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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
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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
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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
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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
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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
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1
CC-BY-NC-SA
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Calgary, Alberta
September 2018
email: pthangarajah@mtroyal.ca
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2
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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
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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
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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.
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1.99999
and
1.999991,
we can find the real number
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1.9999905
which certainly lies in between
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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,
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{x ∈ R : a ≤ x ≤ b}}
which is denoted by [a, b].
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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 < b and b < c then a < c.
2. If a < b , then for any real number c , a + c < b + c .
3. If a < b and c > 0 then ac < bc.
4. If a < b and c < 0 then ac > bc.
Example 0.1.6 :
2 <3
and 4 < 6.
Example 0.1.7 :
2 <3
and −4 > −6.
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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
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)
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
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− 4ac = 0
) then the quadratic equation has exactly one solution (root) and that solution is
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−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.
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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 =
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−
−
−
−
−
−
√ a2 + b2
.
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(0.2.4)
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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
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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)
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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
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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.
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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).
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⎪
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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
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y
Values
(x,y) = (1.000, 0.000)
angle = 0.0°
cosθ =
x
= 1.000
1
x
degrees
radians
cosθ
Trig Tour
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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
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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.
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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
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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
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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.
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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
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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
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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
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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
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−−−
−
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
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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
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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
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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
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−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
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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
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−
−−−−−−−−
−
√ 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
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−
−
−
−
−
−
−
−
−
−
−−−−
−
√ 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
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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
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10/21/2020 0E.19 CC-BY-NC-SA
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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
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−
−
−
−
−
−
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
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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
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−
−−−−
−
√ 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
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|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 .
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10/21/2020 0E.24 CC-BY-NC-SA
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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.
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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 and both a and b are positive real numbers, then a
2
2
<b
Answer
1. False. a < b and b < c implies that a < c .
2. True.
3. False. a < b implies that
1
1
<
a
b
.
4. False. a < b implies that −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
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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:
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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
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− 10x + 1
+ 6x = 2x(8x + 3)
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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) (
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11
, c) 7
)
2
, c) 61
, −30)
2
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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
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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.
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3.
4.
5.
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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.
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¯
¯¯¯¯¯¯
¯
−−
−
¯
¯¯¯¯¯¯
¯
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
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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
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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.
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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
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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
∘
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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:
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Answer
Exercise 0E. 7
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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)
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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
.
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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
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−
−
−
−
−
−
−
–
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
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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
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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.
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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.
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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
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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)
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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
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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 )
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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
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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.
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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
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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
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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
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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
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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 )
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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
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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
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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
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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
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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.
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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
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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
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−
−
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.
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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
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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:
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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 .
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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
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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
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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.
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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
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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
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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
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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}.
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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
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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
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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
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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
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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).
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1.0.7 CC-BY-NC-SA
(1/2, 0)
and
, we can
(0, 2)
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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
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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,
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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:
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(f + g)(x) = f (x) + g(x)
Sum
(f − g)(x) = f (x) − g(x)
Difference
(f ⋅ g)(x) = f (x)g(x)
Product
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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
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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
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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
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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 )
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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.
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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,
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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
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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
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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
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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.
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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
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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
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by f (x) = x . Then D(f ) = R .
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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
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: R → R
by f (x) =
x
−1
x −1
. Then D(f ) = R ∖ {1} .
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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
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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
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: R → R
−
−
−
−
−
by f (x) = √1 − x . Then D(f ) = [0, ∞) .
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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
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by f (x) = |x| + 1
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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
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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
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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
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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
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x
y
x
y
3
3
15
1
1.0E.1 CC-BY-NC-SA
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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.
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1
17
d. −
x+2
−3x+7
e
a−2
3a+7
f.
a+h−2
3a+3h+7
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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.
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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.
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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
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√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
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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.
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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
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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
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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
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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) =
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x
−1
,x ≠1
x −1
,
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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+
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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
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and
lim f (x) = L
.
+
x→a
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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
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x
(where
x >a
) approach the number a , then we say that the limit as
1.2.4 CC-BY-NC-SA
x
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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.
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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
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n
= +∞.
(1.2.16)
(x − a)
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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)
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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
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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.
