Derivatives and Integration: Limits, Rates of Change, Calculus

Now Why? You learned about limits and rates of change. In Chapter 11, you will: BUNGEE JUMPING The basic tools of calculus, derivatives and integrals, are very useful when working with rates that are not constant. A bungee jumper experiences varying rates of descent and ascent, as well as changing acceleration, depending on her position during the jump. E valuate limits of polynomial and rational functions. F ind instantaneous rates of change. F ind and evaluate derivatives of polynomial functions. pproximate the area under a A curve. F ind antiderivatives, and use the Fundamental Theorem of Calculus. PREREAD Use the Mid-Chapter Quiz to write two or three questions about the first three lessons that will help you to predict the organization of the first half of Chapter 11. See students’ work. Chapter Sourced From: 11. Limits and Derivatives, from Precalculus Chapter 12 © 2014 Then Copyright © McGraw-Hill Education Vitalii Nesterchuk/Shutterstock.com 11 Derivatives and Integration Get Ready for the Chapter Take the Quick Check Below NewVocabulary QuickCheck one-sided limit 1–4. See Chapter 11 Answer Appendix. Use the graph of each function to describe its end behavior. 1. q (x ) = -_x 2 3. p (x ) = _   x - 4 x+5 2. f (x ) = _   x 7 7 - 10x tangent line 5. MUSIC The average cost to produce x CDs can be represented by A (x ) = _   x + 1200. Find the limit as x approaches positive infinity. 1200 1700 Find the average rate of change of each function on the given interval. 2 7. f (x ) = -2x 3 - 5x 2 + 6; [-4, -1] -17 8. f (x ) = 4x 3 - x 2 + 9x - 1; [-2, 4] 55 11. h (x) = _   2 2x - 8 x - 10 x 2 - 16 __ 13. g (x) =   (x - 2)(x + 4) Find the next four terms of each arithmetic or geometric sequence. -12, -17, -22, -27 differentiation differential equation lower limit upper limit right Riemann sum integration antiderivative indefinite integral Fundamental Theorem of Calculus 14. 3, 7, 11, 15, … 15. 8, 3, -2, -7, … ReviewVocabulary 16. 5, -1, -7, -13, … 17. -4, 12, -36, 108, … limit the unique value that a function approaches asymptotes a line or curve that a graph approaches 19, 23, 27, 31 -324, 972, -2916, 8748 18. 5, -10, 20, -40, … 19. -28, -21, -14, -7, … 80, -160, 320, -640 0, 7, 14, 21 -19, -25, -31, -37 Copyright © McGraw-Hill Education derivative definite integral Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any. 10–13. See margin. 4x 2x + 1 (x - 1)(x + 5) 12. f (x ) = __   (x + 2)(x - 4) instantaneous velocity regular partition -AED 10 10. f (x ) = _   2 instantaneous rate of change differential operator 9. BOOKS The profit associated with producing x books per week can be represented by C(x) = -2x2 + 140x + 25. Find the average rate of change of the cost if 50 books are produced instead of 25 books. 2 direct substitution indeterminate form 4. m (x ) = _   2x + 7 6. g (x ) = 2x 2 + 4x - 1; [-2, 1] two-sided limit holes removable discontinuities on the graph of a function that occur when the numerator and denominator of the function have common factors 641 11-1 Then You estimated limits to determine the continuity and end behavior of functions. NewVocabulary one-sided limit two-sided limit Estimating Limits Graphically Now 1 2 Why? Estimate limits of functions at a point. Are there limits to world records set by athletes? At the 2008 Beijing Olympics, Elena Isinbaeva of Russia won the pole vault, setting a new world record by vaulting 5.05 meters. E stimate limits of functions at infinity. The logistic function f(x) =  __    -0.129x , where x is 5.334 1 + 62548.213e the number of years since 1900, models the world record heights in women’s pole vaulting from 1996 to 2008. You can use the limit of this function as x goes to infinity to predict a limiting height for this track and field event. 1• Estimate Limits at a Point • Calculus centers around two fundamental problems: finding the equation of a line tangent to the graph of a function at a point, and finding the area between the graph of a function and the x-axis. Fundamental to the solutions of each of these problems is an understanding of the concept of a limit. Recall that if f (x) approaches a unique value L as x approaches c from either side, then the limit of f (x) as x approaches c is L, written limf (x) = L. x→c You can apply this description to estimate the limit of a function f (x) as x approaches a fixed value c or limf (x) by using a graph or making a table of values. x→c Example 1 Estimate a Limit = f (c ) Estimate      lim (-3x + 1) using a graph. Support your conjecture using a table of values. x→2 Analyze Graphically The graph of f (x) = -3x + 1 suggests that as x gets closer to 2, the corresponding function value gets closer to -5. Therefore, we can estimate that lim (-3x + 1) is -5. x→2 x approaches 2 x f(x ) x approaches 2 1.9 1.99 1.999 -4.7 -4.97 -4.997 2 2.001 2.01 2.1 -5.003 -5.03 -5.3 The pattern of outputs suggests that as x gets closer to 2 from the left or from the right, f (x) gets closer to -5. This supports our graphical analysis. GuidedPractice Guided Practice 1A–B. See Chapter 11 Answer Appendix for graphs. Estimate each limit using a graph. Support your conjecture using a table of values. 1A.  lim (1 - 5x) x→-3 642 | Lesson 11-1 16 1B.lim (x 2 - 1) x→1 0 Copyright © McGraw-Hill Education Make a table of values for f, choosing x-values that approach 2 by using some values slightly less than 2 and some values slightly greater than 2. Graphic Compressor/Shutterstock.com Support Numerically In Example 1, lim (-3x + 1) is the same as the value of f (2). However, the limit of a function is not x→2 always equal to a function value. Example 2 Estimate a Limit ≠ f (c ) x - 9 _ 2 Estimate      lim   x - 3 using a graph. Support your conjecture using a table of values. x→3 Analyze Graphically The graph of f (x) = _   x - 3 suggests that as x gets closer to 3, x 2 - 9 the corresponding function value approaches 6. Therefore, we can estimate that lim _   x - 3 is 6. x 2 - 9 x→2 TechnologyTip Tables To help create a table using a graphing calculator, enter the function using the menu. Then use the Table function by pressing [TABLE]. To approach a specific value, change the starting point and interval for x in your table by pressing [TBLSET] and adjusting the TBLSET options. Support Numerically Make a table of values, choosing x-values that approach 3 from either side. x approaches 3 x 2.9 2.99 2.999 f(x ) 5.9 5.99 5.999 x approaches 3 3 3.001 3.01 3.1 6.001 6.01 6.1 The pattern of outputs suggests that as x gets closer to 3, f (x) gets closer to 6. This supports our graphical analysis. GuidedPractice Guided Practice 2A–B. See Chapter 11 Answer Appendix for graphs and tables. Estimate each limit using a graph. Support your conjecture using a table of values. 2A.  lim _   x→-2 x 2 - 4 x+2 2B.lim _   x→5 x-5 x 2 - 4x - 5 -0.25 6 In Example 2, notice that the limit as x approaches 3 is 6, even though f (3) ≠ 6. In fact, f (3) does not   x - 3 is not defined when x = 3. This illustrates an important point even exist, because the expression _ about limits. x 2 - 9 KeyConcept Independence of Limit from Function Value at a Point Words The limit of a function f (x ) as x approaches c does not depend on the value of the function at point c. Copyright © McGraw-Hill Education Symbols lim f (x ) = Llim g (x ) = Llim h (x ) = L x→c f (c ) is undefined. x→c g (c ) = n x→c h (c ) = L It is important to understand that a limit is not about what happens at the number that x is approaching. Instead, a limit is about what happens near or close to that number. 643 In finding limits using a table or a graph, we have looked at the value of f (x) as x approaches c from each side. We can describe the behavior of a graph from the left and right of x more concisely in terms of one-sided limits. KeyConcept One-Sided Limits ReadingMath One-Sided Limits The notation lim f (x ) can also be read as the x→c limit of f (x ) as x approaches c from below. The notation lim+ f(x) x→c can also be read as the limit of f (x ) as x approaches c from above. Left-Hand Limit Right-Hand Limit If the value of f (x ) approaches a unique number L 1 as x approaches c from the left, then If the value of f (x ) approaches a unique number L 2 as x approaches c from the right, then   lim- f (x ) = L1, which is read   lim+ f (x ) = L2 , which is read The limit of f ( x ) as x approaches c from the left is L 1. The limit of f ( x ) as x approaches c from the right is L 2 x→c x→c Using these definitions, we can state more concisely what it means for a two-sided limit to exist. KeyConcept Existence of a Limit at a Point The limit of a function f (x ) as x approaches c exists if and only if both one-sided limits exist and are equal. That is, if   lim f (x ) =   lim+ f (x ) = L, then lim f (x ) = L. x→c x→c x→c Example 3 Estimate One-Sided and Two-Sided Limits Estimate each one-sided or two-sided limit, if it exists. |2x| _ |2x| _ |2x| _ a.  lim-   x ,  lim+   x , and lim  x x→0  x→0 x→0  |2x| The graph of f (x) = _   x suggests that |2x|   lim- _   x = -2 x→0  |2x|   lim _   = 2. and x→0 + x Because the left- and right-hand limits of f (x) as x |2x| approaches 0 are not the same, lim _   x does not exist. x→0 ⎧ b.  lim- g (x),   lim+ g (x), and   lim g (x), where g (x) = ⎨ 4 if x ≠ - 3 ⎩ -2 if x = - 3 x→-3  x→-3 x→-3  The graph of g(x) suggests that   lim-g(x) = 4 x→-3  x→1  Because the left- and right-hand limits of g (x) as x approaches 3 are the same,   lim g (x) exists and is 4. x→-3 x→1  f(x ) = 3, lim f (x) = 3 x→1 3B. lim-g (x ) = 3, x→-2    lim g (x ) = -4, x→-2 +   lim g (x ) x→-2   lim g(x) = 4. x→-3 + does not exist GuidedPractice Guided Practice 3A.  lim- f (x),  lim+ f (x), and lim f (x), x→1  x→1  x→1 ⎧ x 3 + 2 if x < 1 where f (x) = ⎨ ⎩ 2x + 1 if x ≥ 1 644 | Lesson 11-1 | Estimating Limits Graphically 3B.  lim-g(x),   lim+g(x), and  lim g(x), x→-2  x→-2 x→-2  ⎧ -0.5x + 2 if x < - 2 where g(x) = ⎨ -x 2 if x ≥ - 2 ⎩ Copyright © McGraw-Hill Education 3A. lim- f (x ) = 3,  lim+ and Another way a limit can fail to exist is when the value of f (x) as x approaches c does not approach a fixed finite value. Instead, the value of f (x) increases without bound, indicated by ∞, or decreases without bound, indicated by -∞. Example 4 Limits and Unbounded Behavior Estimate each limit, if it exists. 1 _ a.lim   x→4 (x - 4)2 Analyze Graphically The graph of f (x) = _   2 suggests that 1 (x - 4)   lim- _ =∞ 2 1 1   lim _   = ∞, and x→4  (x - 4) 2 x→4 + (x - 4) because as x gets closer to 4, the function values of the graph increase. ReadingMath Without Bound For f (x ) to increase or decrease without bound as x → c means that by choosing an x-value arbitrarily close to c, you can obtain a function value with an absolute value that is as great as you want. The closer to c this x-value is chosen, the greater |f (x )| is. Neither one-sided limit at x = 4 exists; therefore, we can conclude that lim _   2 does not exist. However, 1 x→4 (x - 4) because both sides agree (both tend to ∞), we describe the behavior of f (x) at 4 by writing lim _   2 = ∞. 1 x→4 (x - 4) Support Numerically x approaches 4 x approaches 4 x 3.9 3.99 3.999 f (x ) 100 10,000 1,000,000 4 4.001 4.01 4.1 1,000,000 10,000 100 The pattern of outputs suggests that as x gets closer to 4 from the left and the right, f (x) grows without bound. This supports our graphical analysis. WatchOut! Infinite Limits It is important to understand that the expressions l im f (x ) = -∞ and lim f (x ) = ∞ x→0 - x→0 are descriptions of why these limits fail to exist. The symbols ∞ and -∞ do not represent real numbers. _1 b.lim   x x→0 Analyze Graphically The graph of f (x) = _   x suggests that 1   lim- _   x = -∞ 1 and x→0    lim _   = ∞, 1 x→0 + x because as x gets closer to 0, the function values from the left decrease and the function values from the right increase. Neither one-sided limit at x = 0 exists; therefore, lim _   x does 1 x→0 not exist. In this case, we cannot describe the behavior of f (x) at 0 using a single expression because the unbounded behaviors from the left and right differ. Support Numerically Copyright © McGraw-Hill Education x approaches 0 x approaches 0 x -0.1 -0.01 -0.001 f (x) -10 -100 -1000 0 0.001 0.01 0.1 1000 100 10 The pattern of outputs suggests that as x gets closer to 0 from the left and the right, f (x) decreases and increases without bound, respectively. This supports our graphical analysis. GuidedPractice GuidedPractice x 2 - 4 does not exist 4A.lim _   x→3 x - 3 2 _ 4B.lim - _ 4 l im - 4 = -∞ x→0 x x→0 x 2 645 A limit can also fail to exist if instead of approaching a fixed value, f (x) oscillates or bounces back and forth between two values. Example 5 Limits and Oscillating Behavior TechnologyTip _1 Estimate lim cos   x , if it exists. x→0 Infinite Oscillations The TRACE feature on a graphing calculator can be useful in estimating limits. However, you cannot always trust what a graphing calculator tells you. In the case of the function in Example 5, the calculator uses a finite number of points to produce the graph, but close to 0 this function has infinite oscillations. The graph of f (x) = cos _   x suggests that as x gets closer to 0, the corresponding function values oscillate between -1 and 1. This means that for an x 1 -value close to 0 such that f (x 1 ) = 1, you can always find an x 2 -value closer to 0 such that f (x 2 ) = -1. Likewise, for an x 3-value close to 0 such that f (x 3 ) = -1, you can always find an x 4-value closer to 0 such that f (x 4 ) = 1. 1 Therefore, lim cos _   x does not exist. 1 x→0 GuidedPractice Guided Practice Estimate each limit, if it exists. 5A.lim sin _   x x→0 1 does not exist 5B.lim (x 2 sin x) x→0 0 The three most common reasons why the limit of a function fails to exist at a point are summarized below. ConceptSummary Why Limits at a Point Do Not Exist The limit of f (x ) as x approaches c does not exist if: • • • 2 f (x) approaches a different value from the left of c than from the right, f (x) increases or decreases without bound from the left and/or the right of c, or f (x) oscillates between two fixed values. Estimate Limits at Infinity To this point, limits have been used to describe how a function f (x) behaves as x approaches a fixed value c. You learned that limits can also be used to describe the end behavior of a function, that is, how a function behaves as x increases or decreases without bound. The notation for such limits is summarized below. KeyConcept Limits at Infinity • If the value of f (x ) approaches a unique number L 1 as x increases, then   lim f (x ) = L 1, which is read the limit of f ( x ) as x approaches infinity is L 1. x→∞ If the value of f (x ) approaches a unique number L 2 as x decreases, then   lim f (x ) = L 2 , which is read the limit of f ( x ) as x approaches negative infinity is L 2. x→-∞ You learned that unbounded behavior that can be described by ∞ or -∞ indicates the location of a vertical asymptote. You also learned that the existence of a limit at infinity indicates the location of a horizontal asymptote. That is, • • x = c is a vertical asymptote of the graph of f (x) if  lim-f (x) = ±∞ or  lim+f (x) = ±∞, and x→c x→c y = c is a horizontal asymptote of the graph of f (x) if   lim f (x) = c or  limf (x) = c. 646 | Lesson 11-1 | Estimating Limits Graphically x→-∞ x→∞ Copyright © McGraw-Hill Education • Example 6 Estimate Limits at Infinity Estimate each limit, if it exists. _1 a.  lim   x x→∞ Analyze Graphically The graph of f (x) = _   x suggests that 1 1   lim _   = 0. As x increases, f (x) gets closer to 0. x→∞ x Support Numerically StudyTip Asymptotes The limit in Example 6a indicates that there is an asymptote at y = 0, while the limit in Example 6b indicates that there is an asymptote at y = 2. x approaches ∞ x 10 100 1000 10,000 100,000 f (x ) 0.1 0.01 0.001 0.0001 0.00001 The pattern of outputs suggests that as x grows increasingly larger, f (x) approaches 0. ✔ ( _3 ) b.  lim - 2 + 2 x→-∞ x Analyze Graphically The graph of f (x) = - _ 2 + 2 ( 3 x ) suggests that   lim - _ 2 + 2 = 2. As x decreases, f (x) x→-∞ gets closer to 2. 3 x Support Numerically x approaches -∞ x f (x ) -100,000 -10,000 -1000 -100 -10 1.99999 1.99999 1.99999 1.9997 1.97 The pattern of outputs suggests that as x decreases, f (x) approaches 2. c.  lim e x sin 3πx and  lim e x sin 3πx x→-∞ x→∞ Analyze Graphically The graph of f (x) = e x sin 3xπ suggests that  lim e x sin 3πx = 0. As x decreases, f (x) x→-∞ oscillates but tends toward 0. The graph suggests that  lime x sin 3πx does not exist. x→∞ As x increases, f (x) oscillates between ever increasing values. Support Numerically Copyright © McGraw-Hill Education WatchOut! Oscillating Behavior Do not assume that because a function f (x ) exhibits oscillating behavior, it has no limit as x approaches either ∞ or -∞. If the oscillations fluctuate between two fixed values or are unbounded, then the limit does not exist. If, however, the oscillations diminish and approach a fixed value, then the limit does exist. x approaches ∞ x approaches -∞ -100 x 3 × 10  f (x ) -50 -10 -2.0 × 10  -44 -0.00005 -22 0 0 10 21966 50 100 21 4.8 × 10  -2.0 × 10 43 The pattern of outputs suggests that as x decreases, f (x) approaches 0 and as x increases, f (x) oscillates. GuidedPractice Guided Practice 1 6A.  lim _ - 3 -3 ( x→∞ x 4 ) 6B.  lim e x x→-∞ 0 6C.   limsin x x→∞ does not exist 647 You can use graphical and numerical approaches to estimate limits at infinity in many real-world situations. Real-World Example 7 Estimate Limits at Infinity a. HYDRAULICS A low-energy swing door uses a spring to close the door and a hydraulic mechanism to dampen or slow down the door’s movement. If the π door is opened to an angle of   4 and then released _ from this resting position, the angle θ of the door t seconds after it is released is given by π θ (t) =   (1 + 2t)e -2 t. Estimate limθ (t), if it exists, 4 _ and interpret your result. Real-WorldLink swing door operator is a device A that opens and closes a swing door at a reduced speed to assist people who use wheelchairs. t→∞ Estimate the Limit Graph θ(t) = _   4 (1 + 2t)e -2 tusing a graphing calculator. π The graph shows that when t = 0, θ (t) ≈ 0.785 or about _   4 . Notice that as t increases, the function values π of the graph tend toward 0. So, we can estimate that  limθ (t) is 0. t→∞ Interpret the Result A limit of 0 in this situation indicates that the angle the door makes with its closed position tends toward a measure of 0 radians. That is, as the number of seconds after the door has been released increases, the door comes closer and closer to being completely closed. b. MEDICINE The concentration in milligrams per milliliter of a certain medicine in a patient’s t bloodstream t hours after the medicine has been administered is given by C(t ) = Ate -0.18 , where A is a positive constant. Estimate lim C(t ), if it exists, and interpret your result. t→∞ Estimate the Limit t Graph the related function C 1(t) = te -0.18 using a graphing calculator. The graph shows that as t increases, the function values of the graph tend toward 0. So, we can estimate that  limC1 (t) is 0. t→∞ exist because as t increases, the height of the graph oscillates between -165 and 165. Over time, the voltage of an electrical outlet will fluctuate between -165 and 165. 