]; the exact solution is 2(√e − 1). Answer L100 = 1.294, R100 = 1.301; 10/21/2020 the exact average is between these values. 4.4E.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/13795 3. y = tanx over the interval [0, π 2ln(2) ; the exact solution is ] 4 4. y = x +1 − −−− − √4 − x2 . π over the interval [−1, 1]; the exact solution is π 6 . Answer 1 L100 × ( 2 1 ) = 0.5178, R100 × ( ) = 0.5294 2 Exercise 4.4E . 12 In the following exercises, compute the average value using the left Riemann sums L accuracy compare with the given exact value? N 1. y = x 2 for N . How does the = 1, 10, 100 8 over the interval [0, 2]; the exact solution is − . −4 3 2. y = xe 2 over the interval [0, 2]; the exact solution is x 1 (e 4 − 1). 4 Answer 1 L1 = 0, L10 × ( 3. y = ( 1 2 1 ) = 8.743493, L100 × ( ) = 12.861728. 2 over the interval [0, 4]; the exact solution is x ) 2 The exact answer ≈ 26.799, so L 15 64ln(2) 100 is not accurate. . 2 4. y = xsin(x 2 ) over the interval [−π, 0]; the exact solution is cos(π ) − 1 2π. Answer 1 L1 × ( π 1 ) = 1.352, L10 × ( accurate to first decimal. π ) = −0.1837, L100 × (1π) = −0.2956. The exact answer ≈ −0.303, so L100 is not Exercise 4.4E . 13 Answer the following: 2π 1. Suppose that A = ∫ 2π 2 si n tdt and B = ∫ 0 2 cos tdt. Show that A + B = 2π and A = B. 0 π/4 2. Suppose that A = ∫ π/4 2 sec tdt = π and B = ∫ −π/4 2 4tan tdt. Show that A − B = −π/ π 2 . Answer π/4 Use tan 2 2 θ + 1 = sec θ. Then, B − A = ∫ −π/4 π 1dx = . 2 3. Show that the average value of si n t over [0, 2π] is equal to average value of si n t over [0, π] is also equal to 1/2. 2 1/2 Without further calculation, determine whether the 2 4. Show that the average value of cos t over [0, 2π] is equal to average value of cos (t) over [0, π] is also equal to 1/2. 2 . Without further calculation, determine whether the 1/2 2 Answer 10/21/2020 4.4E.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/13795 2π 2 ∫ cos tdt = π, so divide by the length 2π of the interval. cos 2 t has period π, so yes, it is true. 0 5. Explain why the graphs of a quadratic function (parabola) p(x) and a linear function ℓ(x) can intersect in at most two b b points. Suppose that p(a) = ℓ(a) and p(b) = ℓ(b) , and that ∫ p(t)dt > ∫ a whenever a ≤ c < d ≤ b. 6. Suppose that parabola 2 p(x) = ax p(t) > ∫ c and has a vertex of (a < 0) d . Explain why ∫ a opens downward + bx + c d ℓ(t)dt ℓ(t)dt c −b y = >0 2a . For which B interval [A, B] is ∫ 2 (ax as large as possible? + bx + c)dx A Answer The integral is maximized when one uses the largest interval on which p is nonnegative. Thus, A = and B = − − − − − − − 2 −b − √b − 4ac 2a − − − − − − − 2 −b + √b − 4ac . 2a 7. Suppose [a, b] can be subdivided into subintervals a = a0 < a1 < a2 < ⋯ < aN = b such that either f ≥0 over of [a, b] ai [ ai−1 , ai ] or f over [a ≤0 i−1 . Set A , ai ] i =∫ f (t)dt. ai−1 b a. Explain why ∫ f (t)dt = A1 + A2 + ⋯ + AN . a b b. Then, explain why ∫ b f (t)dt ∣∣∣≤ ∫ a |f (t)|dt. a d 8. Suppose f and g are continuous functions such that ∫ d f (t)dt ≤ ∫ c Explain why f (x) ≤ g(x) for all values of x. for every subinterval g(t)dt [c, d] . c Answer If for some f (t0 ) > g(t0 ) t0 ∈ [a, b] , then since f −g is continuous, there is an interval containing d f (t) > g(t) over the interval [c, d], and then ∫ t0 such that d f (t)dt > ∫ d g(t)d over this interval. c 9. Suppose the average value of f over [a, b] is 1 and the average value of f over [b,c] is 1 where a < c < b . Show that the average value of f over [a, c] is also 1. Exercise 4.4E . 