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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)
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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.
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y = f (x)
to find the values, if possible. Estimate when
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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.
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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
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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.
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y = h(x)
shown here to find the values, if possible. Estimate
1.2E.4 CC-BY-NC-SA
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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
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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
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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
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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.
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Answer
Under construction.
3.
Answer
Under construction.
4.
Answer
Under construction.
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Answer
Under construction.
5..
Answer
Under construction.
6.
Answer
Under construction.
7.
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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) =
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(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
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(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
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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.
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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
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x→a
(1.3.2)
x→a
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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.
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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 ,
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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.
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g(x) = x + 1
x −1
are identical for all values of
x ≠1
. The graphs of these two functions are shown in Figure
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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
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− 3x
2
2x
x
=
− 5x − 3
2x + 1
− 3x
x(x − 3)
= lim
− 5x − 3
x→3
(x − 3)(2x + 1)
. Therefore,
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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
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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
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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.
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−
−
−
−
−
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
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and
lim f (x) = 1
+
x→2
= 1.
−
x→2
, we conclude that lim f (x) does not exist.
x→2
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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
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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)
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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
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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
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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
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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.
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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
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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 .
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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
.
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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)
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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.
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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
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. Note how, as
1.4.4
x
approaches 0,
1.4.2 CC-BY-NC-SA
f (x)
grows very, very large. It seems
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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.
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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→∞
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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→∞
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1.4.5 CC-BY-NC-SA
= ∞
lim
n
x
.
x→−∞
and
lim
n
x
= −∞.
x→−∞
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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
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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
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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
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(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
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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.
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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
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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→∞
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−
−
−
−
−
−
−
√ 4 + 5/x2
3
=
–
√4
1.4.10 CC-BY-NC-SA
3
=
2
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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
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3
+
x
)
x
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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 :
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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 < b < 1, y = b is decreasing over (−∞, ∞). For the natural
exponential function f (x) = e , e ≈ 2.718 > 1 . Therefore, f (x) = e is increasing on `(−∞, ∞) and the range is `(0, ∞).
x
x
x
x
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x
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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→∞
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e
x
7
= 0 = lim
x→∞
e
x
.
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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
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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
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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
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=
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→∞
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2s
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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→−∞
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−−
−
2
√7y
=
5 −y
lim
y→−∞
−y
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√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
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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.
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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.
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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,
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|(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 .
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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.
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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)
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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| .
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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
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x
does not exist. The graph of f (x) = |x|/x is shown here:
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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 < b . Then
lim f (x) = L
(1.5.2)
+
x→a
if for every ε > 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
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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 .
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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,
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−−
−
1
0 < |x − 3| < √
M
. From this we choose
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−
1
δ =√
M
.
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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.
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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)
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= f (a)
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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?
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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
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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.
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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
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+
x+1
x→−1
x+1
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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,+∞)
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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
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x→2
(x − 3) = 2 − 3 = −1
exists, li m
x→2
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|x − 3| = | − 1| = 1
.
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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
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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
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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
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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) =
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sin θ − cos θ
tan θ
at θ = π
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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
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2
+x
− 2x − 1
. Then f (x) is continuous and f (1) = −1 and f (3) = 29.
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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 .
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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:
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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
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2
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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
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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
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.
lim f (x) = f (c)
+
x→c
x→c
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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
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⎧ |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
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⎧ 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)
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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
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1
secθ =
1.7.1 CC-BY-NC-SA
x
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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(
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)
3
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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
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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).
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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:
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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.
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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
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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.
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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
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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
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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
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(x) =
π
2
.
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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.
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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
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x)
=
y
fo
=
x
fo
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(−
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
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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
.
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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
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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)
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1.8E: Exercises
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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
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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
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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
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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
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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).
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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
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(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
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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
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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.
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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
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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.
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1.9E: Exercises
Exercise 1.9E. 1
lim ln x
x→0
Answer
Add texts here. Do not delete this text first.
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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→−∞
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x
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−
−
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
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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
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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
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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
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√t
√t
and the factor the denominator by t:
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√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
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:
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= 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
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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)
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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
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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
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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
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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 =
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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 .