7B. lim P (t ) = 230. t→∞ Over time, the fruit fly population will approach a maximum of 230. Because A is a positive constant, the graph of C (t) = Ate -0.18t will be the graph of 1(t) = te -0.18texpanded vertically by a factor of A but will still tend to 0 as t increases C without bound. So, we can estimate that  limC (t) is also 0. t→∞ Interpret the Result A limit of 0 in this situation indicates that all of the medicine will eventually be eliminated from the patient’s bloodstream. GuidedPractice Guided Practice 7A. ELECTRICITY The typical voltage V supplied by an electrical outlet in the United States can be modeled by the function V(t) = 165 sin 120πt, where t is the time in seconds. Estimate   limV(t), if it exists, and interpret your result. t→∞ 7B. BIOLOGY Fruit flies are placed in a half-pint milk bottle with a piece of fruit and a yeast      . EstimatelimP(t), plant. The fruit fly population after t days is given by P(t) = __ t→∞ 1 + 56.5e -0.37t if it exists, and interpret your result. 648 | Lesson 11-1 | Estimating Limits Graphically 230 Realistic Reflections t→∞ Copyright © McGraw-Hill Education 7A. lim V (t ) does not Exercises 1–10. See Chapter 11 Answer Appendix for graphs and tables. Estimate each limit using a graph. Support your conjecture using a table of values. (Examples 1 and 2) 1.lim(4x - 10) x→5 1 2. lim _   2 x 5 - 2x 3 + 3x 2 10 x→2 3.  lim (x 2 + 2x - 15) x→-2 3 5.lim(2x - 10x + 1) x→3 ) -15 x 3 + 8 4.   lim _   2 25 x cos x 6. lim _   2 1 0 7.lim[5(cos 2 x - cos x)] x→0 9.lim(x + sin x) x→6 ( x→-2 x - 4 -3 8. lim _   √ x-4 x - 2  x→4   x→0  sin x - x x x→∞ 4 x→-5 -9 |4x| 12.   lim- _   -4 x→0  x 2x 2 13.lim _   0 3 -   √x 14.   lim+ _   x - 9 -0.1667 x 2 - 5x + 6 15.  lim- _   x - 3 1 16.   lim-_   0 x 1 x→-_  x→0 |x| x→9  |2x + 1| x→3  2 18.   lim _   x + 7 -15 19.  lim- (  √ -x - 7) 20. lim _   x - 5 10 x 2 + 5x + 6 |x + 2| does not exist -7 x→0  21.  lim+ _   2 x→4  x - x - 12 |x - 4| |3x| 23.lim _   x→0 2x 7 43.  lim _   x -x -1 x→-∞ 3  - 3  44. lim _   45.lim x 2 cos _   0 x x→0 46.   lim _   2 x→-∞ 2x + 8 3 x→∞ 3 x+ 3 -x x→0 sin |x| x does not exist does not exist 4x - 13 1 47. MEDICINE A vaccine was quickly administered to combat a small outbreak of a minor infection. The number of reported cases of the infection w weeks since the vaccine was administered is shown. (Example 7) x→5 3 |x + 1| 24.   lim _   2 does not exist x→-1 x - 1 ⎧ if x < 3 if x ≥ 3 0 9 a. Use the graph to estimate  lim f (w) and  lim f (w). w→1 w→3 b. Use the graph to estimate  lim f (w) if it exists, and w→∞ interpret your results. 48. TRACK AND FIELD The logistic function ⎧ x - 5 if x < 0 does not exist 27.lim f (x) where f (x) = ⎨⎩ 2 x→0 x + 5 if x ≥ 0 ⎧ -x 2 + 2 if x < 0 2 28.lim f (x) where f (x) = ⎨⎩ 2x x→0  _ if x ≥ 0 x For each function below, estimate each limit if it exists. (Examples 1–4) Copyright © McGraw-Hill Education 42.   limx cos x x→∞ 9x + 3 x→-3 x 2 - 25 √-x if x < 0 25.lim f (x) where f (x) = ⎨⎩  x→0 √ x if x ≥ 0 ⎧ 1 3x - 4 _ 41.  lim_     x 2 - x - 56 x→0  3x 26.lim f (x) where f (x) = ⎨⎩ 2 x→3 x 4x 3 - 13 x→-7 22.   lim+ (  √ x + 2x + 3) does not exist -∞ x 2 + 9x + 20 __ 40.   lim   does not exist x+3 a–b. See margin. 17.  lim _   x→-2 38.   lim (x 5 - 7x 4 - 4x + 1) x→-∞ x 2 + x - 22 39.  lim _   0 x 2 + x - 20 0 x 2 -17 34.   lim __      ∞ 2 x→-4 x + 8x + 16 x→5 x - 10x + 25 |x| 35.lim _   does not exist 36.   lime 2x - 5 ∞ x→∞ x→4 x - 4 5 Estimate each one-sided or two-sided limit, if it exists. (Example 3) 11.  lim+ _   33.  lim __   2 -∞ 37.lim _   ∞ x→6 (x - 6) 2 x→0 x + x 10.   lim _   x + 5 5.72 12 Estimate each limit, if it exists. (Examples 4–6) f (x) = __      -0.129x , where x is the number of years 5.334 1 + 62548.213e since 1900, models the world record heights in meters for women’s pole vaulting from 1996 to 2008. (Example 7) a. Graph the function for 96 ≤ x ≤ 196. See margin. ≈5.334 b. Estimate  lim__      -0.129x , if it exists. 5.334 x→∞ 1 + 62548.213e c. Explain the relationship between the limit of the function and the world record heights. See margin. 49. INTERNET VIDEO A group of friends created a video parody of several popular songs and posted it online. As word of the video spread, interest grew. A model that can be used to estimate the number of people p that viewed the video is p (d) = 12(1.25012) d- 12, where d is days since the video was originally posted. (Example 7) a. Graph the function for 0 ≤ d ≤ 20. 4 31. limg(x) -6 32.   lim g(x) 29.  lim f (x) x→-6 30.limf (x) x→4 x→4 x→-6 ∞ does not exist a–c. See margin. b. Estimate the number of people who viewed the video by the ends of the 5th, 10th, and 20th days. How many people will have viewed it after 2 months? (Use d = 60.) c. Estimate  limp (d), if it exists, and interpret your d→∞ results. 649 50. TECHNOLOGY The number of cell phone owners between the ages of 18 and 25 has steadily increased since the 1990s. A sequence model that can estimate the number of people ages 18–25 per cell phone is a n= 64.39(0.82605) n + 1, where n represents years since 1993. (Example 7) a,d. See a. Graph the function for the years 1993 to 2011. margin. b. Use the graph to estimate the amount of people per cell phone for 1998, 2007, and 2011. 25.77; 5.44; 3.07 c. Use your graph to estimate  lima n. 1 n→∞ d. Explain the relationship between the limit of the function and the number of people per cell phone. 51. CHEMICALS An underground pipeline is leaking a toxic chemical. After the leak began, it spread as shown below. The distance the chemical spreads each year can be defined as d(t) = 2000(0.7) t - 1, for t ≥ 1, where t is years since the leak began. (Example 7) See Chapter 11 Answer Appendix. a. Graph the function for 1 ≤ t ≤ 15. b. Use your graph to find values of d for t = 5, 10, and 15 years. 480.2; 80.71; 13.56 c. Use your graph to estimate  limd(t). t→∞ 0 d. Will the chemical ever spread to a hospital that is located 7000 meters away from the leak? Recall that the sum of an infinite geometric series can be found See Chapter 11 Answer Appendix. a. Graph the function for 0 ≤ t ≤ 10. c. Use your graph to estimate  limv (t). t→∞ 0 AED 707.18 d. Explain the relationship between the limit of the function and the value of Saeed’s motorcycle. If Saeed keeps his motorcycle, it will eventually be worth AED 0. For the function below, estimate each limit if it exists. -1 54.  lim+ f (x) 4 x→0  x→0  55.lim f (x) x→0 does not exist 56.  lim- f (x) 1 x→2  57.  lim+ f (x) x→2  58.lim f (x) x→1 60. lim _   2 x 2 - 1 x 2 + x x→2 x - x - 2 |x + 5| no; different 62.   lim _   x + 5 one-sided x→-5 no; verticalπ asymptote 61.lim3 cos _   x x→0 limits   √ 2 - x - 3 _ 64.   lim   x + 4 no; oscillations 7 63.lim _   x→0 1 _   x→-4 1 + e x no; vertical asymptote no; vertical asymptote H.O.T. Problems Use Higher-Order Thinking Skills 65. ERROR ANALYSIS Mazen and Ayoub are finding the limit of the function below as x approaches -6. Mazen says that the limit is -4. Ayoub disagrees, arguing that the limit is 3. Is either of them correct? Explain your reasoning. Neither;sample answer: If f (x ) approaches a different value from the left than from the right, then the limit at that point does not exist. 66. OPEN ENDED Give an example of a function f such that   lim f (x) exists but f (0) does not exist. Give an example of a x→0 function g such that g (0) exists but lim g (x) does not exist. x→0 See Chapter 11 Answer Appendix. x 2 + 1 x+1 67. CHALLENGE Suppose f (x) = _   x - 1 and g (x) = _   2 . x - 4 Estimate lim f (x) and lim g (x). If g(a) = 0 and f (a) ≠ 0, x→1 x→2 f (x) what can you assume about lim_   ? Explain your x→a g(x) reasoning. 68. REASONING Determine whether the following statement is always, sometimes, or never true. Explain your reasoning. If f (c) = L, then limf (x) = L. Answer Appendix. b. Use your graph to estimate the value of the motorcycle for t = 3, 7, and 10 years.AED 4828.74; AED 1610.97; 53.  lim- f (x) 59.lim _   x→1 x 2 - 2x + 1 See Chapter 11 Answer Appendix. 52. DEPRECIATION Saeed purchases a motorcycle for AED 11,000, and it depreciates each year that he owns it. The value v of the motorcycle after t years can be estimated by the model v (t) = 11,000(0.76) t. (Example 7) a. See Chapter 11 B 60. no; vertical asymptote GRAPHING CALCULATOR Determine whether each limit exists. If not, describe what is happening graphically at the limit. 6 2.5 650 | Lesson 11-1 | Estimating Limits Graphically x→c See Chapter 11 Answer Appendix. 69. OPEN ENDED Sketch the graph of a function such that lim f (x) = -3, f (0) = 2, f (2) = 5, and lim f (x) does not exist. x→0 x→2 See Chapter 11 Answer Appendix. 70. CHALLENGE For the function below, estimate each limit if it exists. ⎧ ⎪ f (x) = ⎨ ⎪ ⎩ 2x + 4 if x < -1 -1 if -1 ≤ x ≤ 0 x 2 if x-3 a.   lim f (x) x→-1 does not exist if 0<x≤2 x>2 b.   lim- f (x) x→0  -1 c.   lim+ f (x) -1 x→2  71. WRITING IN MATH Explain what method you could use to estimate limits if a function is continuous. Explain how this differs from methods used to estimate functions that are not continuous. See Chapter 11 Answer Appendix. Copyright © McGraw-Hill Education a 1 by _   1 - r . C Spiral Review a–d. See Chapter 11 Answer Appendix. 72. FUEL ECONOMY The table shows various engine sizes available from an auto manufacturer and their respective fuel economies. a. Make a scatter plot of the data, and identify the relationship. Engine Size (liters) Highway Mileage (km/L) b. Calculate and interpret the correlation coefficient. Determine whether it is significant at the 10% level. c. If the correlation is significant at the 10% level, find the least-squares regression equation and interpret the slope and intercept in context. d. Use the regression equation that you found in part c to predict the expected kilometers per liter that a car would get for an engine size of 8.0 liters. State whether this prediction is reasonable. Explain. 32 16 2 4 _ _ 73. Use Pascal’s triangle to expand 3a + _   3 b . 81a 4 + 72a 3b + 24a 2b 2 +   ab 3 +   b 4 ( ) 81 9 Write and graph a polar equation and directrix for the conic with the given characteristics. 74. 75. 4 _ e = 1; vertex at (0, -2) r =   1 - sin θ 12 _ e = 3; vertices at (0, 3) and (0, 6) r =   74–75. See margin for graphs. 1 + 3 sin θ 1.6 34 2.2 37 2.0 30 6.2 26 7.0 24 3.5 29 5.3 24 2.4 33 3.6 26 6.0 24 4.4 23 4.6 24 Find the angle between each pair of vectors to the nearest tenth of a degree. 76. u = 〈2, 9, -2〉, v = 〈-4, 7, 6〉 63.0° 77. m = 3i - 5j + 6k and n = -7i + 8j + 9k Use a graphing calculator to graph the conic given by each equation. 2 2 78. 7 x - 50xy + 7y = -288 2 93.4° 78–79. See margin. 2 79. x - 2  √ 3 xy + 3y + 16  √ 3 x + 16y = 0 Skills Review for Standardized Tests 83. SAT/ACT What is the area of the shaded region? A 5 C 7 B 6 D 8 A E 9 Copyright © McGraw-Hill Education 84. REVIEW Which of the following best describes the end behavior of f (x) = x 10 - x 9 + 5x 8 ? G F f (x) → ∞ as x → ∞, f (x) → -∞ as x → ∞ G f (x) → ∞ as x → ∞, f (x) → ∞ as x → -∞ H f (x) → -∞ as x → ∞, f (x) → ∞ as x → -∞ J f (x) → -∞ as x → -∞, f (x) → ∞ as x → ∞ 85. According to the graph of y = f (x), lim f (x) = x→0 C A 0 C 3 B 1 D The limit does not exist. 86. REVIEW Which of the following describes the graph of g(x) = _   2 ? 1 x F I It has an infinite discontinuity. II It has a jump discontinuity. III It has a point discontinuity. F I only G II only J I and III only K I, II and III H I and II only 651 11-2 Then You estimated limits using graphical and numerical methods. Now 1 2 NewVocabulary direct substitution indeterminate form Evaluating Limits Algebraically Why? E valuate limits of polynomial and rational functions at selected points. E valuate limits of polynomial and rational functions at infinity. 1 Suppose the width in millimeters of an animal’s pupil is given by d (x ) =  __    , where x is the illuminance of -0.45 152x -0.45 + 85 4x + 10 the light shining on the pupils measured in lux. You can evaluate limits to find the width of the animal’s pupils when the light is at its minimum and maximum intensity. Compute Limits at a Point In the lesson 11-1 You learned how to estimate limits by using a graph or by making a table of values. In this lesson, we will explore computational techniques for evaluating limits. KeyConcept Limits of Functions Limits of Constant Functions WordsThe limit of a constant function at any point c is the constant value of the function. Symbols      lim k = k x→c Limits of the Identity Function Words The limit of the identity function at any point c is c. Symbols x→c      lim x = c If k and c are real numbers, n is a positive integer, and     lim f (x ) and     lim g (x ) both exist, then the following are true. x→c x→c Sum Property    lim [f (x ) + g (x )] =      lim f (x ) +      lim g (x ) x→c x→c x→c Difference Property    lim [f (x ) - g (x )] =      lim f (x ) -      lim g (x ) x→c x→c x→c Scalar Multiple Property    lim [kf (x )] = k      lim f (x ) x→c x→c Product Property    lim [f (x ) · g (x )] =      lim f (x ) ·      lim g (x ) x→c x→c x→c   lim f (x )    f (x ) x→c Quotient Property    lim _   =_   , if      lim g (x ) ≠ 0 x→c g (x )      lim g (x ) x→c x→c n ⎤ ⎡ x→c   lim f (x )⎦ Power Property    lim [f (x )  ] = ⎣    x→c n n  n th Root Property    lim   √ f (x )  =   √      lim f (x ) , if      lim f (x ) > 0 when n is even. n x→c 652 | Lesson 11-2 x→c x→c Copyright © McGraw-Hill Education KeyConcept Properties of Limits Purestock/Getty Images When combined with the following properties of limits, these constant and identity function limits become very useful. StudyTip Limit Properties Each of the limit properties given on the previous page also hold for one-sided limits and limits at infinity, so long as each limit exists. Example 1 Use Limit Properties Use the properties of limits to evaluate each limit. a. lim    ( x 2 - 6x + 3) x→4  lim lim x 2 -      lim 6x +      lim 3    (x 2 - 6x + 3) =    x→4 x→4 x→4 Sum and Difference Properties x→4 = (     lim x)  2 - 6 ·      lim x +      lim 3 Power and Scalar Multiple Properties = 4 2 - 6 · 4 + 3 Limits of Constant and Identity Functions = -5 Simplify. x→4 x→4 x→4 CHECK The graph of f (x) = x 2 - 6x + 3 supports this result. ✔ 4x + 1 _ 3 b.    lim   x→-2 x - 5     lim (4x 3 + 1) __     lim _   x - 5 =      4x 3 + 1 x→-2 Quotient Property     lim (x - 5) x→-2 x→-2     lim 4x 3 +      lim 1 x→-2 x→-2 = __          lim x -      lim 5    x→-2 Sum and Difference Properties x→-2 (x→-2 ) x→-2 __ =           lim x -      lim 5    4      lim x   3 +      lim 1 x→-2 Scalar Multiple and Power Properties x→-2 4(-2) 3 + 1 =_   -2 - 5 Limits of Constant and Identity Functions =_   7 Simplify. 31 CHECK Make a table of values, choosing x-values that approach -2 from either side. ✔ x approaches -2 x approaches -2 x -2.1 -2.01 -2.001 f(x) 5.08 4.49 4.43 -2 -1.999 -1.99 -1.9 4.42 4.37 3.83 c. lim 8 - x      √ x→3 √x→3   lim 8 - x =        lim (8 - x)       √ Copyright © McGraw-Hill Education x→3 n th Root Property =   √        lim 8 -     lim x  Difference Property  =   √8 - 3  =   √ 5  Limits of Constant and Identity Functions x→3 x→3 Simplify. GuidedPractice Guided Practice 1A. lim    (-x 3 + 4) x→2 -4 1B. lim   2    __ x-3 x→2 2x - x - 15 1 _   9 1C.      lim   √ x + 3  x→-1   √ 2  Notice that for each of the functions in Example 1, the limit of f (x) as x approaches c is the same value that you would get if you calculated f (c). While this is not true of every function, it is true of polynomial functions and of rational functions as described at the top of the next page. 653 StudyTip Well-Behaved Functions Continuous functions such as polynomial functions are considered well-behaved, since limits of these functions at any point can be found by direct substitution. The limits of functions that are not well-behaved over their entire domain can still be found using this method, so long as the function is continuous at the domain value in question. KeyConcept Limits of Functions Limits of Polynomial Functions If p(x ) is a polynomial function and c is a real number, then     lim p(x ) = p(c ). x→c Limits of Rational Functions p(c ) p (x ) If r (x ) =   _ is a rational function and c is a real number, then     lim r (x ) = r (c ) = _     if q(c ) ≠ 0. q (x) x→c q (c ) More simply stated, limits of polynomial and rational functions can be found by direct substitution, so long as the denominator of the rational function evaluated at c is not 0. Example 2 Use Direct Substitution Use direct substitution, if possible, to evaluate each limit. If not possible, explain why not. a.    lim (-3x 4 + 5x 3 - 2x 2 + x + 4) x→-1 Since this is the limit of a polynomial function, we can apply the method of direct substitution to find the limit.     lim (-3x 4 + 5x 3 - 2x 2 + x + 4) = -3(-1) 4 + 5(-1) 3 - 2(-1) 2 + (-1) + 4 x→-1 = -3 - 5 - 2 - 1 + 4 or -7 CHECK The graph of f (x) = -3x 4 + 5x 3 - 2x 2 + x + 4 supports this result. ✔ 2x - 6 _ 3 b. lim      x→3 x - x 2 This is the limit of a rational function, the denominator of which is nonzero when x = 3. Therefore, we can apply the method of direct substitution to find the limit. 2(3) 3 - 6 _  lim      _ 2 = 2 2x 3- 6 x→3 x - x x - 1 _ 3 - 3 48 =_   -6 or -8 2 c. lim      x - 1 x→1 This is the limit of a rational function. Since the denominator of this function is 0 when x = 1, the limit cannot be found by direct substitution. 2C. Not possible; when x = -8 the function x + 6  f(x ) =   √ is   √ -2 , which is not a real number. 2A. lim    (x 3 - 3x 2 - 5x + 7) x→4 _ x+1 3 2B.    lim _   2 - 7 1 x→-5 x + 3 2C.      lim   √ x + 6  x→-8 Suppose you incorrectly applied the Quotient Property of Limits or direct substitution to   lim _   x - 1 . evaluate    x 2 - 1 x→1     lim (x 2 - 1) x 2 - 1 x→1 0 1 2 - 1   lim _   =_   =_   or _      x→1 x - 1     lim (x - 1) x→1 654 | Lesson 11-2 | Evaluating Limits Algebraically 1-1 0 This is incorrect because the limit of the denominator is 0. Copyright © McGraw-Hill Education GuidedPractice Guided Practice It is customary to describe the resulting fraction _  0 as having an 0 indeterminate form because you cannot determine the limit of the function with 0 in the denominator. A limit of this type may exist and have a real number value, or it may not exist, possibly diverging to ∞ or -∞. In this case, from the graph of f (x) = _   x - 1 , x 2 - 1   x - 1 does exist and has a value of 2. it appears that     lim _ 2 x - 1 x→1 While an indeterminate form limit results from an incorrect application of limit properties or theorems, an analysis of this form can provide a clue as to what technique should be applied to find a limit. If you evaluate the limit of a rational function and reach the indeterminate form _  0 , you should try 0 to simplify the expression algebraically by factoring and dividing out a common factor. Example 3 Use Factoring Evaluate each limit. a.    lim   x→-4 x - x - 20 _ 2 x+4 (-4) 2 - (-4) - 20 By direct substitution, you obtain __      or _   0 . Since this is an indeterminate form, try -4 + 4 0 factoring and dividing out any common factors. (x - 5)(x + 4)     lim _   x + 4 =      lim __   x+4 x 2 - x - 20 x→-4 Factor the numerator. x→-4 (x - 5)(x + 4) x+4 =      lim __      Divide out the common factor. =      lim (x - 5) Simplify. = (-4) - 5 or -9 Apply direct substitution and simplify. x→-4 x→-4 CHECK The graph of f (x) = _   x + 4 supports this result. ✔ x 2 - x - 20 WatchOut! Factoring If the entire expression in the numerator is divided out, the result is a 1, not a 0. x-3 __    b. lim      x→3 x 3 - 3x 2 - 7x + 21 By direct substitution, you obtain __        or _   . 