14 Answer the following: 1. Suppose that subinterval [a to 1. [a, b] i−1 can be partitioned. taking a = a < a < ⋯ < a = b such that the average value of f over each is equal to 1 for each i = 1, … , N . Explain why the average value of f over [a, b] is also equal 0 1 N , ai ] = 1 Answer The integral of f over an interval is the same as the integral of the average of f over that interval. Thus, b ∫ a1 f (t)dt = ∫ a a0 a2 f (t)dt + ∫ a aN 1f (t)dt + ⋯ + ∫ a1 f (t)dt = ∫ aN + 1 = (a1 − a0 ) + (a2 − a1 ) + ⋯ + (aN − aN −1 ) = aN − a0 = b − a a2 1dt + ∫ a0 a1 aN 1dt + ⋯ + ∫ 1dt aN + 1 . Dividing through by b − a gives the desired identity. 10/21/2020 4.4E.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/13795 i N 2. Suppose that for each i such that 1 ≤ i ≤ N one has ∫ f (t)dt = i . Show that ∫ i−1 N (N + 1) f (t)dt = i N 3. Suppose that for each i such that 1 ≤ i ≤ N one has ∫ . 2 0 2 f (t)dt = i . Show that ∫ i−1 N (N + 1)(2N + 1) f (t)dt = 6 0 . Answer N N ∫ N i 2 f (t)dt = ∑ ∫ 0 f (t)dt = ∑ i i−1 i=1 N (N + 1)(2N + 1) = 6 i=1 4. Compute the left and right Riemann sums L10 and 1 that ∫ 2 t dt = 1/3 , to how many decimal places is and their average R10 L10 + R10 L10 + R10 2 for 2 f (t) = t over . Given [0, 1] accurate? 2 0 5. Compute the left and right Riemann sums, L 10 and R , and their average L10 + R10 10 for f (t) = (4 − t 2 ) over . [1, 2] 2 2 Given that ∫ 2 (4 − t )dt = 1.66 , to how many decimal places is 1 L10 + R10 2 accurate? Answer L10 = 1.815, R10 = 1.515, 5 6. If ∫ L10 + R10 so the estimate is accurate to two decimal places. = 1.665, 2 − − − − − 4 √ 1 + t dt = 41.7133..., 5 what is ∫ 1 − −−− − 4 √ 1 + u du? 1 1 7. Estimate ∫ tdt using the left and right endpoint sums, each with a single rectangle. How does the average of these left 0 1 and right endpoint sums compare with the actual value ∫ tdt? 0 Answer The average is 1/2, which is equal to the integral in this case. 1 8. Estimate ∫ tdt by comparison with the area of a single rectangle with height equal to the value of t at the midpoint 0 1 t = 1 . How does this midpoint estimate compare with the actual value ∫ 2 tdt? 0 9. From the graph of sin(2πx) shown: 1 a. Explain why ∫ sin(2πt)dt = 0. 0 a+1 b. Explain why, in general, ∫ sin(2πt)dt = 0 for any value of a. a 10/21/2020 4.4E.10 CC-BY-NC-SA https://math.libretexts.org/@go/page/13795 Answer a. The graph is antisymmetric with respect to t = 1 2 over [0, 1], so the average value is zero. b. For any value of a, the graph between [a, a + 1] is a shift of the graph over [0, 1], so the net areas above and below the axis do not change and the average remains zero. 1 83) If f is 1-periodic (f (t + 1) = f (t)) , odd, and integrable over [0, 1], is it always true that ∫ f (t)dt = 0? 0 1+a 84) If f is 1-periodic and ∫ 10f (t)dt = A, is it necessarily true that ∫ f (t)dt = A for all A? a Answer Yes, the integral over any interval of length 1 is the same. 10/21/2020 4.4E.11 CC-BY-NC-SA https://math.libretexts.org/@go/page/13795 4.5: The Fundamental Theorem of Calculus Let f (t) be a continuous function defined on [a, b]. The definite integral ∫ turn this concept into a function by letting the upper (or lower) bound vary. b a f (x)dx is the "area under f " on . We can [a, b] x Let F (x) = ∫ f (t)dt . It computes