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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
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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
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1
2
2
1
1
1
4
x
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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].
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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
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−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.
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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 =
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t−a
.
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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)
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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
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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
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, P (0, −1)
0.5x
, P (0, 10) (Round to 4 decimal places.)
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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. \((i)25.22 m/s, (ii)24.12 m/s, (iii)24.01 m/s, (iv)24 m/s;
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c. 24 m/s\)
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.
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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
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2
− 2x
− 11x + 12
, do the following.
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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.
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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.
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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
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lim
Δx→0
Δx
.
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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
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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
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= 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
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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).
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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
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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).
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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
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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.
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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+
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4
x+
5
−5
2
x+10
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−(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
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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
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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
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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)
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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)
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3)
4)
Answers to even numbered questions
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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
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+ 2, a = 2
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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.
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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)
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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.
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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
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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?
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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
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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
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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
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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
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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.
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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,
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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
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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
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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
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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
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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
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d
3
.Use
(2 x k(x))
dx
.
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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
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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
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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
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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.
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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.
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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
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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}
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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 = −
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2
x+
15
2
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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).
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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.
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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,
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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)
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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
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. Is the particle moving from
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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) .
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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
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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.
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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
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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
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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.
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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
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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.
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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
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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
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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
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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
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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
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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.
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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.
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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
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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
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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 .
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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
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3
3
− 5tan(x)sec (x) − tan (x)sec(x)
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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.
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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
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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))
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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 )
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−3
(3) = −6(3x − 5 )
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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).
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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)
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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
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3
+ 1))(14x)
2
+ 1)cos (7 x
+ 1)
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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
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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
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du
⋅
du
dx
.
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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
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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.
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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)
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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
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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
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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
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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.
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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
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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.
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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
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4
2
+ 4x y
3
.
2.6.3 CC-BY-NC-SA
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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 .
′
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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.)
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–
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
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2.6.6 CC-BY-NC-SA
′
n−1
(x) = n ⋅ x
.
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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
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2.6.7 CC-BY-NC-SA
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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)
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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
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) = 2x
2x
2y
x
y
+ 3y
2
= 12
2.6E.1 CC-BY-NC-SA
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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
=
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7
1
7
1
7
+y
+y
−x
2.6E.2 CC-BY-NC-SA
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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 = −
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2
9
x−
x+
3π+38
π+12
15
2
2.6E.3 CC-BY-NC-SA
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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
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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
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2.6E.5 CC-BY-NC-SA
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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.
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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' = −
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1
√1−x2
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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))
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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
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2.7.2 CC-BY-NC-SA
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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
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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
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m
) =
(m/n)−1
x
n
.
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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
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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
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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
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−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
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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)
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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)
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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
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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
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+ 1 ) , P (16, 1)
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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
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−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
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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.
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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.
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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:
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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
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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
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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) =
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x
3
x
3 +2
.
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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
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tanx
+ 1)
Step 1. Take the natural logarithm of both sides.
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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
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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.
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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
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5
+ 3)
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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 =
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−1
5+5ln5
x + (5 +
1
5+5ln5
)
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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)
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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.
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N (0)
and N (4). Briefly describe what these values indicate about the disease in New
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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
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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.?
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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)
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.
(2.9.9)
g(a) + g'(a)(x − a)
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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
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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
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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→∞
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= 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
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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
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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
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+ 1)
and g(x) = 3x .
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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
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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.
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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
) =
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ln x
ln x =
x
.
(2.9.46)
x
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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
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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
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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
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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
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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
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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. \(\lim_{x \to -\ infty} |x|^a e^x=0, a>0\).
3. \(\lim_{x \to \infty} \frac{\ln(x)}{x^a}=0, a>0\).
4. \(\lim_{x \to 0^+} {\ln(x)}{x^a}=0, a>0\).
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
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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.
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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}\
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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
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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
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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
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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) =
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+ x + 1) + x
(x
2
t
+5
+x −3
+ 1) cot x
3 − cos x csc x
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−
−−−−−−−
−
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
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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:
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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
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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)
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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
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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.
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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
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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
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, 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,
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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.
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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
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−
−
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.