3 2 3-3 3 - 3(3) - 7(3) + 21  lim      =      lim  __       __ 3 2 2 x-3 Copyright © McGraw-Hill Education x→3 x - 3x - 7x + 21 x-3 x→3 (x - 7)(x - 3) 0 0 Factor the denominator. =      lim __      2 Divide out the common factor. =      lim  _ 2 Simplify. =_   2 or _   x-3 x→3 (x - 7)(x - 3) 1 x→3 x - 7 1 (3) - 7 GuidedPractice Guided Practice x 3 - 3x 2 - 4x + 12 __ 3A.    lim      20 x→-2 x+2 1 2 Apply direct substitution and simplify. __ 3B. lim            2 x 2 - 7x + 6 x→6 3x - 11x - 42 1 _   5 655 While this method of dividing out a common factor is valid, it requires some justification. In Example 3a, dividing out a common factor from f (x) resulted in a new function, g (x), where f(x) = _   x + 4 and g(x) = x - 5. x 2 - x - 20 These two functions yield the same function values for every x except when x = -4. If two functions differ only at a value c in their domain, their limits as x approaches c are the same. This is because the value of a limit at a point is not dependent on the value of the function at that point.   x + 4 =      lim (x - 5). Therefore,      lim _ x 2 - x - 20 x→-4 x→-4 Another technique to find limits that have indeterminate form is to rationalize the numerator or denominator of a function and then divide out any common factors. Example 4 Use Rationalizing   √ x - 3 _ Evaluate      lim   . x→9 x - 9 By direct substitution, you obtain  _ or _   0 . Rationalize the numerator of the fraction, and 9-9   √ 9  - 3 0 then divide common factors.   √ x  + 3 _     lim _   x- 9 =      lim _   x - 9 ·     √x - 3 x→9   √x - 3 x→9   √x  +3 =      lim __      x-9 x + 3) x→9 (x - 9)(  √ Multiply the numerator and denominator by  √x + 3, the conjugate of  √x - 3. Simplify. x-9 =      lim  __    Divide out the common factor. 1 =      lim _   √ Simplify. =_   Apply direct substitution. x + 3) x→9 (x - 9) (  √ x→9   x  +3 1   √ 9  + 3 1 =_   6 Simplify. CHECK The graph of f(x) = _   x- 9 in Figure 11.2.1 supports this result. ✔   √x - 3 Figure 11.2.1 GuidedPractice Guided Practice Evaluate each limit. 4A.    lim _   √ x→25   x - 25 x- 5 10 4B. lim      _ x→0 2 -   √ x + 4  x -_ 1 4 2 Compute Limits at Infinity You learned that all even-degree power functions have the same end behavior, and all odd-degree power functions have the same end behavior. This can be described in terms of limits as shown below. KeyConcept Limits of Power Functions at Infinity lim x n= ∞. •    x→∞ lim x n= ∞ if n is even. •     x→-∞ lim x n= -∞ if n is odd. •     x→-∞ You also learned that the end behavior of a polynomial function is determined by the end behavior of the power function related to its highest-powered term. This can also be described using limits. 656 | Lesson 11-2 | Evaluating Limits Algebraically Copyright © McGraw-Hill Education For any positive integer n, StudyTip Products with Infinity Since a limit of ∞ means that function values are increasingly large positive numbers, multiplying these numbers by a positive constant does not change this trend. However, multiplying a limit of ∞ by a negative constant changes the sign of all the outputs suggested by this notation. Thus, -1(∞) = -∞. KeyConcept Limits of Polynomial Functions at Infinity Let p be a polynomial function p(x) = a n x n+ … + a 1x + a 0 . Then      lim p(x ) =     lim a nx nand       lim p (x) =       lim a n x n. x→∞ x→∞ x→-∞ x→-∞ You can use these properties to evaluate limits of polynomial functions at infinity. Remember, notating that the limit of a function is ∞ or -∞ does not indicate that the limit exists but instead describes the behavior of the function as increasing or decreasing without bound, respectively. Example 5 Limits of Polynomial Functions at Infinity Evaluate each limit. a.  lim (x 3 - 2x 2 + 5x - 1) x→-∞   lim x 3     lim (x 3 - 2x 2 + 5x - 1) =    Limits of Polynomial Functions at Infinity Limits of Power Functions at Infinity x→-∞ x→-∞ = -∞ b.  lim (4 + 3x - x 2 ) x→∞     lim (4 + 3x - x 2 ) =      lim -x 2 x→∞ Limits of Polynomial Functions at Infinity x→∞ = -   lim x 2 Scalar Multiple Property = -∞ Limits of Power Functions at Infinity x→∞ c.  lim (5x 4 - 3x) x→-∞     lim (5x 4 - 3x) =      lim 5x 4 x→-∞ Limits of Polynomial Functions at Infinity x→-∞ 4 = 5      lim x Scalar Multiple Property = 5 · ∞ or ∞ Limits of Power Functions at Infinity x→-∞ GuidedPractice Guided Practice 3 Evaluate each limit. 2 5A.    lim (-x - 4x + 9) x→∞ -∞ 5B.    lim (4x 6 + 3x 5 - x) ∞ x→-∞ 5C.      lim (2x - 6x 2 + 4x 5 ) -∞ x→-∞ To evaluate limits of rational functions at infinity, we need another limit property. KeyConcept Limits of Reciprocal Function at Infinity Words The limit of a reciprocal function at positive or negative infinity is 0. Copyright © McGraw-Hill Education 1 1 Symbols    lim _   =       lim _   = 0 x→∞ x x→-∞ x Corollary For any positive integer n,       lim _   n = 0. x→±∞ 1 x If we divide the numerator and denominator of a rational function by the highest power of x that occurs in the function, we can use this property to find limits of rational functions at infinity. 657 Example 6 Limits of Rational Functions at Infinity Evaluate each limit. 4x + 5 _ a.  lim   8x - 3 x→∞ 4x 5 _   + _   x x 4x + 5     lim _   =      lim _   8x 3 _ x→∞ 8x - 3 x→∞ _   Divide each term by highest-powered term, x. -   x x 4+_   5 x =      lim _   x→∞   lim 4 + 5  lim _   1 Quotient, Sum, Difference and Scalar Multiple Properties     lim 8 - 3    lim   x Evaluating Limits Using a calculator is not a foolproof way of evaluating      lim f (x) or       lim f(x ). x→∞ x→∞ =_   8 - 3 · 0 or _   2 Limits of Reciprocal Function at Infinity 4+5·0 x→±∞ You may only analyze the values of f (x) for a few values of x near c or for a few large values of x. However, the function may do something unexpected as x gets even closer to c or as x gets even larger or smaller. You should use algebraic methods whenever possible to find limits. 3 x       x x→∞ x→∞ = __         _1 TechnologyTip x→c Simplify. 8-_   Limit of a Constant Function and 1 CHECK The graph of f (x) = _   8x - 3 in supports this result. ✔ 4x + 5 6x - x _ b.     lim   2 3 x→-∞ 3x + 1 _   3 - _   3 6x 2 - x x x _ _     lim   3 =      lim   3 x→-∞ 3x + 1 x→-∞ _ 3x _ 1   +   6x 2 x x 3 Divide each term by highest-powered term, x 3 . x 3 6 1 _   - _   x =      lim _   1 _ x 2 Simplify. x→-∞ 3 +   x 3 6    lim _   x -      lim _   2 1 1 = __        1 _ x→-∞ x x→-∞     lim 3 +      lim   3 Quotient, Sum, Difference and Scalar Multiple Properties x→-∞ x x→-∞ Limit of a Constant Function and =_   3 + 0 or 0Limits of Reciprocal Function at Infinity 6·0-0 5x _ c.    lim   4 x→∞ 9x 3 + 2x     lim _   3 =      lim _   9 2 Divide each term by highest-powered term, x 4 . Then simplify. 5x 4 x→∞ 9x + 2x 5 x→∞ _   + _   x x 3     lim 5 = __      1 _1 _ x→∞ 9    lim   + 2    lim   3 x x→∞ Quotient, Sum, and Scalar Multiple Properties x→∞ x Limit of a Constant Function and 5 5 Because the limit of the denominator is 0, we know that we have not correctly applied the Quotient Property of Limits. However, we can argue that as 5 is divided by increasingly smaller values approaching 0, the value of the fraction gets increasingly larger. Therefore, the limit can be described as approaching ∞. GuidedPractice Guided Practice 5 6A.    lim _   0 x→-∞ x - 10 658 | Lesson 11-2 | Evaluating Limits Algebraically 6B.    lim _   -3x 2 + 7 x→∞ 5x + 1 -∞ 6C.     lim __     3 7x 3 - 3x 2 + 1 x→∞ 2x + 4x 3.5 Copyright © McGraw-Hill Education =_   9 · 0 + 2 · 0 or _   0 Limits of Reciprocal Function at Infinity You learned that since a sequence is a function of the natural numbers, the limit of a sequence is the limit of a function as n → ∞. If this limit exists, then its value is the number to which the sequence 1 1 _ 1 1 converges. For example, the sequence a n= 1, _   , _   ,   , … can be described as f (n) = _   n , where n is a 2 3 4 positive integer. Because     lim _   n = 0, the sequence converges to 0. 1 n→∞ Example 7 Limits of Sequences Write the first five terms of each sequence. Then find the limit of the sequence, if it exists. 3n + 1 _ a. a n=   n + 5 3 (1) + 1 3 (2) + 1 3 (3) + 1 3 (4) + 1 3(5) + 1 The first five terms of this sequence are  _ ,   _ ,   _ ,   _ , and _   5 + 5 or 1+5 2+5 3+5 4+5 approximately 0.667, 1, 1.25, 1.444, and 1.6. To find the limit of the sequence, find  lim   n + 5 .   _ 3n + 1 n→∞ 3+_   1 n     lim _   n + 5 =      lim _   _5 3n + 1 n→∞ n→∞ Divide each term by highest-powered term, n, and simplify. 1 +   n     lim 3 +      lim _   1 n →∞ n →∞ n __ =         _1 Quotient, Sum, and Scalar Multiple Properties =_   1 + 5 · 0 or 3 Limit of a Constant Function and Limits of Reciprocal Function at Infinity     lim 1 + 5    lim   n →∞ n n →∞ StudyTip Check for Reasonableness To check the reasonableness of the results in Example 7, find the 100th, 1000th, and 10,000th terms in each sequence. In Example 7a, these terms are 2.867, 2.986, and 2.999, respectively. Since these values appear to be approaching 3, a limit of 3 is reasonable. 3+0 So, the limit of the sequence is 3. That is, the sequence converges to 3. CHECK The graph of a n= _   n + 5 supports this result. ✔ 3n + 1 _ _⎤⎥ 2 5 ⎡ n (n + 1) 2 b. b n=   4 ⎢⎣ 4 n ⎦ The first five terms of this sequence are approximately 5, 2.813, 2.222, 1.953, and 1.8. Now find the limit of the sequence. 2 n 2(n 2 + 2n + 1) ⎤ 5 ⎡ n (n + 1) 2⎤ 5 ⎡__ _ ⎥ =    ⎢     lim _   4 ⎣⎢ _ ⎥⎦ ⎦   lim   4 ⎣   4 n→∞ n Square the binomial. 4 n→∞ n =      lim __     4 n→∞ n→∞ n n→∞ n 2 = ___         Limit of a Constant Function and Limits of 5 =_   or 1.25 5n 4 + 10n 3 + 5n 2 n→∞ 4n Multiply.     lim 5 + 10     lim _ + 5    lim _   1 1     lim 4 n→∞ Divide each term by highest-powered term. Then use Quotient, Sum, and Scalar Multiple Properties. Reciprocal Function at Infinity 4 Copyright © McGraw-Hill Education So, the limit of b nis 1.25. That is, the sequence converges to 1.25. 7A. Approximately 2, 0.8, 0.4, 0.235, 0.154; the limit of {a n} is 0. 7B. Approximately 0.182,1.143, 3.177, 6.4, 10.87; {b n} has no limit. 7C. Approximately 9, 5.625, 4.667, 4.219, 3.96; the limit of {c n} is 3. CHECK Make a table of values, choosing large n-values that grow increasingly larger. ✔ n approaches ∞ n 10 100 1000 10,000 100,000 a n 1.51 1.28 1.25 1.25 1.25 GuidedPractice Guided Practice 4 7A. a = _   n 2 n + 1 2n 7B. b n= _   3 3n + 8 9 ⎡ n(n + 1)(2n + 1) ⎤ ⎥ 7C. c n= _   3 ⎢⎣__    ⎦ n 6 659 Exercises Use the properties of limits to evaluate each limit. (Example 1) 2.     lim __   x-3 x 2 + 4x + 13 -25 1.    lim (5x - 10) x→-3 x→5 Evaluate each limit. (Examples 3 and 4) __   23. lim       x-4 2x2 - 5x - 12 29 _1 1 5. lim   x + 2x +   √ x ) 21     (_ 4.     lim __         _ 10 3 2 25. lim      _ 2 6.      lim [x (x + 1) + 2] - 46 __ 27.    lim         2 7.    lim _   8.     lim __   x+3   29.    lim _ 3.    lim (7x 2 - 6x - 3) x→-1 10 9 x→9 x 2 - 10x x→12   √ x + 4  x→2 3 2 x→-4 x 3 + 2x - 11 6 9. lim    (26 - 3x) 2x 5 - 4x 3 - 2x - 12 x→-2 x + 5x - x→1 1 0.     lim _   2 x 4 - x 3 x→-6 x 20 x→1 x 2 + 4x - 5 x - 1 x→16   x + 9 x - 4  13.  lim   2    _ x 3 + 9x + 6 x→3 x + 5x + 6 15.     lim _   x + 5 5x 5 - 16x 4 x→-4 2   x - 6 33. lim    _ 1 _   34.     lim __   x→5   √ x + 4  - 5 2x + 11 19. lim      _ x→5 x 2 - x - 20 2x 2 + 7x - 17 37.    lim __        x→∞ 3x 5 + 4x 2 + 2 x+4 - 9216 16.     lim _   x - 4 2 x→4 2 18.      lim (3x - 10x + 35) x→9 188 2 20.      lim (-x + 3x +   √ x ) x→1 x 2 - 2x - 15 x+3   √ 16 + x - 4 x -8 1 _   8 ∞ 3 21. PHYSICS According to the special theory of relativity developed by Albert Einstein, the mass of an object m 0   , where c is traveling at speed v is given by m = _  v 2 1-_ 2 c the speed of light and m 0is the initial, or rest mass, of the object. (Example 2) a–b. See margin. a. Find      lim m. Explain the relationship between this limit 0 __ 36.     lim      3 2 x→∞ 3x 3 - 10x + 2 4x + 20x 3 _   4 38.      lim (10x + 14 + 6x 2 - x 4 ) x 6 + 12x 39.    lim __      x→∞ 3x 6 + 2x 2 + 11x 1 _   3 -∞ 14x 3 - 12x __ 40.      lim      ∞ 2 x→∞ 41.    lim (7x 3 + 4x 4 + x) ∞ 42.     lim __        5 3 6x 5 - 12x 2 + 14x 2x + 13x 3 __ 43.     lim (x 3 - 6x 7 + 2x 6 ) -∞ 44.     lim         5 3 0 x→∞ x→∞ 4x + 13x - 8 x→∞ 6x 3 + 2x - 11 x→∞ -x + 17x + 4x x→∞ 45.     lim __      4 3 10x 4 - 2 x→∞ 5x + 3x - 2x √ x→0 x→∞ x→3 _ 6 35.    lim (5 - 2x 2 + 7x 3 ) 14.      lim   √ 2 - x 1 x 3 17.  lim   - 62     _ 2 2 x→-3 1 10 Evaluate each limit. (Examples 5 and 6) 11, 14, 16, 19. See margin. 12.      lim 4x 3 - 3x 2 + 10 30 x→2   √ 11.    lim _ 1 _   8x 2 + 2x - 3 1 12x + 8x - 7 _ x→ 32.     lim  __   √ x + 3  - 3 -_ 30.     lim __      2 4 -12 2x x→6 2 6 13 6 x-9 √  5   18 + x 28.      lim __   x-7 x→7 x→9 31. lim      _ x→0 3 -   √ x + 9  Use direct substitution, if possible, to evaluate each limit. If not possible, explain why not. (Example 2) 1 _ 4x 2 + 21x + 5 8 1 _     √x - 3 26.      lim _    x→-5 3x + 17x + 10 x→-2   √ 6 + x - 2 42 4x x→0   √x  + 1  - 1 3 x+2 -2 24.     lim _   11 x→4 2 46.      lim (2x 5 - 4x 2 + 10x - 8) -∞ x→-∞ 47. SPONGE A gel capsule contains a sponge animal. When the capsule is submerged in water, it immediately dissolves, allowing the sponge to absorb water and quickly grow in size. The length ℓ in millimeters of the sponge animal after being submerged in water for   t seconds can be defined as ℓ(t) = _ 2 + 25. (Example 6) 105t 2 10 + t v→0 and m 0. b. What happens to the mass of an object if its speed were able to approach the speed of light? to the center and c is the distance from the foci to the center. (Example 2) b. See margin. a. 20π or 62.83 units 2 a. What is the area of an ellipse for a = 5 and c = 3? a. What is the length of the capsule before it is submerged in water? 25 mm b. What is the limit of this function as t → ∞? 130 mm c. Explain how the limit of this function relates to the length of the sponge animal. The sponge animal will not grow larger than 130 mm in length. 48. KITTENS Suppose the weight w in kilograms of a kitten d days after birth can be estimated by w(d ) = __   . d (Example 6) 25 2 + 98(0.85) b. What happens to the eccentricity of an ellipse as the foci move closer to the center of the ellipse? a. What is the weight of the kitten at birth? c. What is the limit of the area of the ellipse as c approaches 0 in terms of a? πa2 b. How much will the kitten eventually weigh (that is, the weight as d → ∞)? 12.5 kg 660 | Lesson 11-2 | Evaluating Limits Algebraically 0.25 kg Copyright © McGraw-Hill Education 22. GEOMETRY The area of an ellipse is defined as 2 A = πa   √a - c 2  , where a is the distance from the vertices 70. BIOLOGY Suppose the width in millimeters of an animal’s Find the limit of each sequence, if it exists. (Example 7) C n 3 - 2 49. a n = _   2 ∞ 8n + 1 50. a n = _   2 0 -4n 2 + 6n - 1 51. a n= __     -4 2 4 - 3n 52. a n = _   3 0 12n + 2 53. a n= _   2 54. a n= __   n n - 3 n + 3n 2 6n - 1 a. Write a limit to describe the width of the animal’s pupils when the light is at its minimum illuminance. Then find the limit, and interpret your results. 2n + 5 8n + 5n + 2 3 + 2n ∞ 5 5 ⎡ n( n + 1) ⎤ _ ⎥   55. a n= _   2 ⎢⎣_ ⎦ n(2n + 1)(n + 1) ⎤ 3 ⎡__ ⎥ 56. a n= _   3 ⎢⎣   ⎦ 6 n 2 1 1 ⎡ n (n + 1) 2⎤ _ ⎥   57. a n= _   4 ⎢⎣_ ⎦ n( 2n + 1)(n + 1) ⎤ 12 ⎡__ ⎥ ∞ 58. a n= _   2 ⎢⎣   ⎦ 2 2 n 4 4 n 1 6 n 152x -0.45 + 85 4x + 10 illuminance of the light shining on the pupils measured in lux. 2 2         , where x is the pupil is given by d(x) = __ -0.45 b. Write a limit to describe the width of the animal’s pupils when the light is at its maximum illuminance. Then find the limit, and interpret your results. a–b. See Chapter 11 Answer Appendix. __ f (x + h) - f (x) Find      lim     for each function. h h→0 71. f (x) = 2x - 1 59. POPULATION After being ranked one of the best cities in which to live by a national publication, Northfield Heights experienced a rise in population that could be 73. f (x) =   √ x 75. f (x) = x 2 __ modeled by p(t) =      , where p is the total rise 3 36t 3 - 12t + 13 3t + 90 Rise in Population 1 ? 2 ? 3 ? √x 1 _   or _ 2x 2  √x 2x 72. f (x) = 7 - 9x 74. f (x) =   √ x + 1  1 _   or 2  √ x + 1   _   √ x + 1  2x + 2 76. f (x) = x 2 + 8x + 4 2x + 8 77. PHYSICS An object that is in motion possesses an energy of motion called kinetic energy because it can do work when it impacts another object. The kinetic energy of an in population in thousands and t is the number of years after 2006. (Example 7) Years Since 2006 2 -9   2 m · [v(t)] 2, where object with mass m is given by k(t) = _ v(t) is the velocity of the object at time t and mass is given 1   in kilograms. Suppose v(t) = _ 2 for all t ≥ 0. What value 50 1 + t does the kinetic energy of an object that has a mass of one kilogram approach as time approaches 100? 0.0000125 a. Complete the table for 2007–2009. See Chapter 11 Answer Appendix. H.O.T. Problems b. What was the total rise in population by 2011? about 9576 people c. What is the limit of the population growth? 12,000 people d. Explain why a city’s population growth may have a limit. B Sample answer: A city’s borders may limit the amount of possible growth and building opportunities. Find each limit, if it exists, by using direct substitution to evaluate the corresponding one-sided limits.  x-3 60.    lim ⎨       x→-2 2x - 1  if x ≤ -2      if x > -2 Copyright © McGraw-Hill Education sin x 0 x→0 68. lim      _ ln x x→1 ln (2x - 1) x→∞ b nx + b n - 1x x→c x→0 + ⋯ + b 2x + b 1x + b 0 where a n≠ 0 and b m≠ 0. (Hint: Consider the cases where m < n, n = m, and m > n.) See Chapter 11 Answer Appendix. 81. REASONING If r(x) is a rational function, is it sometimes, lim r(x) = r(c)? Explain your always, or never true that     x→c reasoning. See margin. 1 82. WRITING IN MATH Use a spreadsheet or a table to summarize the properties of limits. Give an example of each. See Chapter 11 Answer Appendix. p(x) 3x - sin 3x 67.      lim _   2 4.5 0.5 x→c See Chapter 11 Answer Appendix. a nx n + a n - 1x n - 1 + ⋯ + a 2x 2 + a 1x + a 0 x→0 x 79. PROOF Use mathematical induction to show that if   lim f (x) = L, then     lim [ f (x)] n= [    lim f (x)] nor L n for        lim       ____      , m m-1 2 65.      