the area under f on [a, x] as illustrated in Figure 4.5.1. We can study this function using our knowledge of the definite integral. For instance, F (a) = 0 since ∫ f (t)dt = 0 . a a a Figure 4.5.1 : The area of the shaded region is F (x) = ∫ x a f (t) dt . We can also apply calculus ideas to F (x); in particular, we can compute its derivative. While this may seem like an innocuous thing to do, it has far--reaching implications, as demonstrated by the fact that the result is given as an important theorem. Theorem 4.5.1 : The Fundamental Theorem of Calculus, Part 1 x Let f be continuous on [a, b] and let F (x) = ∫ a . Then F is a differentiable function on (a, b), and f (t) dt ′ F (x) = f (x). (4.5.1) Initially this seems simple, as demonstrated in the following example. Example 4.5.1 : Using the Fundamental Theorem of Calculus, Part 1 Let F (x) = ∫ x −5 2 (t + sin t) dt . What is F ′ ?} (x) Solution Using the Fundamental Theorem of Calculus, we have F ′ 2 (x) = x + sin x . This simple example reveals something incredible: F (x) is an antiderivative of x + sin x ! F (x) = x − cos x + C for some value of C . (We can find C , but generally we do not care. We know that which allows us to compute C . In this case, C = cos(−5) + .) 2 1 3 3 Therefore, , F (−5) = 0 125 3 We have done more than found a complicated way of computing an antiderivative. Consider a function f defined on an open interval containing a , b and c . Suppose we want to compute ∫ f (t) dt . First, let F (x) = ∫ f (t) dt . Using the properties of the definite integral found in Theorem 5.2.1, we know b x a b ∫ a c c b f (t) dt = ∫ f (t) dt + ∫ a f (t) dt a = −∫ b f (t)dt + ∫ c 10/21/2020 (4.5.2) c f (t)dt (4.5.3) c = −F (a) + F (b) (4.5.4) = F (b) − F (a). (4.5.5) 4.5.1 CC-BY-NC-SA https://math.libretexts.org/@go/page/10978 We now see how indefinite integrals and definite integrals are related: we can evaluate a definite integral using antiderivatives! This is the second part of the Fundamental Theorem of Calculus. Theorem 4.5.2 : The Fundamental Theorem of Calculus, Part 2 Let f be continuous on [a, b] and let F be any antiderivative of f . Then b ∫ f (x)dx = F (b) − F (a). (4.5.6) a Example 4.5.2 : Using the Fundamental Theorem of Calculus, Part 2 We spent a great deal of time in the previous section studying Calculus, evaluate this definite integral. ∫ 4 0 2 (4x − x )dx . Using the Fundamental Theorem of Solution We need an antiderivative of simplicity, choose C = 0 . 2 f (x) = 4x − x . All antiderivatives of f have the form 2 F (x) = 2 x − 1 3 3 x +C ; for The Fundamental Theorem of Calculus states 4 2 ∫ 2 (4x − x )dx = F (4) − F (0) = (2(4 ) 1 − 64 3 4 ) − (0 − 0) = 32 − 3 0 = 32/3. (4.5.7) 3 This is the same answer we obtained using limits in the previous section, just with much less work. Notation: A special notation is often used in the process of evaluating definite integrals using the Fundamental Theorem of b Calculus. Instead of explicitly writing F (b) − F (a) , the notation F (x)∣∣ is used. Thus the solution to Example 4.5.2 would be a written as: 4 2 ∫ 2 (4x − x )dx = (2 x 1 − ∣ 3 x )∣ 3 0 4 2 = (2(4 ) 1 − 3 4 ) − (0 − 0) = 32/3. (4.5.8) 3 ∣0 The Constant C : Any antiderivative F (x) can be chosen when using the Fundamental Theorem of Calculus to evaluate a definite integral, meaning any value of C can be picked. The constant always cancels out of the expression when evaluating F (b) − F (a) , so it does not matter what value is picked. This being the case, we might as well let C = 0 . Example 4.5.3 : Using the Fundamental Theorem of Calculus, Part 2 Evaluate the following definite integrals. 