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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 :
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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)
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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?
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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?
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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?
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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
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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
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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
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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
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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.
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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.
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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.
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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.
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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
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3
]8.1
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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
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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
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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).
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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
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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
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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
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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.
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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 +
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−
−
−
−
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.
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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.
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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.
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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
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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
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′
(c) = 0
or f
′
(c)
is undefined.
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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
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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
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2
2
− 1 ) (2x) = 6x(x
2
− 1) .
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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.
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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.
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1
3
4
2
. The absolute
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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.
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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
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absolute maximum
–
√5
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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.
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Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)
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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
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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)
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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
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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 =
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2
33
2
; absolute minimum: x = 1, y = 3
,y =4
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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
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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
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given that a > 0
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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
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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.
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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.
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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)).
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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
′
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′
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.
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4
.
(9)
3
. At this point, the slope of the tangent line equals the slope of the line joining
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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.
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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 .
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b−a
f'(c) ≠ 0
7
c ∈ (a, b)
such
.
, which contradicts the assumption that
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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
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Since f(a) ≥ f(b) , we know that
conclude that
f(b) − f(a) ≤ 0
. Also,
a < b
tells us that
b − 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.
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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
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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
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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
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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\)?
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y = cot x
have the same derivative. What can you say about \
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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.
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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 .
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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
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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.
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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 .
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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
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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
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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
′′
′′
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′′
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(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 < b and f'(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 and f is increasing 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
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–
–
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
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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.
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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)
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Answer
Increasing for −2 < x < −1 and x > 2 ; decreasing for x < −2 and −1 < x < 2
2)
3)
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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)
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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)
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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)
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3)
Answer
Concave up on all x , no in ection points
4)
5)
Answer
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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,
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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
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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 = ±
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π
2
3π
4
,x <−
π
4
3π
4
, concave down for x < −
π
2
,x >
π
2
π
2
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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.
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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).
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\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}
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\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
$$
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\displaystyle{f(t)=e^{-t^2}}
$$
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
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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.
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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.
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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.
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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 +
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−
−
−
−
−
−
−
−
−.
√ x2 + 10002
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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
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inches, the length is
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inches, and the width is
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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
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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
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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
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−
−
−
−
−
−
−
−
−
−
−
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
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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
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−
−−−
−
′
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
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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
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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.
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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.
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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)
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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
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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?
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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.
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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.
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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
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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)
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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)
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Conclusion
− 3 = 3(x − 1)(x + 1)
(+)(−)(−) = +
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is increasing
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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
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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.
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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.
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3.7.4 CC-BY-NC-SA
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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)
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x = 0
+/− = −
3.7.5 CC-BY-NC-SA
f
is concave dow
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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)
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x = 0
+/− = −
3.7.6 CC-BY-NC-SA
f
is decreasing
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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
.
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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)
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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
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1
2
0
1
3.8.1 CC-BY-NC-SA
2
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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 )
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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
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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
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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.
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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
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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
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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
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x +4
= 0
1
2
3
, then
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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.
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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.
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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
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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
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n
f (x)
f '(x)
]
2
2
0
where the function F (x) = x − [
1
n−1
for a given function f .
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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.
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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
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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
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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
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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.
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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
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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
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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
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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 .
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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
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x
x
x
) = a
a
∫ a dx =
ln(a)
+C
ln(a)
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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.
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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.
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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
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+
2
x
− 7x + C
2
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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' .
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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)
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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
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1
2
, x0 = 0
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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
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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
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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
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−t
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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:
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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.
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CHAPTER OVERVIEW
1
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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
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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
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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.
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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
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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:
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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:
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v(t)
is an antiderivative of
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′
v (t) = −32
. So we begin by finding the
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∫ (−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.
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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
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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
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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
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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.
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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
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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.
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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:
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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.
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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
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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
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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
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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 ∫
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√16−x2
dx
.
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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 =
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2
−5
1
2
2
1
(2z) dz = z dz.
(4.1.97)
2
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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
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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
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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
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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
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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
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∫ 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)
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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
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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
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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
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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
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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.
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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?
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. How can we use
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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.
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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
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4.2.4 CC-BY-NC-SA
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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
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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