lim (1 + x + 2 x- cos x) tan 2x 66. lim   2    _ See Chapter 11 Answer Appendix. 80. CHALLENGE Find Find each limit, if it exists, using any method. x→π x→c any integer n.  4x + 2 if x ≤ 0 61.  lim      2    ⎨    x→0 2 - x 2 if x > 0   5 - x 2 if x ≤ 0 62. lim      5    ⎨    x→0 5 - x if x > 0   (x - 2) 2 + 1 if x ≤ 2 63. lim      No limit exists.    ⎨      x→2 x - 6 if x > 2  64.    lim _   x 78. PROOF Use the properties of limits to show that for any p(x) = a nx n + a n - 1x n - 1 + ⋯ + a 2x 2 + a 1x + a 0and a real number c,      lim p(x) = p(c). x→c -5 Use Higher-Order Thinking Skills x sin x  69.     lim _   x→1 x - 1 1 -   √x -_ 1 2 83. WRITING IN MATH Consider      lim _   =_   . Suhaila says that x→a q(x) ∞ ∞ this answer means the limit is 1. Why is she incorrect? What further analysis could be used to determine the limit, if it exists? See Chapter 11 Answer Appendix. 661 Spiral Review Use the graph of y = f (x) to find each value. 84.    lim f (x) and f(-2) x→-2 -1; -1 85. lim    f (x) and f(0) 0; undefined 86. lim    f (x) and f(3) 4; 2 x→0 x→3 87. HEALTH The table shows the average life expectancy for people born in various years in the United States. a–d. See margin. a. Make a scatter plot of the data, and identify the relationship. b. Calculate and interpret the correlation coefficient. Determine whether it is significant at the 5% level. c. If the correlation is significant at the 5% level, find the least-squares regression equation and interpret the slope and intercept in context. d. Use the regression equation that you found in part c to predict the average life expectancy for 2080. State whether this prediction is reasonable. Explain. 88. ACOUSTICS Polar coordinates can be used to model the shape of a concert amphitheater. Suppose the performer is placed at the pole and faces the direction π π of the polar axis. The seats have been built to occupy the region with - _ ≤θ≤_   3 3 and 0.1 ≤ r ≤ 1, where r is measured in hundreds of meters. See margin. a. Sketch this region in the polar plane. Years Since 1900 Life Expectancy 10 50 20 54.1 30 59.7 40 62.9 50 68.2 60 69.7 70 70.8 80 73.7 90 75.4 100 76.9 17,278 b. How many seats are there if each person has 0.6 square meters of space? 89. Write the pair of parametric equations, x = 2 sin t and y = 5 cos t, in rectangular form. Then sketch the graph. 2 x 2 y _ +_   = 1; See margin for graph. 4 25 Skills Review for Standardized Tests 90. SAT/ACT According to the data in the table, by what percent did the number of applicants to Green College increase from 1995 to 2000? B Year Applicants 18,000 1995 20,000 24,000 2005 25,000 A 15% C 25% B 20% D 27% F 3 G 4 C -_π B -_ D 0 1 2 93. REVIEW Consider the graph of y = f (x) shown. What is the      lim f (x)? G x→2 + E 29% __ 91. REVIEW What is     lim      ? h→0 A -π 2h 3 - h 2 + 5h h H H 5 J The limit does not exist. 662 | Lesson 11-2 | Evaluating Limits Algebraically F 0 G 1 H 5 J The limit does not exist. Copyright © McGraw-Hill Education 2000 x+π 3 4 Number of Applicants to Green College 1990 92. What value does g(x) = _   approach as x cos (x + π) approaches 0? A 11-3 Objective Use TI-Nspire technology to estimate the slope of a curve Graphing Technology Lab The Slope of a Curve The slope of a line as a constant rate of change is a familiar concept. General curves do not have a constant rate of change because the slope is different at every point on the graph. However, the graphs of most functions are locally linear. That is, if you examine the graph of a function on a very small interval, it will appear linear. By looking at successive secant lines, it is possible to apply slope to curves. Activity Secant Lines Estimate the slope of the graph of y = (x - 2) 3 + 1 at (3, 2). Step 1 Enter y = (x - 2) 3 + 1 in f1. Then calculate the slope of the line secant to the graph of y = (x - 2) 3 + 1 through x = 2 and x = 4. The slope of the secant line is 4. Step 2 Find the slope of the line secant to the graph of y = ( x - 2)3 + 1 through x = 2.5 and x = 3.5. The slope of the secant line is 3.25. Step 3 Find the slope of the line secant to the graph of y = ( x - 2)3 + 1 through x = 2.8 and x = 3.2. The slope of the secant line is 3.04. Copyright © McGraw-Hill Education Step 4 Find the slope of 3 more secant lines on decreasing intervals around (3, 2). 5. Sample answer: As the points through which a secant line passes approach (a, b), the secant line gets closer and closer to the line tangent to the function at (a, b). 6. Sample answer: Find the limit of the slopes of the secants as they approach the tangent to the curve at the given point. As the interval around (3, 2) decreases, the slope of the secant line approaches 3. So, the slope of y = (x - 2) 3 + 1 at (3, 2) is about 3. Exercises Estimate the slope of each function at the given point. 1. y = (x + 1) 2 ; (-4, 9) -6 3. y = 4x 4 - x 2 ; (0.5, 0) 1 2. y = x 3 - 5; (2, 3) 4. y =   √ x ; (1, 1) 12 0.5 Analyze the Results 5. ANALYZE Describe the change to a line secant to the graph of a function as the points of intersection approach a given point (a, b). 6. MAKE A CONJECTURE Describe how you could determine the exact slope of a curve at a given point. 663 11-3 Then You found average rates of change using secant lines. Now 1 2 NewVocabulary tangent line instantaneous rate of change difference quotient instantaneous velocity Tangent Lines and Velocity Why? Find instantaneous rates of change by calculating slopes of tangent lines. When a skydiver exits a plane, gravity causes the speed of his or her fall to increase. For this reason, the velocity of the sky diver at each instant before terminal velocity is achieved or the parachute is opened varies. Find average and instantaneous velocity. 1 Tangent Lines You calculated the average rate of change between two points on the graph of a nonlinear function by finding the slope of the secant line through these points. In this lesson, we develop a way to find the slope of such functions at one instant or point on the graph. The graphs below show successively better approximations of the slope of y = x 2at (1, 1) using secant lines. Figure 11.3.2 Figure 11.3.1 Figure 11.3.3 Notice as the rightmost point moves closer and closer to (1, 1), the secant line provides a better linear approximation of the curve near that point. We call the best of these linear approximations the tangent line to the graph at (1, 1). The slope of this line represents the rate of change in the slope of the curve at that instant. To define each of these terms more precisely, we use limits. (x + h) - x h This expression is called the difference quotient. Figure 11.3.4 As the second point approaches the first, or as h → 0, the secant line approaches the tangent line at (x, f (x)). We define the slope of the tangent line at x, which represents the instantaneous rate of change of the function at that point, by finding the limits of the slopes of the secant lines as h → 0. KeyConcept Instantaneous Rate of Change The instantaneous rate of change of the graph of f (x ) at the point (x, f (x )) is the slope m of the tangent line at (x, f (x )) given f (x + h) - f (x )   , provided the limit exists. by m = lim __ h h→0 664 | Lesson 11-3 Copyright © McGraw-Hill Education f__ (x + h) - f (x) f__ (x + h) - f (x) m =      or      . Joggie Botma/Alamy To define the slope of the tangent line to y = f (x) at the point (x, f (x)), find the slope of the secant line through this point and one other point on the curve. Let the x-coordinate of the second point be x + h for some small value of h. The corresponding y-coordinate for this point is then f (x + h), as shown in Figure 11.3.4. The slope of the secant line through these two point is given by You can use this expression to find the slope of the tangent line to a graph for a specified point. StudyTip Instantaneous Rate of Change When calculating the limit of the slopes of the secant lines as h→0, any term containing a value of h that has not been divided out will become 0. Example 1 Slope of a Graph at a Point Find the slope of the line tangent to the graph of y = x 2 at (1, 1). f (x + h) - f (x) m = lim   __    Instantaneous Rate of Change Formula h h→0 f (1 + h) - f (1) h = lim   __    x=1 h→0 (1 + h) 2 - 1 2 h h→0 = lim  __ f (1 + h) = (1 + h) 2 and f (1) = 1 2 = lim __      h→0 1 + 2h + h 2 - 1 h Multiply. h(2 + h) h = lim _   Simplify and factor. h→0 = lim (2 + h) Divide by h. h→0 Sum Property of Limits and = 2 + 0 or 2 Limits of Constant and Identity Functions The slope of the graph at (1, 1) is 2, as shown. GuidedPractice Guided Practice Find the slope of the line tangent to the graph of each function at the given point. 1A. y = x 2 ; (3, 9) 6 1B. y = x 2 + 4; (-2, 8) -4 The expression for instantaneous rate of change can also be used to find an equation for the slope of the tangent line to a graph at any point x. Example 2 Slope of a Graph at Any Point _4 Find an equation for the slope of the graph of y =   x at any point. f (x + h) - f (x) h h→0 m = lim __      4 _4 _   - x x+h = lim _   h h→0 - 4h _   Instantaneous Rate of Change Formula f (x + h ) = _   x + h and f (x ) = _   4 4 x x (x + h) = lim   _ Add fractions in the numerator and simplify. = lim _   Simplify. = lim _   2 Divide by h and multiply. =_   2 Quotient and Sum Properties of Limits and Limits of Constant and Identity Functions h→0 h -4h h→0 xh (x + h) -4 h→0 x + xh -4 x + x(0) =_   2 Simplify. Copyright © McGraw-Hill Education -4 x An equation for the slope of the graph at any point is 4 m = - _ 2 , as shown. x GuidedPractice Guided Practice Find an equation for the slope of the graph m of each function at any point. 2A. y = x 2 - 4x + 2 m = 2x - 4 2B. y = x 3 m = 3x 2 665 2 Instantaneous Velocity You calculated the average speed of a dropped object by dividing the distance traveled by the time it took for the object to cover that distance. Velocity is speed with the added dimension of direction. You can calculate average velocity using the same approach that you used when calculating average speed. KeyConcept Average Velocity If position is given as a function of time f (t ), for any two points in time a and b, the average velocity v is given by change in distance f (b ) - f (a) v  avg = __      change    =_   b - a . in time Real-World Example 3 Average Velocity of an Object MARATHON The distance in kilometers that a runner competing in the Boston Marathon has traveled after a certain time t in hours can be found by f(t) = -1.3t 2 + 12t. What was the runner’s average velocity between the second and third hour of the race? First, find the total distance traveled by the runner for a = 2 and b = 3. Real-WorldLink f (t) = -1.3t 2 + 12t Original equation f (t) = -1.3t 2 + 12t f (2) = -1.3(2) 2 + 12(2) a = 2 and b = 3 f (3) = -1.3(3) 2 + 12(3) f (2) = 18.8 Simplify. f (3) = 24.3 Now use the formula for average velocity. f(b) - f(a) b-a Robert K. Cheruiyot of Kenya completed the 2008 Boston Marathon in less than two hours eight minutes. On average, he completed a mile every four minutes fifty seconds. v avg =_   Source: B oston Athletic Association The average velocity of the runner during the third hour was 5.5 kilometers per hour forward. Average Velocity Formula =_   3 - 2 f (b) = 24.3, f (a ) = 18.8, b = 3, and a = 2 = 5.5 Simplify. 24.3 - 18.8 GuidedPractice Guided Practice 3. WATER BALLOON A water balloon is propelled straight up using a launcher. The height of the balloon in meters t seconds after it is launched can be defined by d(t) = 2 + 20t - 5t 2 . What was the balloon’s average velocity between t = 1 and 2? 5 m/s Looking more closely at Example 3, we can see that the velocity was found by calculating the slope of the secant line that connects the two points (2, 18.8) and (3, 24.3), as shown in the graph. The velocity that was calculated is the average velocity traveled by the runner over a period of time and does not represent the instantaneous velocity, the velocity or speed the runner achieved at a specific point in time. KeyConcept Instantaneous Velocity If the distance an object travels is given as a function of time f (t ), then the instantaneous velocity v (t )at a time t is given by f (t + h ) - f ( t ) v (t ) = lim__   , h h→0 provided the limit exists. 666 | Lesson 11-3 | Tangent Lines and Velocity Copyright © McGraw-Hill Education To find the actual velocity of the runner at a specific time t, we find the instantaneous rate of change of the graph of f (t)at t. Example 4 Instantaneous Velocity at a Point A football is dropped from the top of a building 600 meters above the ground. The height of the football in meters after t seconds is given by f (t ) = 600 - 5t 2 . Find the instantaneous velocity v (t ) of the football at 5 seconds. WatchOut! Substitution Remember to distribute the negative sign that precedes f (t ) to each term that is substituted. To find the instantaneous velocity, let t = 5 and apply the formula for instantaneous velocity. f (t + h) - f (t) h v(t)= lim __     h→0 Instantaneous Velocity Formula ___ v(5) = lim          f (t + h ) = 600 - 5(5 + h )  2 and f (t ) = 600 - 5(5) 2 = lim -_ Multiply and simplify. 600 - 5(5 + h)2 h→0 h→0 - [600 h 5(5)2] 50h - 5h2 h h(-50 - 5h) h = lim _   Factor. = lim (-50 - 5h) Divide by h. h→0 h→0 = -50 - 5(0) or -50 Difference Property of Limits and Limits of Constant and Identity Functions The instantaneous velocity of the football at 5 seconds is 50 meters per second. The negative sign indicates that the height of the ball is decreasing. GuidedPractice Guided Practice 4. A window washer accidentally drops his lunch off his scaffold 420 meters above the ground. The position of the lunch in relation to the ground is given as d(t) = 420 - 5t 2 , where time t is given in seconds and the position of the lunch is given in meters. Find the instantaneous velocity v ( t)of the lunch at 7 seconds. -70 m/s Equations for finding the instantaneous velocity of an object at any time t can also be determined. Example 5 Instantaneous Velocity at Any Point The distance a particle moves along a path is given by s(t ) = 18t - 3t 3 - 1, where t is given in seconds and the distance of the particle from its starting point is given in centimeters. Find the equation for the instantaneous velocity v(t) of the particle at any point in time. Apply the formula for instantaneous velocity. s(t + h) - s(t) h     v ( t)= lim __ h→0 Instantaneous Velocity Formula 18(t + h) - 3(t + h) 3- 1 - [18t - 3t 3- 1] s (t + h ) = 18(t + h ) - 3(t + h ) 3 - 1 h and s (t ) = 18t - 3t 3 - 1 18h - 9t 2 h - 9th 2 - 3h 3 = lim __      Multiply and simplify. h h→0 h(18 - 9t 2 - 9th - 3h 2 ) = lim __      Factor. h h→0 = lim ____          h→0 = lim (18 - 9t 2 - 9th - 3h 2) h→0 = 18 - 9t 2 - 9t(0) - 3(0) 2 Copyright © McGraw-Hill Education = 18 - 9t 2 Divide by h. Difference Property of Limits and Limits of Constant and Identity Functions Simplify. The instantaneous velocity of the particle at time t is v (t)= 18 - 9t 2 . GuidedPractice Guided Practice 5. The distance in meters of a water rocket from the ground after t seconds is given by s(t) = 30t - 5t 2 . Find the expression for the instantaneous velocity v ( t)of the rocket at any time t. v (t )= 30 - 10t 667 Exercises 1. y = x 2 - 5x; (1, -4) and (5, 0) -3; 5 3. y = x 2 + 7; (3, 16) and (6, 43) 4. y = _   x ; (1, 3) and (3, 1) 3 25. d(t) = 100 - 16t 2 ; t = 3 -3; -3 2. y = 6 - 3x; (-2, 12) and (6, -12) 26. d(t) = 38t - 16t ; t = 0.8 12; 3 28. d(t) = 500 - 30t - 16t 2 ; t = 4 1 3 1 27. d(t) = -16t 2 - 47t + 300; t = 1.5 -_ ; -1 1 16 m = -2 9. y = x 2 + 3 m = 2x 10. y = x 3 11. y = 8 - x 2 m = -2x 12. y = 2x 2 15. y = _   √ 1   x 2 m = - _ √x 2x m = 3x 2 m = 4x m = 2x + 2 14. y = x 2 + 2x - 3 2 _ 1 16. y = _   2 m = - 3 x x 17. SLEDDING A person’s vertical position on a sledding hill after traveling a horizontal distance x units away from the top of the hill is given by y = 0.06x 3 - 1.08x 2 + 51.84. (Example 2) 63.6 ft/s 31. d(t) = 1275 - 16t 2; t = 3.8 -121.6 ft/s 32. d(t) = 853 - 48t - 16t 2; t = 1.3 a. Find an equation for the hill’s slope m at any distance x. m = 0.18x 2 - 2.16x b. Find the hill’s slope for x = 2, 5, and 7. -3.6, -6.3, -6.3 The position of an object in kilometers after t minutes is given by s(t). Find the average velocity of the object in kilometers per hour for the given interval of time. Remember to convert from minutes to hours. (Example 3) 45 km/h 64.8 km/h 20. s(t) = 0.2t for 2 ≤ t ≤ 4 72 km/h 21. s(t) = 0.01t 3 - 0.01t 2 for 4 ≤ t ≤ 7 49.2 km/h 19. s(t) = 1.08t - 30 for 4 ≤ t ≤ 8 2 22. s(t) = -0.5(t + 3 for 4 ≤ t ≤ 4.5 45 km/h 23. s(t) = 0.6t + 20 for 3.8 ≤ t ≤ 5.7 36 km/h 5) 2 a. 46 words/min 24. TYPING The number of words w a person has typed after t minutes is given by w(t) = 10t 2 - _   2 t 3 . (Example 3) 1 a. What was the average number of words per minute the person typed between the 2nd and 4th minutes? b. What was the average number of words per minute the person typed between the 3rd and 7th minutes? 60.5 words/min 668 | Lesson 11-3 | Tangent Lines and Velocity -89.6 ft/s Find an equation for the instantaneous velocity v(t) if the path of an object is defined as s(t) for any point in time t. (Example 5) 33. s(t) = 14t 2 - 7 34. s(t) = t - 3t 2 35. s(t) = 5t + 8 36. s(t) = 18 - t 2 + 4t 37. s(t) = t 3 - t 2 + t 38. s(t) = 11t 2 - t 39. s(t) =   √t - 3t 2   √t _ 40. s(t) = 12t 2 - 2t 3 v (t ) = 28t v (t ) = 1 - 6t v (t ) = -2t + 4 v (t ) = 5 v (t ) = 3t 2 - 2t + 1 v (t ) =   1 -512 ft/s 30. d(t) = 150t - 16t ; t = 2.7 2 8. y = -x 2 + 4x 18. s(t) = 0.4t 2 - _   20 t 3for 3 ≤ t ≤ 5 -158 ft/s 29. d(t) = -16t - 400t + 1700; t = 3.5 m = -2x + 4 7. y = 4 - 2x -95 ft/s 2 Find an equation for the slope of the graph of each function at any point. (Example 2) m = -6x 2 12.4 ft/s 6; 12 6. y = _   x + 2 ; (2, 0.25) and (-1, 1) 13. y = -2x 3 -96 ft/s 2 -3; - _ 5. y = x 3 + 8; (-2, 0) and (1, 9) The distance d an object is above the ground t seconds after it is dropped is given by d(t). Find the instantaneous velocity of the object at the given value for t. (Example 4) 2t v (t ) = 22t - 1 v (t ) = 24t - 6t 2 - 6t 41. SKYDIVER Refer to the beginning of the lesson. The position d of the skydiver in meters relative to the ground can be defined by d(t) = 5,000 - 5t 2 , where t is seconds passed after the skydiver exited the plane. (Example 5) a. What is the average velocity of the skydiver between the 2nd and 5th seconds of the jump? -35 m/s b. What was the instantaneous velocity of the sky diver at 2 and 5 seconds? -20 m/s; -50 m/s c. Find an equation for the instantaneous velocity v(t) of the skydiver. v (t ) = -10t 42. DIVING A cliff diver’s distance d in meters above the surface B of the water after t seconds is given. t 0.5 0.75 1.0 1.5 2.0 2.5 3.0 d 43.7 42.1 40.6 33.8 25.3 14.2 0.85 a. Calculate the diver’s average velocity for the interval 0.5 ≤ t ≤ 1.0. -6.2 m/s b. Use quadratic regression to find an equation to model d(t). Graph d(t) and the data on the same coordinate plane. See margin. c. Find an expression for the instantaneous velocity v(t) of the diver and use it to estimate the velocity of the diver at 3 seconds. v (t ) = -9.82t - 0.04; -29.5 m/s Copyright © McGraw-Hill Education Find the slope of the lines tangent to the graph of each function at the given points. (Example 1) 43. FOOTBALL A goal keeper can kick a ball at an upward velocity of 25 meters per second. Suppose the height d of the ball in meters t seconds after it is kicked is given by d(t) = -5t 2 + 25t + 1. 