2 π 3 1. ∫ x dx 2. ∫ −2 5 sin xdx 3. ∫ 0 9 t e dt 4. ∫ 0 5 − √u du 4 5. ∫ 2dx (4.5.9) 1 Solution 1. ∫ 2 2. ∫ π 3 −2 0 x dx = 2 1 1 4 x ∣ ∣−2 = ( 4 4 4 2 )−( 1 4 4 (−2 ) ) = 0. π sin xdx = − cos x ∣ ∣ = − cos π − ( − cos 0) = 1 + 1 = 2. 0 (This is interesting; it says that the area under one "hump" of a sine curve is 2.) 3. ∫ 5 0 4. ∫ 9 4 5. ∫ 5 1 t e dt = e 5 t∣ ∣ =e 5 −e 0 =e 5 − 1 ≈ 147.41. 0 9 − √u du = ∫4 u 1 2 du = 2 3 3 u ∣ ∣ 2 9 = 4 2 3 3 (9 2 3 −4 2 ) = 2 3 (27 − 8) = 38 3 . 5 2dx = 2x ∣ = 2(5) − 2 = 2(5 − 1) = 8. ∣1 This integral is interesting; the integrand is a constant function, hence we are finding the area of a rectangle with width (5 − 1) = 4 and height 2. Notice how the evaluation of the definite integral led to 2(4) = 8 . In general, if c is a constant, then ∫ b a 10/21/2020 cdx = c(b − a) . 4.5.2 CC-BY-NC-SA https://math.libretexts.org/@go/page/10978 Understanding Motion with the Fundamental Theorem of Calculus We established, starting with Key Idea 1, that the derivative of a position function is a velocity function, and the derivative of a velocity function is an acceleration function. Now consider definite integrals of velocity and acceleration functions. Specifically, if v(t) is a velocity function, what does ∫ b a v(t)dt mean? The Fundamental Theorem of Calculus states that b ∫ v(t)dt = V (b) − V (a), (4.5.10) a where V (t) is any antiderivative of v(t) . Since v(t) is a velocity function, V (t) must be a position function, and V (b) − V (a) measures a change in position, or displacement. Example 4.5.4 : Finding displacement A ball is thrown straight up with velocity given by interpret, ∫ v(t)dt. } v(t) = −32t + 20 ft/s, where is measured in seconds. Find, and t 1 0 Solution Using the Fundamental Theorem of Calculus, we have 1 ∫ 1 v(t)dt = ∫ 0 (−32t + 20)dt (4.5.11) 0 2 = −16 t 1 + 20t ∣ ∣ (4.5.12) 0 = 4. (4.5.13) Thus if a ball is thrown straight up into the air with velocity v(t) = −32t + 20 , the height of the ball, 1 second later, will be 4 feet above the initial height. (Note that the ball has traveled much farther. It has gone up to its peak and is falling down, but the difference between its height at t = 0 and t = 1 is 4ft. Integrating a rate of change function gives total change. Velocity is the rate of position change; integrating velocity gives the total change of position, i.e., displacement. Integrating a speed function gives a similar, though different, result. Speed is also the rate of position change, but does not account for direction. So integrating a speed function gives total change of position, without the possibility of "negative position change." Hence the integral of a speed function gives distance traveled. As acceleration is the rate of velocity change, integrating an acceleration function gives total change in velocity. We do not have a simple term for this analogous to displacement. If a(t) = 5 miles/h and t is measured in hours, then 2 3 ∫ a(t)dt = 15 (4.5.14) 0 means the velocity has increased by 15m/h from t = 0 to t = 3 . The Fundamental Theorem of Calculus and the Chain Rule x Part 1 of the Fundamental Theorem of Calculus (FTC) states that given F (x) = ∫ f (t)dt , F (x) = f (x). Using other notation, (F (x)) = f (x). While we have just practiced evaluating definite integrals, sometimes finding antiderivatives is impossible and we need to rely on other techniques to approximate the value of a definite integral. Functions written as F (x) = ∫ f (t)dt are useful in such situations. ′ a d dx x a It may be of further use to compose such a function with another. As an example, we may compose F (x) with g(x) to get g(x) F (g(x)) = ∫ f (t)dt. (4.5.15) a What is the derivative of such a function? The Chain Rule can be employed to state 10/21/2020 4.5.3 CC-BY-NC-SA https://math.libretexts.org/@go/page/10978 d ′ ′ ′ (F (g(x))) = F (g(x))g (x) = f (g(x))g (x). (4.5.16) dx An example will help us understand this. Example 4.5.4 : The FTC, Part 1, and the Chain Rule Find the derivative of F (x) = ∫ 2 x 2 ln tdt . Solution x We can view F (x) as being the function G(x) = ∫ ln tdt composed with g(x) = x ; that is, Fundamental Theorem of Calculus states that G (x) = ln x . The Chain Rule gives us 2 2 F (x) = G(g(x)) . The ′ ′ ′ ′ F (x) = G (g(x))g (x) (4.5.17) ′ = ln(g(x))g (x) (4.5.18) 2 = ln(x )2x (4.5.19) 2 = 2x ln x (4.5.20) Normally, the steps defining G(x) and g(x) are skipped. Practice this once more. Example 4.5.5 : The FTC, Part 1, and the Chain Rule Find the derivative of F (x) = ∫ 5 cos x 3 t dt. Solution Note that F (x) = − ∫ cos x 5 3 t dt . Viewed this way, the derivative of F is straightforward: ′ 3 F (x) = sin x cos x. (4.5.21) Area Between Curves Consider continuous functions f (x) and g(x) defined on [a, b], where f (x) ≥ g(x) for all Figure 4.5.2. What is the area of the shaded region bounded by the two curves over [a, b]? x in , as demonstrated in [a, b] Figure 4.5.2 : Finding the area bounded by two functions on an interval; it is found by subtracting the area under g from the area under f . The area can be found by recognizing that this area is "the area under f area is b ∫ a − the area under g ." Using mathematical notation, the b f (x)dx − ∫ g(x)dx. (4.5.22) a Properties of the definite integral allow us to simplify this expression to 10/21/2020 4.5.4 CC-BY-NC-SA https://math.libretexts.org/@go/page/10978 b ∫ (f (x) − g(x))dx. (4.5.23) a Theorem 4.5.3 : Area Between Curves Let f (x) and g(x) be continuous functions defined on [a, b] where f (x) ≥ g(x) for all bounded by the curves y = f (x), y = g(x) and the lines x = a and x = b is x in . The area of the region [a, b] b ∫ (f (x) − g(x))dx. (4.5.24) a Example 4.5.6 : Finding area between curves Find the area of the region enclosed by y = x 2 +x −5 and y = 3x − 2 . Solution It will help to sketch these two functions, as done in Figure 4.5.3. Figure 4.5.3 : Sketching the region enclosed by y = x 2 +x −5 and y = 3x − 2 in Example 4.5.6 The region whose area we seek is completely bounded by these two functions; they seem to intersect at x = −1 and x = 3 . To check, set x + x − 5 = 3x − 2 and solve for x: 2 2 x 2 (x +x −5 + x − 5) − (3x − 2) = 3x − 2 (4.5.25) =0 (4.5.26) − 2x − 3 = 0 (4.5.27) (x − 3)(x + 1) = 0 (4.5.28) 2 x x = −1, 3. (4.5.29) Following Theorem \4.5.3, the area is 3 ∫ 3 2 (3x − 2 − (x + x − 5))dx −1 2 =∫ (−x + 2x + 3)dx (4.5.30) −1 1 = (− 3 x 2 +x 3 ∣ + 3x) ∣ ∣ 1 =− 3 (4.5.31) −1 1 (27) + 9 + 9 − ( 3 2 = 10 + 1 − 3) (4.5.32) 3 ¯ ¯ ¯ = 10. 