53. PROJECTILE When an object is thrown straight down, the total distance y the object falls can be modeled by y = 16t 2 + v0t, where time t is measured in seconds and the initial velocity v0is measured in feet per second. a. If an object thrown straight down from a height of 816 feet takes 6 seconds to hit the ground, what was the initial velocity of the object? -40 ft/s b. What was the average velocity of the object? -136 ft/s c. What was the object’s velocity when it hit the ground? a. Find an equation for the instantaneous velocity v(t) of the ball. v(t ) = -10t + 25 b. How fast is the ball traveling 0.5 second after it is kicked? 20 m/s c. If the instantaneous velocity of the ball is 0 when the ball reaches its maximum height, at what time will the ball reach its maximum height? t = 2.5 s 32.25 m 44–47. See Chapter 11 Answer Appendix for graphs. d. What is the maximum height of the ball? Find an equation for a line that is tangent to the graph of the function and perpendicular to the given line. Then use a graphing calculator to graph the function and both lines on the same coordinate plane. 44. f (x) = x 2 + 2x; y = - _ x + 3 1 2 45. g (x) = -4x 2 ; y = _   4 x + 5 1 y = 2x y = -4x + 1 _3 1 46. f (x) = - _ x 2 ; y - x = 2 y = -x +   6 2 1 1 47. g (x) = _   2 x 2 + 4x; y = - _ x + 9 y = 6x - 2 6 48. PHYSICS The distance s of a particle moving in a straight line is given by s(t ) = 3t 3 + 8t + 4, where t is given in seconds and s is measured in meters. a. Find an equation for the instantaneous velocity v(t) of the particle at any given point in time. v (t ) = 9t 2 + 8 b. Find the velocity of the particle for t = 2, 4, and 6 seconds. 44 m/s, 152 m/s, 332 m/s C Each graph represents an equation for the slope of a function at any point. Match each graph with its original function. 49. f (x) = _   x a c 51. h(x) = ax 4 a. d 52. j(x) = a√x a 50. g(x) = ax 5 b b. 54. -232 ft/s MULTIPLE REPRESENTATIONS In this problem, you will investigate the Mean Value Theorem. The theorem states that if a function f is continuous and differentiable on (a, b), then there exists a point c in (a, b) such that the tangent line is parallel to the line passing through (a, f (a)) and (b, f(b)). a. ANALYTICAL Find the average rate of change for f (x) = -x 2 + 8x on the interval [1, 6], and find an equation for the related secant line through (1, f (1)) and (6, f(6)). 1; y = x + 6 b. ANALYTICAL Find an equation for the slope of f (x) at any point. m = -2x + 8 c. ANALYTICAL Find a point on the interval (1, 6) at which the slope of the tangent line to f (x) is equal to the slope of the secant line found in part a. Find an equation for the line tangent to f (x) at this point. (3.5, 15.75); y = x + 12.25 d. VERBAL How are the secant line in part a and the tangent line in part b related? Explain. e. GRAPHICAL Using a graphing calculator, graph f (x), the secant line, and the tangent line on the same screen. Does the graph verify your answer in part d? Explain. d–e. See margin. H.O.T. Problems Use Higher-Order Thinking Skills 55. ERROR ANALYSIS Yasmin and Wafa were asked to find an equation for the slope at any point for f (x) = ⎜x⎟. Yasmin thinks the graph of the slope equation will be continuous because the original function is continuous. Wafa disagrees. Is either of them correct? Explain your reasoning. See margin. 56. CHALLENGE Find an equation for the slope of f (x) = 2x 4 + 3x 3 - 2x at any point. m = 8x 3 + 9x 2 - 2 57. REASONING True or false: The instantaneous velocity of an object modeled by s (t) = at + b is always a. See margin. f (a + h) - f (a) h h→0 58. REASONING Show that lim   __    = Copyright © McGraw-Hill Education c. d. f (x) - f (a) 2 lim   _ x - a for f (x) = x + 1. x→a See Chapter 11 Answer Appendix. 59. WRITING IN MATH Suppose that f (t) represents the balance in dirhams in a bank account t years after the initial deposit. Interpret each of the following. See Chapter 11 Answer Appendix. f (4) - f (0) a.   _ ≈ 41.2 4 f (4 + h) - f (4) h __≈ 42.9 b. lim      h→0 669 Spiral Review Evaluate each limit. 60.lim (x 2 + 2x - 2) x→4 22 61.  lim (-x 4 + x 3 - 2x + 1) x→-1 1 62. lim (x + sin x) x→0 0 63. HYDRAULICS The velocity, in centimeters per second, of a molecule of liquid flowing through a pipe is given by v (r) = k (R 2 - r 2 ), where R is the radius of the pipe in centimeters, r is the distance of the molecule from the center of the pipe in centimeters, and k is a constant. Suppose for a particular liquid and a particular pipe that k = 0.65 and R = 0.5. a. Graph v (r). See margin. b. Determine the limiting velocity of molecules closer and closer to the wall of the pipe. 0 cm 64. EDUCATION A college professor plans to grade a test on a curve. The mean score on the test is 65, and the standard deviation is 7. The professor wants the grades distributed as shown in the table. Assume that the grades are normally distributed. a. What is the lowest score for an A? 72 b. If a D is the lowest passing letter grade, find the lowest passing score. c. What is the interval for the Bs? 68–71 58 Find the specified nth term of each geometric sequence. 65. a 4 = 50, r = 2, n = 8 800 1 67. a 6for a n= _   5 an - 1, a 1 = -2 -_ 2 3125 66. a 4 = 1, r = 3, n = 10 729 68. a 5for a n= (-3)an - 1, a 1 = 11 Grade Percent of Class A 15 B 20 C 30 D 20 F 15 891 Find the indicated arithmetic means for each set of nonconsecutive terms. 69. 7 means; 62 and -2 54, 46, 38, 30, 22, 14, 6 70. 4 means; 17.2 and 47.7 23.3, 29.4, 35.5, 41.6 71. 3 means; -5.6 and 8 -2.2, 1.2, 4.6 72. 9 means; -45 and 115 -29, -13, 3, 19, 35, 51, 67, 83, 99 Skills Review for Standardized Tests A 2π B 4π C 8π D 12π E 16π 74. REVIEW Which of the following best describes the point at (0, 0) on f (x) = 2x 5 - 5x 4 ? G F absolute maximum G relative maximum H relative minimum J absolute minimum 670 | Lesson 11-3 | Tangent Lines and Velocity 75. When a bowling ball is dropped, the distance d(t) that it falls in t seconds is given by d(t) = 5t 2 . Its velocity after d(2 + h) - d(2) h 2 seconds is given by lim__    . What is the h→0 velocity of the bowling ball after 2 seconds? C A 14 meters per second C 20 meters per second B 18 meters per second D 23 meters per second 76. REVIEW The monthly profit P of a manufacturing company depends on the number of units x manufactured and can be described by 1 P(x) = _   3 x 3 - 34x 2 + 1012x, 0 ≤ x ≤ 50. How many units should be manufactured monthly in order to maximize profits? G F 15 H 37 G 22 J 46 Copyright © McGraw-Hill Education 73. SAT/ACT If the radius of the circle with center O is 4, what is the length of arc RST? A 11 Mid-Chapter Quiz Lessons 11-1 through 11-3 Estimate each one-sided or two-sided limit, if it exists. (Lesson 11-1) 1.  lim+ _ sin x x→0  x |x| 2. lim_ x→0 x 1 2x 2 - 18 3.  lim- _ x→3  x - 3 does not exist cos x - 1 4.   lim- _ x x→0  12 0 _3 2x 7.  lim e 2x + 3 x→-2   √ x + 20  -1 1 1; -5 18. y = x 2 - 3x; (2, -2) and (-1, 4) 400t - 2 19. y = 2 - 5x; (-2, 12) and (3, -13) -5; -5 20. y = x 3 - 4x 2 ; (1, -3) and (3, -9) -5; 3 21. FIREWORKS Fireworks are launched with an upward velocity of 30 meters per second. Suppose the height d of the fireworks in meters t seconds after they are launched is defined as d (t ) = -5t 2 + 30t + 1.5. (Lesson 11-3) See margin. b. Use your graph to estimate the value of the stamps for t = 2, 5, and 10 years. 42; 80; 114 t→∞ B _ 2 Find the slope of the line tangent to the graph of each function at the given points. (Lesson 11-3) 9. COLLECTIBLES The value of Yousif’s stamps has been increasing every year. The value v of the stamps after t years can be represented by the model c. Use your graph to evaluate limv (t ). A 1 2 8.   lim _   x x→-4 a. Graph the function for 0 ≤ t ≤ 10. 10 - e   x 1 x→1 v (t ) = _   2t + 15 . (Lesson 11-1) A does not exist x→0 D _ 10 6. lim   √ x 3 + 3  0.3679 2x 2 + 5 C _ 5 Estimate each limit, if it exists. (Lesson 11-1) 5.lim _     x→3 x 2 + 1 5 17. MULTIPLE CHOICE Evaluate l im  _ 16 . (Lesson 11-1) _ 200 d. Explain the relationship between the limit of the function and the value of Yousif’s stamps. Yousif’s baseball card will be worth no more than AED 200. Use direct substitution, if possible, to evaluate each limit. If not possible, explain why not. (Lesson 11-2) a. Find an equation for the instantaneous velocity v(t) of the fireworks. v (t ) = -10t + 30 b. How fast are the fireworks traveling 0.5 seconds after they are lit? 25 m /s c. What is the maximum height of the fireworks? ≈46.5 m 22. MULTIPLE CHOICE Find an equation for the slope of the graph of y = 7x 2 - 2 at any point. (Lesson 11-3) H Not possible; when x = 9, the denominator is 0. 3 11.  lim (2x + x 2 - 8) -20 10.lim  _ x - 3 x→9   √ x 2 + 1 F m = 7x G m = 7x - 2 x→-2 12. WILDLIFE A deer population P in hundreds at a national park after t years can be estimated by the model P (t ) =   __ , where t ≥ 3. The population for 3 H m = 14x J m = 14x - 2 3 10t - 40t + 2 2t + 14t + 12 5 years is shown below. What is the maximum number of deer that can live at the national park? (Lesson 11-2) 500 deer The position of an object in kilometers after t minutes is given by s (t ). Find the average velocity of the object in kilometers per hour given two values of time t. Remember to convert from minutes to hours. (Lesson 11-3) 23. s (t ) = 12 + 0.7t for t = 2 and 5 42 km/h 24. s (t ) = 2.05t - 11 for t = 1 and 7 123 km/h 25. s (t ) = 0.9t - 25 for 3 ≤ t ≤ 6 Copyright © McGraw-Hill Education 26. s (t ) = 0.5t 2 - 4t for 4 ≤ t ≤ 8 54 km/h 120 km/h Find an equation for the instantaneous velocity v (t ) if the position of an object is defined as s (t ) for any point in time t. (Lesson 11-3) Evaluate each limit. (Lesson 11-2) 13.  lim(15 - x 2 + 8x 3 ) x→∞ 15.  lim __    4 2 2x 2 + 5x - 1 x→∞ 2x - 14x + 2 ∞ 0 14.   lim __ 3 2 2x 3 - x - 2 x→∞ 4x + 5x 1 _   2 27. s (t ) = 4t 2 - 9t 28. s (t ) = 2t - 13t 2 -∞ 16.   lim(10x 3 - 4 + x 2 - 7x 4) x→∞ v (t ) = 2 - 26t 2 v (t ) = 2 - 10t 3 v (t ) = 12t - 3t 2 29. s (t ) = 2t - 5t 2 v (t ) = 8t - 9 30. s (t ) = 6t - t 671 11-4 Then You calculated the slope of tangent lines to find the instantaneous rate of change. NewVocabulary derivative differentiation differential equation differential operator Derivatives Now 1 F ind instantaneous rates of change by calculating derivatives. 2 the Product and Use Quotient Rules to calculate derivatives. 1 Why? Nasser, who lives on the sixth floor of an apartment building, accidentally drops a ball out of his window. Mansour, standing on the ground outside of Nasser’s building, retrieves the ball and attempts to throw it back up to Nasser. If Mansour can throw the ball at a speed of 20 meters per second, can he get it to Nasser’s window 21 meters above the ground? Basic Rules In Lesson 11-3 You used limits to determine the slope of a line tangent to the graph of a function at any point. This limit is called the derivative of a function. The derivative of f (x) is f ′(x), which is given by f (x + h) - f (x) h __   lim      , f ′(x) =    h→0 provided the limit exists. The process of finding a derivative is called differentiation and the result is called a differential equation. Example 1 Derivative of a Function at Any Point Find the derivative of f (x) = 4x 2 - 5x + 8. Then evaluate the derivative at x = 1 and 5. f (x + h) - f (x) h 4(x + h) 2 - 5(x + h) + 8 - (4x 2 - 5x + 8) f (x + h ) = 4(x + h ) 2 - 5(x + h ) + 8 and ____ =      lim          h f (x ) = 4x 2 - 5x + 8 h→0 __   lim      Definition of a derivative f ′(x) =    h→0 __ =      lim      Expand and simplify. 8xh + 4h 2 - 5h h h→0 h(8x + 4h - 5) h h→0 __ =      lim      Factor. =      lim (8x + 4h - 5)Divide by h. h→0 Sum and Difference Properties of Limits and Limits of Constant and Identity Functions = 8x + 4(0) - 5 or 8x - 5 Original equation x = 1 and 5 f ′(1) = 8(1) - 5 Simplify. f ′(1) = 3 f ′(x) = 8x - 5 f ′(5) = 8(5) - 5 f ′(5) = 35 GuidedPractice Guided Practice Find the derivative of f (x). Then evaluate the derivative for the given values of x. 1A. f (x) = 6x 2 + 7; x = 2 and 5 f ′(x ) = 12x ; f ′(2) = 24, f ′(5) = 60 1B. f (x) = -5x 2 + 2x - 12; x = 1 and 4 f ′(x ) = -10x + 2; f ′(1) = -8, f ′(4) = -38 dy df The derivative of the function y = f (x) may also be denoted y′, _   , or _   . If a function is preceded by dx dx d   , then you are to take the derivative of the function. a differential operator _ dx 672 | Lesson 11-4 Copyright © McGraw-Hill Education f ′(x) = 8x - 5 Exactostock/SuperStock The derivative of f (x) is f ′(x) = 8x - 5. Evaluate f ′(x) for x = 1 and 5. To this point, you have had to evaluate limits as they approach 0 in order to calculate derivatives, slopes of tangent lines, and instantaneous velocity. A very helpful rule for simplifying this process and for reducing calculation errors is the Power Rule. It allows for the evaluation of derivatives without the need to calculate limits. ReadingMath Derivatives The notation for a derivative f ′(x ) is read f prime of x or the derivative of f with respect to x. KeyConcept Power Rule for Derivatives Words The power of x in the derivative is one less than the power of x in the original function, and the coefficient of the power of x in the derivative is the same as the power of x in the original function. Symbols If f (x ) = x n and n is a real number, then f ′(x ) = nx n - 1. Example 2 Power Rule for Derivatives Find the derivative of each function. a. f (x) = x 9 f (x) = x 9 Original equation f ′(x) = 9x 9 - 1 Power Rule = 9x 8 Simplify. 5 b. g(x) =   √ x 7   5 g(x) =   √ x7  g(x) = Original equation 7 _   x  5 Rewrite using a rational exponent. 7 7 _   g′(x) = _   5 x  5 - 1 Power Rule _2 7   75 2 =_   5 x  5 or _   5   √ x   Simplify. _1 c. h(x) =   8 x 1 h(x) = _   Original equation x8 WatchOut! Negative Derivatives The derivative of f (x ) = x - 4 is not f ′(x ) = - 4x -3. Remember that 1 must be subtracted from the exponent and that - 4 - 1 = - 4 + (-1) or -5. Therefore f ′(x ) = - 4x -5. h(x) = x -8 Rewrite using a negative exponent. -8 - 1 h′(x) = -8x 8 = -8x -9 or -_9 x GuidedPractice Guided Practice 3 2A. j(x) = x 4 j ′(x) = 4x Power Rule Simplify. 1 3 _   2 k ′(x ) = _   x 2 or _     √x 2B. k(x) =  √ x 3   3 2 2C. m(x) = _   5 1 x m′(x ) = -_6 5 x Copyright © McGraw-Hill Education There are several other rules of derivatives that are useful when finding the derivatives of functions that contain several terms. KeyConcept Other Derivative Rules Constant The derivative of a constant function is zero. That is, if f (x ) = c, then f ′(x ) = 0. Constant Multiple of a Power If f (x ) = cx  n, where c is a constant and n is a real number, then f ′(x ) = cnx n - 1. Sum or Difference If f (x ) = g (x ) ± h (x ), then f ′(x ) = g ′(x ) ± h ′(x ). 673 Example 3 Derivative Rules Find the derivative of each function. a. f (x) = 5x 3 + 4 StudyTip Derivatives If f (x ) = x, then f ′(x ) = 1, and if f (x ) = cx, then f ′(x ) = c. f (x) = 5x 3+ 4 Original equation f ′(x) = 5 · 3x 3 - 1 + 0 Constant, Constant Multiple of a Power, and Sum Rules = 15x 2 Simplify. b. g (x) = x 5 (2x 3 + 4) g(x) = x 5(2x 3 + 4) Original equation g(x) = 2x 8+ 4x 5 Distributive Property g′(x) = 2 · 8x 8 - 1 + 4 · 5x 5 - 1 Constant Multiple of a Power and Sum Rules = 16x 7 + 20x 4 Simplify. 5x - 12x + 6  √ x   __ 3 5 c. h(x) =      x 5x 3 - 12x + 6  √ x 5   __ h(x) =    x Original equation 6  √ x 5   5x 3 12x h(x) = _   x - _   x + _   x Divide each term in the numerator by x. 3 _   _5 _3 h(x) = 5x 2- 12 + 6x 2 x  2 · x - 1 = x  2 _3 _    - 1 3 h′(x) = 5 · 2x 2 - 1  + 0 + 6 ·   x 2 Constant, Constant Multiple of a Power, and Sum and Difference Rules 2 1 _   4 2 3A. f ′(x ) = 10x - 3x 3B. g ′(x ) = 15x 4 + 24x 3 3C. h ′(x ) = 12x 2 - 3 Simplify. = 10x + 9x 2 or 10x + 9  √ x GuidedPractice Guided Practice 3A. f (x) = 2x 5 - x 3 - 102 3B. g(x) = 3x 4 (x + 2) __ 3C. h(x) =      x 4x 4- 3x 2+ 5x Now that you are familiar with the basic rules of derivatives, problems involving the slopes of tangent lines and instantaneous velocity can be calculated in just a few steps. Example 5 in Lesson 11-3 involved finding an expression for the instantaneous velocity of a particle. Notice the simplification of the problem as a result of the derivative rules. Example 4 Instantaneous Velocity The distance a particle moves along a path is defined by s(t) = 18t - 3t 3  - 1, where t is given in seconds and the distance of the particle is given in centimeters. Find the expression for the instantaneous velocity v(t) of the particle. The instantaneous velocity v(t) is equivalent to s′(t). s(t) = 18t - 3t 3 - 1 Original equation = 18 - 9t 2 Constant, Constant Multiple of a Power, and Difference Rules Simplify. The instantaneous velocity is v(t) = 18 - 9t 2 . Notice that this is the same result as found in Example 5 of Lesson 11-3. GuidedPractice Guided Practice 4. A soccer ball is kicked straight up. The height of the ball is defined by h(t) = 18t - 5t 2 , where time t is given in seconds and the height of the ball is given in meters. Find the expression for the instantaneous velocity v(t) of the ball at any point in time. v (t ) = 18 - 10t 674 | Lesson 11-4 | Derivatives Copyright © McGraw-Hill Education s′(t) = 18 · 1t 1 - 1 - 3 · 3t 3 - 1 - 0 You found relative and absolute extrema of functions graphically and numerically. On a closed interval, these values can be found using the derivative and the following theorem. KeyConcept Extreme Value Theorem If a function f is continuous on a closed interval [a, b ], then f (x ) attains both a maximum and a minimum on [a, b ]. Relative extrema occur only at critical points where the slope of the tangent line, the derivative of the function, is 0 or undefined. To locate the maximum and minimum of a polynomial function f (x) on [a, b], evaluate the function at a, b, and at any values x in [a, b] for which f ‘(x) = 0. Real-World Example 5 Maximum and Minimum ROLLER COASTER The height h in meters of a car travelling along the track of a roller coaster can be _1 11 _ _4 modeled by h(t) = - t 3 + t 2 +   9 on the interval [1, 12], where time t is given in seconds. Find 9 9 the maximum and minimum heights of the car. Find the derivative of h(t). h(t) =-_t 3 + _   3 t 2 + _   11 4 1 9 9 4 1 _ _ 3-1 h′(t) =- · 3t +   3 · 2t 2 - 1 + 0 9 8 1 2 _ = -_ t +   3 t 3 Original equation Constant, Constant Multiple of a Power, and Sum and Difference Rules Simplify. Solve h′(t) = 0 to find where the critical points of h(x) occur. 8 1 -t 2 + 8t = 0 h ′(t ) = - _t 2 + _   t 3 Factor. -t(t - 8) = 0 Real-WorldLink oller coasters have recently R achieved speeds that exceed 193 kmph and heights over 137 meters. Source: Guinness World Records 3 Critical points for this function occur when t = 0 and 8. Note that although t = 0 is a critical point of the function h(t), it does not lie on the interval [1, 12]. To find the maximum and minimum of the function on [1, 12] evaluate h(t) for 1, 8, and 12. h(1) =-_(1) 3 +  _ (1) 2 + _   9 or 2.44 3 4 11 1 9 4 2 _ 11 1 h(8) =-_(8) 3 +  _ (8)  +   9 or 30 3 9 4 11 1 h(12) = -_(12) 3 +  _ (12) 2 + _   9 or 1.22 3 9 maximum minimum The car will achieve a maximum height of 30 meters 8 seconds into the ride and a minimum height of about 1.2 meters 12 seconds into the ride. Copyright © McGraw-Hill Education WatchOut! Interpreting Graphs The graph in Example 5 shows the height of the car over time. It does not show the shape of the roller coaster. 5. max.: 100 m, min.: 4 m CHECK The graph of h(t) = -_t 3 + _   3 t 2 + _   9 shows that 1 9 4 11 h(t) has a maximum of 30 at t = 8 and a minimum of about 1.2 at t = 12 on the interval [1, 12]. ✔ GuidedPractice Guided Practice 5. BUNGEE JUMPING A bungee jumper’s height h in meters relative to the ground can be modeled by h(t) = 6t 2 - 48t + 100 on the interval [0, 6], where time t is given in seconds. Find the maximum and minimum heights of the jumper. 675 2 Product and Quotient Rules Earlier, you learned that the derivative of the sum of functions is equal to the sum of the individual derivatives. Is the derivative of a product of functions equal to the product of the derivatives? Consider the functions f (x) = x and g(x) = 3x 3. Derivative of Product Product of Derivatives d d _ [ f (x) · g(x)] =  _[x · 3x 3 ] d d d d _ f (x) · _   g(x) =  _(x) · _   (3x 3 ) dx dx d =_   (3x 4 ) dx dx dx dx dx 2 = 1 · 9x = 12x 3 = 9x 2 It is clear that the derivative of the product is not necessarily the product of the derivatives. The following rule can be applied when calculating the derivative of products. KeyConcept Product Rule for Derivatives If f and g are differentiable at x, then _  dx [f (x )g (x )] = f ′(x )g (x ) + f (x )g′(x ). d You will prove the Product Rule for Derivatives in Exercise 64. Example 6 Product Rule Find the derivative of each product. a. h(x) = (x 3 - 2x + 7)(3x 2 - 5) Let f (x) = x 3 - 2x + 7 and g(x) = 3x 2 - 5. So, h(x) = f (x)g(x). f (x) = x 3 - 2x + 7 Original equation 2 Power, Constant Multiple of a Power, Constant, and Sum and Difference Rules f ′(x) = 3x - 2 g(x) = 3x 2 - 5 Original equation g′(x) = 6x Constant Multiple of a Power, Constant, and Difference Rules Use f (x), f ′(x), g(x), and g′(x) to find the derivative of h(x). Product Rule h′(x) = f ′(x)g(x) + f (x)g′(x) 2 2 3 Substitution = (3x - 2)(3x - 5) + (x - 2x + 7)(6x) 4 2 Distribute and simplify. = 15x - 33x + 42x + 10 StudyTip Product Rule The Product Rule results in an answer that can still be simplified. Unless there is an easy simplification to make or a reason to do so, you may leave the answer as is. b. h(x) = (x 3 - 4x 2 + 48x - 64)(6x 2 - x - 2) Let f (x) = x 3 - 4x 2 + 48x - 64 and g(x) = 6x 2 - x - 2. f (x) = x 3 - 4x 2 + 48x - 64 Original equation 2 f ′(x) = 3x - 8x + 48 Power, Constant Multiple of a Power, Constant, and Sum and Difference Rules g(x) = 6x 2 - x - 2 Original equation g′(x) = 12x - 1 Constant Multiple of a Power, Power, Constant, and Difference Rules Product Rule h′(x) = f ′(x)g(x) + f (x)g′(x) 2 2 3 2 = (3x - 8x + 48)(6x - x - 2) + (x - 4x + 48x - 64)(12x - 1) 4 3 2 = 30x - 100x + 870x - 848x - 32 Substitution Distribute and simplify. GuidedPractice Guided Practice 6A–B. See margin. 6A. h(x) = (x 5 + 13x 2 )(7x 3 - 5x 2 + 18) 676 | Lesson 11-4 | Derivatives 6B. h(x) = (x 2 + x 3 + x)(8x 2 + 3) Copyright © McGraw-Hill Education Use f (x), f ′(x), g(x), and g′(x) to find the derivative of h(x). The same reasoning used for the derivatives of products can be applied to quotients. The following rule can be applied when calculating the derivative of quotients. KeyConcept Quotient Rule for Derivatives f ′(x )g (x ) - f (x )g ′(x ) d ⎡ f (x ) ⎤ __ If f and g are differentiable at x and g (x ) ≠ 0, then _  dx ⎢⎣ _ ⎥⎦ =      . 2 g (x ) [g (x )]  You will prove the Quotient Rule for Derivatives in Exercise 67. Example 7 Quotient Rule Find the derivative of each quotient. 5x - 3 _ a. h(x) =   2 x 2 - 6 f (x) g(x) Let f (x) = 5x 2 - 3 and g(x) = x 2 - 6. So, h(x) = _   . f (x) = 5x 2 - 3 Original equation Constant Multiple of a Power, Constant, and Difference Rules f ′(x) = 10x StudyTip Quotient Rule Simplification with the Quotient Rule tends to be more significant and useful. However, it is not necessary to expand the denominator if doing so does not result in further simplification. g(x) = x 2 - 6 Original equation g′(x) = 2x Power, Constant, and Difference Rules Use f (x), f ′(x), g(x), and g ′(x) to find the derivative of h(x). f ′(x)g(x) - f(x)g ′(x) __ h′(x) =      2 [g(x)]  10x(x 2 - 6) - (5x 2 - 3)(2x) Quotient Rule ___ =         2 2 Substitution = __        2 2 Distributive Property =_   2 2 Simplify. (x - 6)  10x 3 - 60x - 10x 3 + 6x (x - 6)  -54x (x - 6) x + 8 _ b. h(x) =   2 x 3 - 2 Let f (x) = x 2 + 8 and g(x) = x 3 - 2. f (x) = x 2 + 8 Original equation Power, Constant, and Sum Rules f ′(x) = 2x g(x) = x 3 - 2 Original equation 2 Power, Constant, and Difference Rules g′(x) = 3x Use f (x), f ′(x), g(x), and g′(x) to find the derivative of h(x). f ′(x)g(x) - f(x)g′(x) __ h′(x) =      2 [ g(x)]  Copyright © McGraw-Hill Education 2x(x 3 - 2) - (x 2 + 8)3x 2 Quotient Rule __ =         3 2 Substitution = __     3 2 Expand and simplify. (x - 2)  -x 4 - 24x 2 - 4x (x - 2)  GuidedPractice Guided Practice 155 7x - 10 _ 7A. j(x) = _     2 12x + 5 (12x + 5)  7B. k(x) = _   2 6x 2x + 4 -12x 2 + 24 __   (2x 2 + 4) 2 677 Exercises Evaluate limits to find the derivative of each function. Then evaluate the derivative of each function for the given values of each variable. (Example 1) 1–6. See margin. Find the derivative of each function. (Example 6) 1. f (x) = 4x 2 - 3; x = 2 and -1 29. g(x) = (3x 4 + 2x)(5 - 3x) 2. g(t) = -t 2 + 2t + 11; t = 5 and 3 30. h(x) = (-7x 2 + 4)(2 - x) ( _ ) 1 3. m( j) = 14 j - 13; j = -7 and -4 31. s(t) = t 2 + 2 (3t 11 - 4t) 4. v(n) = 5n 2 + 9n - 17; n = 7 and 2 3 ( 5. h(c) = c + 2c - c + 5; c = -2 and 1 33. c(t) = (t 3 + 2t - t 7 )(t 6 + 3t 4 - 22t) 6. r(b) = 2b 3 - 10b; b = -4 and -3 34. p(r) = (r 2.5 + 8r)(r - 7r 2 + 108) Find the derivative of each function. (Examples 2 and 3) 7–16. See margin. 8. z(n) = 2n 2 + 7n 1 _   9. p(v) = 7v + 4 2 _   11. b(m) = 3m 3 - 2m 2 1 _   3 _   1 _   3 _   10. g(h) = 2h 2 + 6h 3 - 2h 2 3 _   12. n(t) = _   + _   2 + _   3 + 4 1 t 1 -_ ) 3 _   32. g(x) = x 2 + 2x (0.5x 4 - 3x) 2 7. y( f ) = -11f 28–37. See Chapter 11 Answer Appendix. 28. f (x) = (4x + 3)(x 2 + 9) 3 t 2 t 13. f (x) = 3x 2 - x 2 + 2x 2 14. q(c) = c 9 - 3c 5 + 5c 2 - 3c 15. p(k) = k 5.2 - 8k 4.8 + 3k 16. f (x) = -5x 3 - 9x 4 + 8x 5 17. TEMPERATURE The temperature in degrees Centigrade over a 24-hour period in a certain city can be defined as f (h) = -0.0036h 3 - 0.01h 2 + 2.04h + 52, where h is the number of hours since midnight. (Example 4) ( 9 _   )( -_ 1 ) 5 _   35. q(a) = a 8 + a 4 a 4 - 13a 36. f (x) = (1.4x 5 + 2.7x)(7.3x 9 - 0.8x 5 ) (1 _ 2 )( _ 2 -_1 5 ) 7 _   37. h(x) = _   8 x 3 + _   5 x 6 x 2 + x 8     a–c. See Chapter 11 Answer Appendix. 38. BASEBALL A pitch is struck by a bat with a mass of m kilograms. Suppose the initial speed of the ball after being struck is given by s(m) = _   m + 0.15 . (Example 7) 185m - 15 a. Find an equation for the instantaneous rate of change for 2 the temperature.f ′(h) = -0.0108h - 0.02h + 2.04 b. Find the instantaneous rate of change for h = 2, 14, and 20. f ′(2) = 1.96; f ′(14) = -0.36; f ′(20) = -2.68 c. Find the maximum temperature for 0 ≤ h ≤ 24. 68.92°F Use the derivative to find any critical points of the function. Then find the maximum and minimum points of each graph on the given interval. (Example 5) 18–26. See Chapter 11 Answer Appendix. 19. g(m) = m 3 - 4m + 10; [-3, 3] 20. r(t) = t 4 + 6t 2 - 2; [ 1, 4] 21. t(u) = u 3 + 15u 2 + 75u + 115; [-6, -3] 22. k( p) = p 4 - 8p 2 + 2; [0, 3] Use the Quotient Rule to find the derivative of each function. (Example 7) 24. z(k) = k 3 - 3k 2 + 3k; [0, 3] 25. a(d) = d 4 - 3d 3 + 2; [-1, 4] 39. f (m) = _   3 + 2m 40. g(n) = _   2n + 3 41. r(t) = _   2 42. m(q) = __   3 43. v(t) = _   3 44. c(m) = _   3 45. f (x) = _   2 46. q(r) =  __    3 47. t(w) = _   2 48. m(x) = __        4 3 3 - 2m 1 1 26. c(n) = _   3 n 3 + _   2 n 2 - 6n + 8; [-5, 5] 27. THROWN OBJECT Refer to the application at the beginning of the lesson. The height h of the ball in meters after t seconds can be defined as h(t) = 20t - 5t 2 + 2 for 0 ≤ t ≤ 4. (Example 5) h ′(t ) = 20 - 10t b. Find the maximum and minimum points of h(t) on the interval. Max. (2,22); Min. (0,2) c. Can Mansour throw the ball up to Nasser’s window? 678 | Lesson 11-4 | Derivatives c. If the mass of the bat varies inversely with the hitter’s control of the swing, is it wise to use a bat that weighs 1.05 kilograms over a bat that weighs 0.80 kilogram? Explain your reasoning. 39–48. See Chapter 11 Answer Appendix. 23. f (x) = -5x 2 - 90x; [-11, -8] a. Find h′(t). b. Use a calculator to graph the equation you found in part a on 0 ≤ m ≤ 2. What is happening to the instantaneous rate of change for the initial speed of the ball as the mass of the bat increases? c. See Chapter 11 Answer Appendix. t 2 + 2 3 - t t 2 - 5t + 3 t - 4t x 3 + 2x -x + 3 w+w 4 w 3n + 2 q 4 + 2q 2 + 3 q - 2 4 + 1 m -m + 2m 1.5r 3 + 5 - r 2 r x 5 + 3x -x - 2x - 2x - 3 Copyright © McGraw-Hill Education 18. f (x) = 2x 2 + 8x; [-5, 0] a. Find an equation for the instantaneous rate of change for the initial speed of the ball. 49. ECONOMICS Mahmoud and Mohammad are selling Their weekly B sweatshirts to raise money for the3 junior class. revenue is given by r(x) = 0.125x - 11.25x 2 + 250x, where x is the cost of one sweatshirt. r ′(x) = 0.375x 2 - 22.5x + 250 b. Find the solutions of r ′(x) = 0. 14.72, 45.28 a. Find r’(x). c. What do the solutions you found in part b represent in terms of the given situation? See margin. 50–54. See Chapter 11 Answer Appendix for graphs. Find the equation of the line tangent to f (x) at the given point. Verify your answer graphically. 50. f (x) = 3x 2 + 2x - 7; (1, -2) y = 8x - 10 51. f (x) = -5x 2 - 10x + 25; (-2, 25) y = 10x + 45 y = -0.5x + 4.25 52. f (x) = -0.2x 2 + 1.5x - 0.75; (5, 1.75) 53. f (x) = 4x 2 - 12x - 35; (-1.2, -14.84) 54. f (x) = 0.8x 2 + 0.64x - 12; (10, 74.4) y = -21.6x - 40.76 y = 16.64x - 92 a–c. See margin. 55. DERIVATIVES Let f ′(x) be the derivative of a function f (x). If it exists, we can calculate the derivative of f ′(x), which is called the second derivative, and is denoted f ′′(x) or f (2)(x). We can continue and find the derivative of f ′′(x), which is called the third derivative and is denoted f ′′′(x) or f (3)(x). These are examples of higher-order derivatives. Find the indicated derivative of each function. a. second derivative of f (x) = 4x 5 - 2x 3 + 6 b. third derivative of g(x) = -2x 7 + 4x 4 - 7x 3 + 10x 2 c. fourth derivative of h(x) = 3x + 2x + 4x -3 -2 C MULTIPLE REPRESENTATIONS In this problem, you will investigate how derivatives relate to some geometric properties. b, c, e. See Chapter 11 Answer Appendix. a. ANALYTICAL Find the derivatives of the formulas for the area A of a circle and the volume V of a sphere in 2 terms of r. A ′ = 2πr ; V ′ = 4πr b. VERBAL Explain the relationship between each formula and its derivative. c. GEOMETRIC Draw a square and its apothem a. Draw a cube and the apothem a for three faces that are joined at a shared vertex.2 3 2 A = 4a ; A ′ = 8a ; V = 8a ; V ′ = 24a d. ANALYTICAL Write formulas for the area A of the square and the volume V of the cube in terms of the apothem a. Find the derivative of each formula in terms of a. e. VERBAL Explain the relationship between each formula and its derivative. H.O.T. Problems Use Higher-Order Thinking Skills 62. ERROR ANALYSIS Hiyam and Hana are finding [ f ′(x)] 2, where f (x) = 6x 2 + 4x. Hana thinks that the answer is 144x 2 + 96x + 16, but Hiyam thinks that the answer is 144x 3 + 144x 2 + 32x. Is either of them correct? Explain your reasoning. See Chapter 11 Answer Appendix. 63. CHALLENGE Find f ′(y) if f ( y) = 10x 2y 3 + 5xz 2 - 6xy 2 + 8x 5 - 11x 8yz 7. f ′(y ) = 30x 2y 2 - 12x y - 11x 8z 7 64. PROOF Prove the Product Rule for Derivatives by showing that Sketch a graph for a function that has the given characteristics. 56–59. See Chapter 11 Answer Appendix. f ′(x)g(x) + f (x)g′(x) =      lim __     . 56. The derivative is 0 at x = -1 and x = 1. (Hint: Work with the right side. Add and subtract f (x)g(x + h) in the numerator.) 57. The derivative is -2 at x = -1, x = 0, and x = 2. 58. The derivative is 0 at x = -1, x = 2, and x = 4. 59. The derivative is undefined at x = 4. 60. STUDYING Huda kept track of the amount of time t in minutes that she studied the night before an exam and the percentage p that she earned on the exam. a–c. See Chapter 11 Answer Appendix. Copyright © McGraw-Hill Education 61. t 30 60 90 120 180 210 240 p 39 68 86 96 90 76 56 a. Find a quadratic equation p(t) that can be used to model the data. Round the coefficients to the nearest ten thousandth. Graph the data and p(t) on the same screen. b. Use p′(t) to find the maximum test score that Huda can earn and the amount of time that she would need to study to achieve this score. c. Explain why more time spent studying does not necessarily result in a higher exam score. f (x + h)g(x + h) - f (x)g(x) h h→0 See Chapter 11 Answer Appendix. 65. REASONING Determine whether the following statement is true or false. Explain your reasoning. If f (x) = x 5n + 3, then f ′(x) = (5n + 3)x 5n + 2. See Chapter 11 Answer Appendix. 66. PREWRITE Use a plot pyramid to map out the process of finding the derivative of f (x) = 4x2 - 2x + 5 evaluated at x = 1. See Chapter 11 Answer Appendix. 67. PROOF Prove the Quotient Rule for Derivatives by showing that f ′(x)g(x) - f (x)g′(x) f (x + h)g(x) - f (x)g(x + h) __      =      lim  __       . hg(x + h)g(x) [g(x)] 2 h→0 (Hint: Work with the right side. Add and subtract f (x)g(x) in the numerator.) See Chapter 11 Answer Appendix. See Chapter 11 Answer Appendix. 68. WRITING IN MATH Can two different functions have the same derivative? Explain why this is or is not possible, and provide examples to support your answer. 679 Spiral Review Find the slope of the lines tangent to the graph of each function at the given points. 69. y = x 2 - 3x; (0, 0) and (3, 0) -3; 3 70. y = 4 - 2x; (-2, 8) and (6, -8) -2; -2 Evaluate each limit. 72.      lim _   x 2 - 16 x→-4 x + 4 73.    lim _   2 x 2 + 5x + 6 x→-2 x + x - 2 -8 -_ 1 3 71. y = x 2 + 9; (3, 18) and (6, 45) 74.     lim __   2 3x + 9 x→2 x - 5x - 24 75. EXERCISE A gym teacher asked his students to track how many days they exercised each week. Use the frequency distribution shown to construct and graph a probability distribution for the random variable X, rounding each probability to the nearest hundredth. See margin. 76. SPORTS The number of hours per week members of the North High School basketball team spent practicing, either as a team or individually, are listed below. a–b. See margin. Days, X Frequency 0 3 1 6 2 7 3 8 4 4 5 2 6; 12 -_ 1 2 15, 18, 16, 20, 22, 18, 19, 20, 24, 18, 16, 18 a. Construct a histogram and use it to describe the shape of the distribution. b. Summarize the center and spread of the data using either the mean and standard deviation or the five-number summary. Justify your choice. Use the fifth partial sum of the trigonometric series for cosine or sine to approximate each value to three decimal places. 77. cos _   11 2π 78. sin _   14 3π 0.841 79. sin _   13 0.623 π Write an explicit formula and a recursive formula for finding the nth term of each geometric sequence. 80–82. See margin. 80. 1.25, -1.5, 1.8, … 81. 1.4, -3.5, 8.75, … 0.239 1 1 _ 1 82. _   , _   ,   , … 8 4 2 Skills Review for Standardized Tests A 18 C 24 B 20 D 36 E 40 A AED 7 C AED 9 B AED 8 D AED 10 3 86. REVIEW Find the derivative of f (x) = 5  √ x 8  . 5 40 _   F f ′(x) = _   3 x 3 84. REVIEW What is the slope of the line tangent to the graph of y = 2x 2 at the point (1, 2)? H F 1 H 4 G 2 J 8 680 | Lesson 11-4 | Derivatives 85. The Better Book Company finds that the cost in dirhams to print x copies of a book is given by C(x) = 1000 + 10x - 0.001x 2 . The derivative C′(x) is called the marginal cost function. The marginal cost is the approximate cost of printing one more book after x copies have been printed. What is the marginal cost when 1000 books have been printed? B 8 40 _   G f ′(x) = _   3 x 3 F 5 _   H f ′(x) = 225x 3 8 _   J f ′(x) = 225x 3 Copyright © McGraw-Hill Education 83. SAT/ACT The figure shows the dimensions, in meters, of a stone slab. How many slabs are required to construct a rectangular patio 24 meters long and 12 meters wide? D 11-5 Then You computed limits algebraically by using the properties of limits. NewVocabulary regular partition definite integral lower limit upper limit right Riemann sum integration Area Under a Curve and Integration Now 1 2 Why? Approximate the area under a curve using rectangles. Approximate the area under a curve using definite integrals and integration. Marginal cost is the approximate cost that a company incurs to produce an additional unit of a product. The marginal cost equation is the derivative of the actual cost equation. The marginal cost function for a particular publisher is f (x ) = 10 - 0.002x, where x is the number of books manufactured and f (x ) is in dirhams. 1 Area Under a Curve In geometry, you learned how to calculate the area of a basic figure, such as a triangle, a rectangle, or a regular polygon. You also learned how to calculate the area of a composite figure, that is, a region comprising basic shapes. However, many regions are not a collection of the basic shapes. As a result, you need a general approach for calculating the area of any twodimensional figure. We can approximate the area of an irregular shape by using a basic figure with a known formula for area, the rectangle. For example, consider the graph of f (x) = -x 2 + 12x on the interval [0, 12]. We can approximate the area between the curve and the x-axis using rectangles of equal width. Example 1 Area Under a Curve Using Rectangles Approximate the area between the curve f (x) = -x 2 + 12x and the x-axis on the interval [0, 12] using 4, 6, and 12 rectangles. Use the right endpoint of each rectangle to determine the height. Copyright © McGraw-Hill Education Mark Dierker/McGraw-Hill Education Using the figures from below for reference, notice that the rectangles were drawn with a height equal to f (x) at each right endpoint. For example, the heights of the rectangles in the first figure are f (3), f (6), f (9), and f (12). We can use these heights and the length of the base of each rectangle to approximate the area under the curve. Area using 4 rectangles R 1 = 3 · f (3) or 81 R 2 = 3 · f (6) or 108 R 3 = 3 · f (9) or 81 R 4 = 3 · f (12) or 0 total area = 270 Area using 6 rectangles R 1 = 2 · f (2) or 40 R 2 = 2 · f (4) or 64 R 3 = 2 · f (6) or 72 R 4 = 2 · f (8) or 64 R 5 = 2 · f (10) or 40 R 6 = 2 · f (12) or 0 total area = 280 The approximation for the area under the curve using 4, 6, and 12 rectangles is 270 square units, 280 square units, and 286 square units, respectively. Area using 12 rectangles R 1 = 1 · f (1) or 11 R 2 = 1 · f (2) or 20 R 3 = 1 · f (3) or 27 R 4 = 1 · f (4) or 32 R 5 = 1 · f (5) or 35 R 6 = 1 · f (6) or 36 R 7 = 1 · f (7) or 35 R 8 = 1 · f (8) or 32 R 9 = 1 · f (9) or 27 = 1 · f (10) or 20 R 10 R 11 = 1 · f (11) or 11 R 12 = 1 · f (12) or 0 total area = 286 681 TechnologyTip Tables To help generate multiple heights of rectangles with your graphing calculator, enter the function using the menu. Then use the TABLE function by pressing [TABLE]. This will generate a list of heights for different values of x. You can also change the interval for the x values in your table by pressing [TBLSET] and adjusting the TBLSET options. GuidedPractice Guided Practice 1. Approximate the area between the curve f (x) = -x 2 + 24x and the x-axis on the interval [0, 24] using 6, 8, and 12 rectangles. Use the right endpoint of each rectangle to determine the height. 