6 (4.5.33) 3 The Mean Value Theorem and Average Value 10/21/2020 4.5.5 CC-BY-NC-SA https://math.libretexts.org/@go/page/10978 Figure 4.5.4 : A graph of a function f to introduce the Mean Value Theorem. Consider the graph of a function f in Figure 4.5.4 and the area defined by ∫ 4 . Three rectangles are drawn in Figure f (x)dx 1 ; in (a), the height of the rectangle is greater than f on [1, 4], hence the area of this rectangle is is greater than ∫ 4.5.5 4 0 In (b), the height of the rectangle is smaller than f on [1, 4], hence the area of this rectangle is less than ∫ 4 1 Figure 4.5.5 : Differently sized rectangles give upper and lower bounds on ∫ exactly. 4 1 . f (x)dx . f (x)dx ; the last rectangle matches the area f (x)dx 4 Finally, in (c) the height of the rectangle is such that the area of the rectangle is exactly that of ∫ f (x)dx. Since rectangles that are "too big", as in (a), and rectangles that are "too little," as in (b), give areas greater/lesser than ∫ f (x)dx, it makes sense that there is a rectangle, whose top intersects f (x) somewhere on [1, 4], whose area is exactly that of the definite integral. 0 4 1 We state this idea formally in a theorem. Theorem 4.5.4 : The Mean Value Theorem of Integration Let f be continuous on [a, b]. There exists a value c in [a, b] such that b ∫ f (x)dx = f (c)(b − a). (4.5.34) a This is an existential statement; c exists, but we do not provide a method of finding it. Theorem 4.5.4 is directly connected to the Mean Value Theorem of Differentiation, given as Theorem 3.2.1; we leave it to the reader to see how. We demonstrate the principles involved in this version of the Mean Value Theorem in the following example. Example 4.5.7 : Using the Mean Value Theorem Consider ∫ π 0 . Find a value c guaranteed by the Mean Value Theorem. sin xdx Solution We first need to evaluate ∫ π 0 10/21/2020 . (This was previously done in Example 4.5.3) sin xdx 4.5.6 CC-BY-NC-SA https://math.libretexts.org/@go/page/10978 π ∫ π sin xdx = − cos x ∣ ∣ = 2. (4.5.35) 0 0 Thus we seek a value c in [0, π] such that π sin c = 2 . π sin c = 2 ⇒ sin c = 2/π ⇒ c = arcsin(2/π) ≈ 0.69. (4.5.36) Figure 4.5.6 : A graph of y = sin x on [0, π] and the rectangle guaranteed by the Mean Value Theorem. In Figure 4.5.6 sin x is sketched along with a rectangle with height sin(0.69). The area of the rectangle is the same as the area under sin x on [0, π]. Let f be a function on [a, b] with c such that f (c)(b − a) = ∫ b a f (x)dx b ∫ . Consider ∫ b (f (x) − f (c))dx = ∫ a b a (f (x) − f (c))dx : b f (x) − ∫ a f (c)dx (4.5.37) a = f (c)(b − a) − f (c)(b − a) (4.5.38) = 0. (4.5.39) When f (x) is shifted by −f (c), the amount of area under f above the x--axis on [a, b] is the same as the amount of area below the x--axis above f ; see Figure 4.5.7 for an illustration of this. In this sense, we can say that f (c) is the average value of f on [a, b]. Figure 4.5.7 : On the left, a graph of y = f (x) and the rectangle guaranteed by the Mean Value Theorem. On the right, y = f (x) is shifted down by f (c); the resulting "area under the curve" is 0. The value f (c) is the average value in another sense. First, recognize that the Mean Value Theorem can be rewritten as b 1 f (c) = ∫ b −a for some value of c in a = x1 < x2 < … < xn+1 = b f (x)dx, . Next, partition the interval [a, b] into n equally spaced subintervals, and choose any c in [x , x ]. The average of the numbers f (c ), f (c ), \ldots, f (c ) is: [a, b] i i i+1 1 n 10/21/2020 (4.5.40) a 1 1 (f (c1 ) + f (c2 ) + … + f (cn )) = n 2 n n ∑ f (ci ). (4.5.41) i=1 4.5.7 CC-BY-NC-SA https://math.libretexts.org/@go/page/10978 Multiply this last expression by 1 in the form of n 1 n (b−a) (b−a) : n 1 ∑ f (ci ) = ∑ f (ci ) i=1 (4.5.42) n i=1 n 1 (b − a) = ∑ f (ci ) (4.5.43) n (b − a) i=1 n 1 = ∑ f (ci )Δx b −a (where Δx = (b − a)/n) (4.5.44) i=1 Now take the limit as n → ∞ : n 1 lim n→∞ b −a b 1 ∑ f (ci )Δx = ∫ b −a i=1 f (x)dx = f (c). (4.5.45) a This tells us this: when we evaluate f at n (somewhat) equally spaced points in f (c) as n → ∞ . , the average value of these samples is [a, b] This leads us to a definition. Definition 4.5.1 : The Average Value of f on [a, b] Let f be continuous on [a, b]. The average value of Mean Value Theorem. I.e., f on [a, b] is , where f (c) is a value in c [a, b] guaranteed by the b 1 Average Value of f on [a, b] = ∫ b −a f (x)dx. (4.5.46) a An application of this definition is given in the following example. Example 4.5.8 : Finding the average value of a function An object moves back and forth along a straight line with a velocity given by measured in seconds and v(t) is measured in ft/s. 2 v(t) = (t − 1) on , where [0, 3] t is What is the average velocity of the object? Solution By our definition, the average velocity is: 3 1 ∫ 3 −0 0 2 3 1 (t − 1 ) dt = ∫ 3 0 2 (t 1 − 2t + 1)dt = 1 ( 3 3 t 3 2 −t ∣ + t) ∣ ∣ 3 = 1 ft/s. (4.5.47) 0 We can understand the above example through a simpler situation. Suppose you drove 100 miles in 2 hours. What was your average speed? The answer is simple: displacement/time = 100 miles/2 hours = 50mph. What was the displacement of the object in Example 4.5.8? We calculate this by integrating its velocity function: ∫ (t − 1 ) dt = 3 ft. Its final position was 3 feet from its initial position after 3 seconds: its average velocity was 1 ft/s. 3 2 0 This section has laid the groundwork for a lot of great mathematics to follow. The most important lesson is this: definite integrals can be evaluated using antiderivatives. Since the previous section established that definite integrals are the limit of Riemann sums, we can later create Riemann sums to approximate values other than "area under the curve," convert the sums to definite integrals, then evaluate these using the Fundamental Theorem of Calculus. This will allow us to compute the work done by a variable force, the volume of certain solids, the arc length of curves, and more. The downside is this: generally speaking, computing antiderivatives is much more difficult than computing derivatives. The next chapter is devoted to techniques of finding antiderivatives so that a wide variety of definite integrals can be evaluated. Before that, the next section explores techniques of approximating the value of definite integrals beyond using the Left Hand, Right Hand and Midpoint Rules. 10/21/2020 4.5.8 CC-BY-NC-SA https://math.libretexts.org/@go/page/10978 Contributors Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/ Integrated by Justin Marshall. 10/21/2020 4.5.9 CC-BY-NC-SA https://math.libretexts.org/@go/page/10978 4.5E: Exercises Exercise 4.5E . 1 1. How are the definite and indefinite integrals related? 2. What constants of integration is most commonly used when evaluating definite integrals? 3. T/F: If f is a continuous function, then F (x) = ∫ x a f (t) dt

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