6 rectangles = 2240 units 2; 8 rectangles = 2268 units 2; 12 rectangles = 2288 units 2 Notice that the thinner the rectangles are, the better they fit the region, and the better their total area approximates the area of the region. Also, the rectangles were drawn so that the right endpoint of each one evaluated at f (x) represents the height. The left endpoints may also be used to determine the height of each rectangle and can produce a different result for the approximated area. Using right or left endpoints may result in adding or excluding area that does or does not lie between the curve and the x-axis. In some cases, better approximations may be obtained by calculating the area using both left and right endpoints and then averaging the results. Example 2 Area Under a Curve Using Left and Right Endpoints Approximate the area between the curve f (x) = x 2 and the x-axis on the interval [0, 4] by first using the right endpoints and then by using the left endpoints of the rectangles. Use rectangles with a width of 1. Using right endpoints for the height of each rectangle produces four rectangles with a width of 1 unit (Figure 11.5.1). Using left endpoints for the height of each rectangle produces four rectangles with a width of 1 unit (Figure 11.5.2). Figure 11.5.1 Area using right endpoints R 1 = 1 · f (1) or 1 R 2 = 1 · f (2) or 4 R 3 = 1 · f (3) or 9 Figure 11.5.2 Area using left endpoints R 1 = 1 · f (0) or 0 R 2 = 1 · f (1) or 1 R 4 = 1 · f (4) or 16 R 3 = 1 · f (2) or 4 R 4 = 1 · f (3) or 9 total area = 30 total area = 14 The area using the right and left endpoints is 30 and 14 square units, respectively. We now have lower and upper estimates for the area of the region, 14 < area < 30. Averaging the two areas would give a better approximation of 22 square units. right endpoint = 15.4 units 2; left endpoint = 25 units 2; average = 20.2 units 2 12 2. Approximate the area between the curve f (x) = _   x and the x-axis on the interval [1, 5] by first GuidedPractice Guided Practice Any point within the width of the rectangles may be used as the heights when approximating the area between the graph of a curve and the x-axis. The most commonly used are the left endpoints, the right endpoints, and the midpoints. 2 Integration As we saw in Example 1, as the rectangles get thinner, their total area approaches the exact area of the region under the curve. We can conclude that the area of the region under a curve is the limit of the total area of the rectangles as the widths of the rectangles approach 0. 682 | Lesson 11-5 | Area Under a Curve and Integration Copyright © McGraw-Hill Education using the right endpoints and then by using the left endpoints. Use rectangles with a width of 1 unit. Then find the average of the two approximations. In the figure, the interval from a to b has been subdivided into n equal subintervals. This is called a regular partition. The length of the entire interval from a to b is b - a, so the width of each of the n rectangles is _   n and is denoted ∆x. The height b-a of each rectangle at the right endpoint corresponds with the value of the function at that point. Thus, the height of the first rectangle is f (x1), the height of the second is f (x 2), and so on, with the height of the last rectangle being f (xn). ReadingMath Sigma Notation n f (x i) ∆x is read the i =1 s ummation of the product of f of x sub i from 1 to n and the change in x. The area of each rectangle can now be calculated by taking the product of ∆x and the corresponding height. The area of the first rectangle is f (x 1)∆x, the area of the second rectangle is f (x 2)∆x, and so on. The total area A of the n rectangles is given by the sum of the areas and can be written in sigma notation. A = f (x1)∆x + f (x2)∆x + ⋯ + f (x n)∆x Add the areas. A = ∆x[ f (x1) + f (x2) + ⋯ + f (xn)] Factor out ∆x. n A = ∆x  f(xi) Write the sum of the heights in sigma notation. i =1 n A =  f(xi) ∆x Commutative Property of Multiplication i =1 To assist in future calculations, we can derive a formula to find any x i. The width ∆x of each rectangle is the distance between successive x i-values. Consider the x-axis. We can see that xi= a + i∆x. This formula will be useful when finding the area under the graph of any function. To make the width of the rectangles approach 0, we let the number of rectangles approach infinity. This limit is called a definite integral and is given special notation. KeyConcept Definite Integral The area of a region under the graph of a function is b n      f (x )dx =   limf (xi )∆x, a n→∞ i =1 b-a _ where a and b are the lower limit and upper limit respectively, ∆x =   n , and xi = a + i∆x. This method is referred to as the right Riemann sum. The Riemann sum is named for the German mathematician Bernhard Riemann (1826–1866). He is credited with formulating the expression for approximating the area under a curve using limits. The expression may be altered to use left endpoints or midpoints. Copyright © McGraw-Hill Education The process of evaluating an integral is called integration. The following summation formulas will assist in evaluating definite integrals. WatchOut! Summations The sum of a constant c is cn, not 0 or ∞. n For instance,  5 = 5n. i =1 n n i =1 i =1 n 2 (n + 1) 2 c= cn, c is a constant  i 3 = _   4 n n(n + 1) n i= _   2   i 4 = __     30 i =1 n 6n 5 + 15n 4 + 10n 3 - n i =1 n(n + 1)(2n + 1) n __ __  i 2 =        i 5 =      12 6 i =1 2n 6 + 6n 5 + 5n 4 - n 2 i =1 683 There are two summation properties that are needed to evaluate some integrals. n n n n (a i+ b i) =  a i±  b i i =1 i =1 n ci= ci, c is a constant i =1 i =1 i =1 Example 3 Area Under a Curve Using Integration Use limits to find the area of the region between the 4     x 2 dx. graph of y = x 2 and the x-axis on the interval [0, 4], or  0 First, find ∆x and x i.   n ∆x = _ b-a Formula for ∆x 4-0 _4 =  _ n or  n b = 4 and a = 0 x i= a + i∆x Formula for x i _4  4in = 0 + in or _ _4 a = 0 and ∆ x =   n Calculate the definite integral that gives the area. n 4      x 2 dx =  lim f (xi) ∆x Definition of a definite integral n→∞ i =1 n 0 =  lim (xi) 2 ∆x f (xi ) = x i 2 n→∞ i =1 n ( ) (_) _ _ 4i 2 4 4i 4 =  lim   _   n      n xi=   n and ∆ x =   n n→∞ i =1 n 4i 2 4 =  lim_    _   n Factor. ( ) n→∞ n i=1 n 16i 2 4 =  lim_    _   2 Expand. n n→∞ n i=1 StudyTip Limits Factor each summation until the expression remaining involves only a constant or i. Then apply the necessary summation formula. ( n ) =  lim_   _ 2   i 2 Factor. 4 16 n i =1 n→∞ n __ __ n n (n + 1)(2n + 1) 4 ⎡ 16 n(n + 1)(2n + 1) ⎤ ⎥  i 2 =     =  lim_   ⎢⎣_   2 ·      6 ⎦ n n→∞ n 6 i =1 16n(2n 2 + 3n + 1)⎤ 4 ⎡__ ⎥ =  lim_   ⎢    Multiply and expand. n→∞ n ⎣ ⎦ 6n 2 64n(2n 2 + 3n + 1) =  lim__     Multiply. 3 6n n→∞ 64(2n 2 + 3n + 1) =  lim__     Divide by n. n→∞ 6n 2 2 64 ⎡ (2n + 3n + 1) ⎤ ⎥ =  lim_   ⎢ __    Factor. n→∞ 6 ⎣ ⎦ n 2 ( ) =  lim_   2 + _   + _   2 Divide each term by n 2. 64 n→∞ 6 ( 3 n ) 1 n ( ) 64 ⎡ 1 1⎤ =   lim_   ⎢⎣ lim2 + (  lim3)   lim _ +  lim_   2 ⎥⎦ n→∞ 6 n→∞ n→∞ n→∞ n n→∞ n Limit theorems = _   6 [2 + 3(0) + 0] or _   3 The area is _   3 or 21 _ square units. 3 64 64 1 GuidedPractice Guided Practice Use limits to find the area between the graph of each function and the x-axis given by the definite integral. 1 3A.     3x 2 dx 0 1 unit 2 684 | Lesson 11-5 | Area Under a Curve and Integration 3 3B.      x dx 0 4.5 units 2 Copyright © McGraw-Hill Education 64 The areas of regions that do not have a lower limit at the origin can also be found using limits. Example 4 Area Under a Curve Using Integration Use limits to find the area of the region between the graph 3     4x 3 dx. of y = 4x 3and the x-axis on the interval [1, 3], or  1 First, find ∆x and x i. ∆x = _   n b-a Formula for Δ x 3-1 2 =_   n or _  n b = 3 and a = 1 xi= a + i∆x Formula for x i _ = 1 + i_ n or 1 +   n 2 2i _2 a = 1 and ∆ x =   n Calculate the definite integral that gives the area. n 3      4x 3 dx =  lim f ( xi)∆x Definition of a definite integral =  lim 4(x i) 3∆x f (xi ) = 4(xi ) 3 n→∞ i =1 n 1 n→∞ i =1 n _2i )3 (_2 ) ( _2i _2 =  lim 4 1 +   n     n xi= 1 +   n and ∆ x =   n n→∞ i =1 n =  lim_   n    1 + _   n 8 n→∞ i =1 ( ) 2i 3 n Factor ⎡ 2i 2i 2 2i 3⎤ 8 =  lim _   ⎢1 + 3 _   n + 3 _   n + _   n ⎥⎦ n→∞ n i =1 ⎣ ( ) n ( ( ) ( ) Expand. ) =  lim _   1 + _   + _   2 + _   3 8 n→∞ n i =1 ( _ ( 12i 2 n 6i n 8i 3 n n n n i =1 i =1 i =1 Simplify. n 6i 12i 2 8 =  lim _  1 +  _   n + _ 2   +  n→∞ n n n n 8i3 i =1 n n ) _   3 ) n =  lim   1+ _ n  i+ _   2   i  2 + _   3   i 3 n→∞ n n n 8 WatchOut! Limits When finding the area under a curve by using limits, evaluate the summation expressions for the given values of i before distributing the width ∆x or any other constants. 6 i =1 12 i =1 8 i =1 _ i =1 Apply summations. Factor constants. _ 2 6 n(n + 1) 8 n (n + 1) 2⎤ 12 n(n + 1)(2n + 1) 8⎡ ⎥ =  lim_ ⎢ n +  _·   + _   2 · __     + _   3 ·   n→∞ n ⎣ ⎦ n n n 6 2 4 Summation formulas ⎡ 8n 48n(n +1) __ 96n(2n 2 + 3n + 1) 64n 2(n 2 + 2n + 1) ⎤ __ ⎥ =  lim⎢_+ _   +      +      2 3 n→∞ ⎣ n ⎦ 4n 4 2n 6n Distribute   n . _8 ⎡ 24(n + 1) 16(2n 2 + 3n + 1) 16(n 2 + 2n + 1) ⎤ __ ⎥ =  lim⎢8 + _   + __      +      Simplify. 2 n→∞ ⎣ ⎦ n n 2 n Factor and ⎡ 3 2 1 1 1 ⎤ =  lim⎢8 + 24 1 + _   + 16 2 + _   +  _2 + 16 1 +  _+ _   2 ⎥⎦ perform n→∞ ⎣ n n n n n ( ( Copyright © McGraw-Hill Education ( ) ) ) ( ( ) ) ( ) division. 3 2 1 1 1 =  lim8 + 24 lim 1 + _   +16 lim 2 + _   + _   2 +16 lim 1 + _   + _   2 n→∞ n→∞ n→∞ n→∞ n n n n n Limit Theorems = 8 + 24(1 + 0) + 16(2 + 0 + 0) + 16(1 + 0 + 0) or 80 Simplify. The area of the region is 80 square units. GuidedPractice Guided Practice Use limits to find the area between the graph of each function and the x-axis given by the definite integral. 3 4A.     x 2 dx 1 26 2 _   or 8 _   units 2 3 3 4 4B.      x 3 dx 2 60 units 2 685 Definite integrals can be used to find the areas of other irregular shapes. Real-World Example 5 Area Under a Curve LANDSCAPING Amer charges AED 2.40 per square meter of mulch to deliver and install it. He is hired to create two identical flower beds in the back corners of a residential lot. If the area 10 of each flower bed can be found by      (10 - 0.1x 2 )dx, how much will Amer charge for the 0 flower beds if x is given in terms of meters? First, find ∆x and x i. ∆x = _   n b-a Formula for Δx =_   n or _   n 10 - 0 10 b = 10 and a = 0 x i= a + i∆x 10 _ 10i _10 a = 0 and ∆x =   n Calculate the definite integral that gives the area. n 10      (10 - 0.1x 2 ) dx =  lim f ( x i)∆x Definition of a definite integral n→∞ i =1 0 n =  lim (10 - 0.1x i  2)∆x n→∞ i =1 n f (xi ) = 10 - 0.1xi 2 (_) _ n ( ) =  lim_      10 - _   2 10 n→∞ n i =1 ( _ ( 10i 2 n _ n Expand and simplify. ) n =  lim_    10 -    _ 2 10i 2 i =1 n 10 n→∞ n i =1 n n Apply summations. ) 10 10  10 - _   2   i 2 n→∞ n i =1 n i =1 =  lim  __) ( 10 n(n + 1)(2n + 1) n 6 =  lim_   10n - _   2 ·      10 n→∞ n ) ( =  lim(100 - __   ) 2 100n(2n + 3n + 1) 100n __ =  lim _ -      2 n→∞ 3n n 50(2n 2 + 3n + 1)    3n 2 n→∞ ( ) ⎡ 50 3 1 ⎤ =  lim⎢ 100 - _   2 + _   + _   2 ⎥⎦ n→∞ ⎣ n 3 n ( Factor. Summation formulas _10 Distribute   n . Divide by n. Factor and perform division. =  lim100 - _     lim 2 + _   + _   2 ) Limit Theorems = 100 - _   3 (2 + 0 + 0) or 66 _ 3 Simplify. 50 3 n→∞ n→∞ 50 5. No; the area of one hill is approximately 16.67 m2. The students would then need 2 · 16.67 or about 33.3 m2 of paint, which they do not have. _ ⎡ 10i 2⎤ 10 10i 10 =  lim  ⎢⎣10 - 0.1   n   ⎥⎦·   n xi=   n and ∆ x =   n n→∞ i =1 3 n 1 n 2 The area of one flower bed is about 66.67 square meters. For Amer to install both flower beds, he 2 would have to charge AED 2.40 · 66 _ · 2 or AED 320. 3 ( ) GuidedPractice Guided Practice 5. PAINTING Mrs. Hidaya’s art class is painting a large mural depicting a winter ski scene. The students want to paint a ski hill at both the beginning and the end of the mural, but they only have enough paint to cover 30 square meters. If the area of each ski hill can be found by 5      (5 - 0.2x 2)dx, will the students have enough paint for both hills? Explain. 0 686 | Lesson 11-5 | Area Under a Curve and Integration Spaces Images/Blend Images Landscape Architect Employment opportunities as a landscape architect are projected to increase by 16% by 2016. Landscape architects design golf courses, college campuses, public parks, and residential areas. A professional license and a bachelor’s degree in landscape architecture are generally required. = 0 + i n or _   n Copyright © McGraw-Hill Education Real-WorldCareer Formula for x i Exercises Approximate the area of the shaded region for each function by first using the right endpoints and then by using the left endpoints. Then find the average for both approximations. Use the specified width for the rectangles. (Example 2) Approximate the area of the shaded region for each function using the indicated number of rectangles. Use the specified endpoints to determine the heights of the rectangles. (Example 1) 2 2 1. 5 rectangles 15 units  2. 4 rectangles 9.25 units  right endpoints left endpoints 8–13. See margin. 8. width = 0.5 9. width = 1.0 10. width = 0.75 2 18.29 3. 8 rectangles 4. 8 rectangles 7.75 units  right endpoints left endpoints units 2 11. width = 0.5 16.23 5. 4 rectangles 6. 5 rectangles 0.65 units  units 2 left endpoints right endpoints 2 12. width = 0.75 13. width = 0.5 7. FLOORING Majed is installing a hardwood floor and has to cover a semicircular section defined by f (x) = (-x 2 + 10x) 0.5 . (Example 1) 37.96 units 2 a. Approximate the area of the semicircular region using left endpoints and rectangles one unit in width. b. Majed decides that using both left and right endpoints may give a better estimate because it would eliminate any area outside the semicircular region. Approximate the area of the semicircular region as shown in the figure. 32.96 units 2 Use limits to find the area between the graph of each function and the x-axis given by the definite integral. (Examples 3 and 4) 4 14.      4x 2 dx 1 2 16.      6x dx 0 84 units 2 2 12 units  75 units  5 18.      (x 2 + 4x - 2) dx 2 4 10 _   units 2 3 24 _   4 units 2 Copyright © McGraw-Hill Education 0 2 3 3 3 22.      (x + x) dx 0 33 _ units 2 5 24.      (x 2 - x + 1) dx 1 3 2 1 3 18.6 units 2 26.      (8 - 0.6x ) dx 0 2 3 52 units 2 23 _ units 2 2 1 3 17.      (2x + 3) dx 1 2 20.      (4x - x 2) dx c. Find the area of the region by using the formula for the area of a semicircle. Which approximation is closer to the actual area of the region? Explain why this estimate gives a better approximation. See margin. 6 15.      (2x + 5) dx 2 19.      8x 3 dx 1 30 units 2 8 _ units 2 2 4 21.      (-x 2 + 6x) dx 3 3 4 12 units 2 23.      (-3x + 15) dx 2 3 25.      12x dx 1 8 48 units 2 27.      0.5x 2 dx 3 80 _ units 2 5 6 687 28. PUBLISHING Refer to the beginning of the lesson. The publisher would like to increase daily production from 1000 books to 1500 books. Find the cost of the increase if it is defined as    1500 AED 3750 (10 - 0.002x) dx. (Example 5) 1000 Use limits to find the area between the graph of each function and the x-axis given by the definite integral. 13 -3 3 2    (-0.2x + 2.8x - 1.8) dx, where x is given in meters. 1 (Example 5) 39.      (-x 3) dx 2 10 _   units 2 3 40.    2   dx 29. ARCHWAY The graduation ceremony committee decided the entrance into the graduation ceremony would be a balloon archway. In addition, the committee wanted to hang streamers that would extend from the top of the arch down to the floor, completely covering the entrance. Find the area beneath the balloon archway if it can be defined as 0 -1 38.      (-2x 2 - 7x) dx -4 -2 -1 14 units 2 41.      -_x + 3   dx -2 ( ) 1 2 3 3 _ units 2 4 42. C MULTIPLE REPRESENTATIONS In this problem, you will investigate the process of finding the area between two curves. a–e. See Chapter 11 Answer Appendix. a. GRAPHICAL Graph f(x) = -x 2 + 4 and g(x) = x 2 on the same coordinate plane and shade the areas 1 67.2 m2 4 units 2 1 represented by    (-x 2+ 4) dx and    x 2dx. 0 30. LOGO Part of a company’s logo is in the shape of the region shown. If this part of the logo is to be sewn on a flag, how much material is required if x is given in meters? (Example 5) 8 m2 0 1 1 b. ANALYTICAL Evaluate      (-x 2 + 4) dx and      x 2 dx. 0 0 c. VERBAL Explain why the area between the two curves is 1 1 equal to      (-x 2 + 4) dx -      x 2 dx. Then calculate this 0 0 value using the values found in part b. d. ANALYTICAL Find f(x) - g(x). Then evaluate 1      [ f (x)-g(x)] dx. 0 e. VERBAL Make a conjecture about the process of finding the area between two curves. 31. NEGATIVE LIMITS Definite integrals can be calculated for both positive and negative limits. B H.O.T. Problems Use Higher-Order Thinking Skills 43. ERROR ANALYSIS Fahd says that when you use the right endpoints of rectangles to estimate the area between a curve and the x-axis, the total area of the rectangles will always be greater than the actual area. Faleh says that the area of the rectangles is always greater when you use the left endpoints. Is either of them correct? Explain. See Chapter 11 Answer Appendix. 1 44. CHALLENGE Evaluate      (5x 4 + 3x 2 - 2x + 1) dx. a. Find the height and length of the base of the triangle. Then calculate the area of the triangle using its height height: 4 units, base: 4 units; 8 units 2 and base. b. Calculate the area of the triangle by evaluating -2 1 -1 -2 2 _   units 2 3 ℓ ∙     f (x) dx, where ℓ is the length of the tunnel. 34.      (-x - 6x) dx 1 3 0 2 8 units  36.    (  2x + 6) dx -2 0 33.    (  x 3 + 2) dx -1 -2 35.      -5x dx -3 -4 See margin. a 17 _ units 2 2 45. REASONING Suppose every vertical cross section of a tunnel can be modeled by f (x) on [a, b]. Explain how the volume of the tunnel can be calculated by using b 8 units 2 Use limits to find the area between the graph of each function and the x-axis given by the definite integral. 32.    x  2 dx 2 0 3 1 12 _ units 2 37.    (  x 3 - 2x) dx -1 1 _ units 2 4 2 3 _   units 2 4 688 | Lesson 11-5 | Area Under a Curve and Integration 46. PREWRITE Write an outline that could be used to describe the steps involved in estimating the area between the x-axis and the curve of a function on a given interval. See students’ work. t 47. CHALLENGE Evaluate      (x 2 + 2) dx. 0 t 3 _ + 2t units 2 3 48. WRITING IN MATH Explain the effectiveness of using triangles and circles to approximate the area between a curve and the x-axis. Which shape do you think gives the best approximation? See margin. Copyright © McGraw-Hill Education 2    (x + 2) dx. 0 Spiral Review 50. f ′(k ) = -119k 16 + 16k 15 - 28k 3 - 39k 2 + 4k Find the derivative of each function. 49. j(x) = (2x 3 + 11x)(2x 8 - 12x 2 ) 50. f(k) = (k 15 + k 2 + 2k)(k - 7k 2 ) 51. s(t) = (  √t - 7)(3t 8 - 5t) 15 51 _   2 Find the slope of the line tangent to the graph of each function at x = 1. 52. y = x 3 3 53. y = x 3 - 7x 2 + 4x + 9 -7 54. y = (x + 1)(x - 2) Evaluate each limit. 55.lim _   x x→0 x 2 + 3x 3 56.lim _   x - 1 x→1 x 2 - 3x + 2 1 5 _   2   t 2 - 168t 7 - _   t 2 + 35 s′(t ) = _ j ′(x) = 44x 10 + 198x 8 - 120x 4 - 396x 2 57. lim _   3 x 2 - 9 x→3 x - 27 -1 1 2 _   9 58. SHOPPING In a recent year, 33% of Americans said that they were planning to go shopping on Friday. What is the probability that in a random sample of 45 people, fewer than 14 people 33.4% plan to go shopping on Friday? Classify each random variable X as discrete or continuous. Explain your reasoning. 59. X represents the number of mobile phone calls made by a randomly chosen student during a given day. Discrete; the number of mobile phone calls is countable and therefore discrete. 60. X represents the time it takes a randomly selected student to run a kilometer. Continuous; the time could be any time between a reasonable interval of time, such as between 5 and 15 minutes. Skills Review for Standardized Tests 61. SAT/ACT If the statement below is true, then which of the following must also be true? If at least 1 bear is sleepy, then some ponies are happy. D A If all bears are sleepy, then all ponies are happy. B If all ponies are happy, then all bears are sleepy. C If no bear is sleepy, then no pony is happy. Copyright © McGraw-Hill Education D If no pony is happy, then no bear is sleepy. E If some ponies are happy, then at least 1 bear is sleepy. x 2 + 3x - 10 62. REVIEW What is lim __   ? J x→3 x 2 + 5x + 6 3 1 F _   15 H _   15 2 4 G _   15 J _   15 63. Find the area of the region between the graph of y = -x 2 + 3x and x-axis on the interval [0, 3] 3 or      (-x 2 + 3x) dx. 0 B A 3 _ units 2 4 3 C 21 _ units 2 4 B 4 _ units 2 2 D 22 _ units 2 2 1 1 1 64. REVIEW Find the derivative of n(a) = _   -  _2 + _   4 + 4a. F n’(a) = 8a - 5a 2 + 3a 4 J 4 a 5 a 3 a H n’(a) = 4a 2 - 5a 3 + 3a 4 + 4 G n’(a) = -_2 + _   3 - _   5 + 4 5 3 4 a a a 10 12 4 J n’(a) = -_2 + _   - _   + 4 a a 3 a 5 689 11-6 Then You used limits to approximate the area under a curve. NewVocabulary antiderivative indefinite integral Fundamental Theorem of Calculus The Fundamental Theorem of Calculus Now 1 2 Why? Find antiderivatives. Use the Fundamental Theorem of Calculus. During the initial ascent of a hot-air balloon ride, Nahla realized that she had her brother’s cell phone in her pocket. Before the balloon rose too high, Nahla dropped the phone over the edge to her brother waiting on the ground. Knowing that the velocity of the phone could be described as v(t ) = -10t, where t is given in seconds and velocity is given in meters per second, Nahla was able to determine how high she was from the ground when she dropped the phone over the edge. 1 Antiderivatives and Indefinite Integrals In Lessons 11-3 and 11-4, You learned that if the position of an object is defined as f(x) = x 2 + 2x, then an expression for the velocity of the object is the derivative of f (x) or f ′(x) = 2x + 2. However, if you are given an expression for velocity, but need to know the formula from which it was obtained, we need a way to work backward or undo the differentiation. In other words, if given f (x), we want to find an equation F(x) such that F′(x) = f (x). The function F(x) is an antiderivative of f (x). Example 1 Find Antiderivatives Find an antiderivative for each function. a. f (x) = 3x 2 We need to find a function that has a derivative of 3x 2 . Recall that the derivative has an exponent that is one less than the exponent in the original function. Therefore, F(x) will be raised to a power of three. Also, the coefficient of a derivative is determined in part by the exponent of the original function. F(x) = x 3 fits this description. The derivative of x 3 = 3x 3 - 1 or 3x 2 . However, x 3 is not the only function that works. The function G(x) = x 3 + 10 is another because its derivative is G′(x) = 3x 3 - 1 + 0 or 3x 2 . Another answer could be H(x) = x 3 - 37. G(x) = x -8 + 3 and H(x) = x -8 - 12 are other antiderivatives. GuidedPractice Guided Practice Find two different antiderivatives for each function. 1A. 2x Sample answers: x 2, x 2 + 5, x 2 - 7, x 2 + 28 Sample answers: x -3 , 1B. -3x -4 -3 x + 33, x -3 - 4, x -3 + 9 In Example 1, notice that adding or subtracting constants to the original antiderivative produced other antiderivatives. In fact, because the derivative of any constant is 0, adding or subtracting a constant C to an antiderivative will not affect its derivative. Therefore, there are an infinite number of antiderivatives for a given function. Antiderivatives that include a constant C are said to be in general form. 690 | Lesson 11-6 Copyright © McGraw-Hill Education x Rewrite f(x) with a negative exponent, -8x -9 . Again, the derivative has an exponent that is one less than the exponent in the original function, so F(x) will be raised to the negative eighth power. We can try F(x) = x -8 . The derivative of x -8 is -8x -8 - 1 or -8x -9 . David H. Carriere/Photodisc/Getty Images _8 b. f ( x ) = - 9 As with derivatives, there are rules for finding antiderivatives. KeyConcept Antiderivative Rules Power Rule If f ( x ) = x n, where n is a rational number other than -1, F ( x ) =  _ + C. n+1 Constant Multiple of a Power If f ( x ) = kx n, where n is a rational number other than -1 and k is a constant, then x n + 1 + C. F ( x ) =   _ n+1 kx n + 1 Sum and Difference If the antiderivatives of f ( x) and g (x ) are F ( x ) and G (x ) respectively, then an antiderivative of f ( x ) ± g ( x ) is F ( x ) ± G (x ). StudyTip Antiderivatives An antiderivative of a constant k is k x. For instance, if f (x ) = 3, then F (x ) = 3x. Example 2 Antiderivative Rules Find all antiderivatives for each function. a. f (x) = 4x 7 f(x) = 4x 7 Original equation F(x) = _   7 + 1 + C Constant Multiple of a Power 4x 7 + 1 =_ x 8 + C 1 2 Simplify. _2 b. f (x) =   4 x 2 f(x) = _   Original equation x 4 = 2x -4 Rewrite with a negative exponent. F(x) = _   -4 + 1 + C -4+1 2x Constant Multiple of a Power = - _ x -3 + C or - _ 3 + C 2 3 2 3x Simplify. c. f (x) = x 2 - 8x + 5 f(x) = x 2 - 8x + 5 2 1 Original equation 0 Rewrite the function so each term has a power of x. = x - 8x + 5x F(x) =   _ -_   1 + 1 + _   0 + 1 + C 2+1 2+1 x 1+1 8x 0+1 5x =_   3 x 3 - 4x 2 + 5x + C 1 Antiderivative Rule Simplify.   5 x 5 + C GuidedPractice Guided Practice A. F (x ) = _ 6 4 Copyright © McGraw-Hill Education 2A. f(x) = 6x F (x ) = x 8 + 3x 2 + 2x + C 10 _ 2B. f (x) = 3 2C. f (x) = 8x 7 + 6x + 2 x F (x ) = -5x -2 + C The general form of an antiderivative is given a special name and notation. KeyConcept Indefinite Integral The indefinite integral of f ( x) is defined by f ( x) dx = F ( x ) + C, where F ( x ) is an antiderivative of f ( x ) and C is any constant. 691 Real-World Example 3 Indefinite Integral EGG DROP Students of Ms. Nisreen’s technology class are participating in an egg drop competition. Each team of students must build a protective device that will keep an egg from cracking after a 9-meter drop. The instantaneous velocity of an egg can be defined as v(t) = -10t, where t is given in seconds and velocity is measured in meters per second. a. Find the position function s(t) of a dropped egg. To find the function for the position of the egg, find the antiderivative of v(t). s(t) =  v(t) dt Real-WorldLink In egg drop competitions, participants attempt to protect their eggs from a two-story drop. Scoring may be based on the weight of the protective device, the number of parts the device includes, whether the device hits a target, and of course, whether the egg breaks. Relationship between position and velocity =  -10t dt v (t ) = -32t = - _ +C Constant Multiple of a Power = -5t 2 + C Simplify. 10t 1 + 1 1+1 Find C by substituting 9 meters for the initial height and 0 for the initial time. s(t) = -5t 2 + C 9= -5(0) 2 Antiderivative of v(t ) s(t ) = 9 and t = 0 +C Simplify. 9=C Source: Salem-Winston Journal The position function for the egg is s(t) = -5t 2 + 9. b. Find how long it will take for the egg to hit the ground. Solve for t when s(t) = 0. s(t) = -5t 2 + 9 2 0 = -5t + 9 -9 = -5t 2 2 1.8 =t 1.341 = t Position function for the egg s(t ) = 0 Subtract 30 from each side. Divide each side by -16. Take the positive square root of each side. The egg will hit the ground in about 1.34 seconds. GuidedPractice Guided Practice 3. FALLING OBJECT A maintenance worker, securely standing on a catwalk in a gymnasium, is fixing a light positioned 36 meters above the floor when his wallet accidentally falls out of his pocket. The instantaneous velocity of his wallet can be defined as v(t) = -10t, where t is given in seconds and velocity is measured in meters per second. A. Find the position function s(t) of the dropped wallet. s (t ) = -5t 2 + 36 B. Find how long it will take for the wallet to hit the floor. The wallet will hit the floor in about 2.68 s 2 Fundamental Theorem of Calculus KeyConcept Fundamental Theorem of Calculus If f is continuous on [a, b ] and F ( x ) is any antiderivative of f ( x ), then b      f ( x ) dx = F (b ) - F(a ). a ∣b The difference F (b) - F ( a ) is commonly denoted by F ( x )∣∣ . a 692 | Lesson 11-6 | The Fundamental Theorem of Calculus Copyright © McGraw-Hill Education Notice that the notation used for indefinite integrals looks very similar to the notation used in Lesson 11-5 for definite integrals. The only difference appears to be the lack of upper and lower limits in indefinite integrals. In fact, finding an antiderivative of a function is a shortcut to calculating a definite integral by evaluating a Riemann sum. This connection between definite integrals and antiderivatives is so important that it is called the Fundamental Theorem of Calculus. A by-product of the Fundamental Theorem of Calculus is that it makes a connection between integrals and derivatives. Integration is the process of calculating antiderivatives, while differentiation is the process of calculating derivatives. Thus, integration and differentiation are inverse processes. We can use the Fundamental Theorem of Calculus to evaluate definite integrals without having to use limits. Example 4 Area Under a Curve Use the Fundamental Theorem of Calculus to find the area of the region between the graph of each function and the x-axis on the given interval. 3 a. y = 4x 3 on the interval [1, 3], or  4x 3 dx. First, find the antiderivative. 1 4x  4x 3 dx = _   3 + 1 + C 3+1 Math HistoryLink Maria Gaetana Agnesi (1718–1799) An Italian linguist, mathematician, and philosopher, Maria Agnesi authored Analytical Institutions, the first book to discuss both differential and integral calculus. She is also known for an equation she developed for a curve called the “Witch of Agnesi.” Constant Multiple of a Power = x 4 + C Simplify. Now evaluate the antiderivative at the upper and lower limits, and find the difference. 3 ∣3 ∣1      4x 3 dx = x 4 + C∣ 1 Fundamental Theorem of Calculus = (3 4 + C) - (1 4 + C) b = 3 and a = 1 = 81 - 1 or 80 Simplify. The area under the graph on the interval [1, 3] is 80 square units. 4 b. y = -x 2 + 4x + 6 on the interval [0, 4], or  (-x 2 + 4x + 6) dx 0 First, find the antiderivative.  (-x 2 + 4x + 6) dx 4x 1 + 1 6x 0 + 1 = - _ +_   1 + 1 + _   0 + 1 + C x 2 + 1 2+1 x 3 3 = - _ + 2x 2 + 6x + C Antiderivative Rule Simplify. Now evaluate the antiderivative at the upper and lower limits, and find the difference. 4 x 3 3 ∣4 ∣0      (-x 2 + 4x + 6) dx = - _ + 2x 2 + 6x + C∣ 0 ( (4) 3 3 Fundamental Theorem of Calculus ) ( ) (0) 3 3 = - _ + 2(4) 2 + 6(4) + C - - _ + 2(0) 2 + 6(0) + C Copyright © McGraw-Hill Education Photo Researchers, Inc/Alamy = 34.67 - 0 or 34.67 b = 4 and a=0 Simplify. The area under the graph on the interval [0, 4] is 34.67 square units. GuidedPractice Guided Practice Evaluate each definite integral. 5 4A.   3x 2 dx 2 117 2 4B.  (16x 3 - 6x 2 ) dx 1 46 Notice that when the antiderivative is evaluated at the upper and lower limits and the difference is found, we do not have an exact value for C. However, regardless of the value of the constant, because it is present in each antiderivative, the difference between the constants is 0. Therefore, when evaluating definite integrals using the Fundamental Theorem of Calculus, you can disregard the constant term when writing the antiderivative. 693 Before evaluating an integral, determine whether it is indefinite or definite. WatchOut! Integrals Although the constant C can be excluded when evaluating definite integrals, it must be included when evaluating indefinite integrals because it is part of the antiderivative. Example 5 Indefinite and Definite Integrals Evaluate each integral. a.  (9x - x 3 ) dx This is an indefinite integral. Use the antiderivative rules to evaluate. 9x x  (9x - x 3 ) dx = _   1 + 1 - _   3 + 1 + C 1+1 3+1 Constant Multiple of a Power =_   2 x 2 - _   4 + C x 4 9 Simplify. 3 b.  (9x - x 3 ) dx 2 This is a definite integral. Evaluate the integral using the given upper and lower limits. 3 ) ( ⎡9 9 3 2 ⎤ = (_   2 (3) - _   4 ) - ⎢⎣_ (2) - _   4 ⎥⎦ 2 9 x 4 | 3      (9x - x 3 ) dx = _   2 x 2 - _   4 || 2 2 4 2 2 4 = 20.25 - 14 or 6.25 Fundamental Theorem of Calculus b = 3 and a = 2 Simplify. The area under the graph on the interval [2, 3] is 6.25 square units. GuidedPractice Guided Practice 5A.  (6x 2 + 8x - 3) dx 3 2x 3 + 4x 2 - 3x + C 5B.    (-x 4 + 8x 3 - 24x 2 + 30x - 4) dx 15.6 1 Notice that indefinite integrals give the antiderivative of a function while definite integrals not only give the antiderivative but also require it to be evaluated for given upper and lower limits. Thus, an indefinite integral gives a function, the antiderivative, for finding the area under a curve for any set of limits. The integral becomes definite when a set of limits is provided and the antiderivative can be evaluated. Example 6 Definite Integrals The work, in joules, required to stretch a certain spring a distance of 0.5 meter beyond its 0.5 natural length is given by  360x dx. How much work is required? 0 Evaluate the definite integral for the given upper and lower limits.      0 0.5 | 0.5 360x dx = 180x 2 || Constant Multiple of a Power and the Fundamental Theorem of Calculus 0 = 180(0.5) 2 - 180(0) 2 Let b = 0.5 and a = 0 and subtract. The work required is 45 joules. GuidedPractice Guided Practice Find the work required to stretch a spring if it is defined by the following integrals. 0.7 6A.    476x dx 0 116.62 J 694 | Lesson 11-6 | The Fundamental Theorem of Calculus 1.4 6B.    512x dx 0 501.76 J Copyright © McGraw-Hill Education = 45 - 0 or 45 Simplify. Exercises Find all antiderivatives for each function. (Examples 1 and 2) 1. 2. 3. _1 6 f (x) = x 5 F (x ) =   x + C 6 _5 2 h (b) = -5b - 3 H(b) = - b - 3b + C 2 4 _3 _  f (z) =   √z F (z ) =   z 3 + C 3 22. SURVEYOR A plot of land has two perpendicular fences and a stream for borders as shown. 4 3 1 5 _ 2 3 _ _ 3 2 1 4. n(t) = _   4 t 4 - _   3 t 2 + _   4 N (t ) =   t -   t +   t + C 4 20 9 4 3 7   15 _ 15 _     2 _ 28 32 3 1 6 _ 1 4 _ 1 2 _ 2 5 _ 2 1 3 _ _ 6. w(u) =   3 u +   6 u -   5 u W (u ) =   u +   u -   u + C 5 24 9 1 5 _   2 3 _   1 _   5. q (r) = _   4 r 5 + _   8 r 3 + r 2 Q (r ) = _   r 5 + _   r 3 + _   r 2 + C 7. g (a) = 8a 3 + 5a 2 - 9a + 3 G (a) = 2a 4 + _   a 3 - _   a 2 + 3a + C 5 12 3 2 _ _ 2 8. u(d ) =   5 +   3 - 6d + 3.5 5 3 d d U (d ) = - _ 4 - _   2 - 2d 3 + 3.5d+ C 5 9. m(t) = 16t 3 - 12t 2 + 20t - 11 d 9 2d M (t )2= 4t 4 - 4t 3 + 10t 2 - 11t + C 10. p(h) = 72h 8 + 24h 5 - 12h + 14 P (h) = 8h 9 + 4h 6 - 4h 3 + 14h + C 11. CELL PHONE Refer to the beginning of the lesson. Suppose the phone took exactly two seconds to drop from the balloon to the ground. (Example 3) a. Evaluate s(t) =  -32t dt. s(t ) = -16t 2 + C b. Solve for C in the position function s(t) by substituting two seconds for t and 0 for s(t). 64 c. How far from the ground is the phone after 1.5 seconds of falling? 28 ft Evaluate each integral. (Examples 4 and 5) 3m 2 + 3m 4 + C 12.  (6m + 12m 3 ) dm 13.  (20n 3 - 9n 2 - 18n + 4) dn 4 14.    2x 3 dx 1 127.5 5 15.    (a 2 - a + 6) da 2 2 16.    (4g + 6g 2 ) dg 1 10 (2 1 _   2 5 _   46.5 20 ) 17.    _   5 p 8 + _   4 p 7 + _   4 dp 2 3 1 22.37 Copyright © McGraw-Hill Education 18.    _   2 h 2 + _   3 h 3 - _   5 h 4 dh 1 (1 2 2 5n 4 - 3n 3 - 9n 2 + 4n + C 1 ) 7.99 18.93 5 0.68t - 0.3t 4 + 1.15t 2 - 5.7t + C 19.    (-v 4 + 2v 3 + 2v 2 + 6) dv 0 20.  (3.4t 4 - 1.2t 3 + 2.3t - 5.7) dt 2w 7.1 - 3w 6.7 + 4w 3.3 + 3w + C 6.1 5.7 2.3 21.  (14.2w - 20.1w + 13.2w + 3) dw Suppose the edge of the stream that borders the plot can be modeled by f (x) = -0.00005x 3 + 0.004x 2 - 1.04x + 40, where the fences are the x and y-axes, and x is given in 40 kilometers. Evaluate  f (x) dx to find the area of the land. (Example 6) 0 821.33 km2 23. INSECTS The velocity of a flea’s jump can be defined as v(t) = -10t + 11, where t is given in seconds and velocity is given in meters per second. (Example 6) a. Find the position function s(t) for the flea’s jump. 2 Assume that for t = 0, s(t) = 0. s (t) = -5t + 11t b. After a flea jumps, how long does it take before the flea lands on the ground? 2.125 seconds 24. NATIONAL MONUMENT A magician wants to make the Gateway Arch in St. Louis disappear. To attempt the illusion, he needs to cover the arch with a large canopy. Before constructing the canopy, the magician wants an approximation of the area under the arch. One equation that can be used to model the shape of the arch is y = - _ 47.25 + 1.3x, where x is given in meters. Find the x2 area under the arch. (Example 6) 24,580 m2 25. TRACK A sprinter needs to decide between starting a B 100-meter race with an2 initial burst of speed, modeled by v 1 (t) = 3.25t - 0.2t , or conserving his energy for more acceleration towards the end of the race, modeled by v 2(t) = 1.2t + 0.03t 2 , where velocity is measured in meters per second after t seconds. a–c. See margin. a. Use a graphing calculator to graph both velocity functions on the same screen for 0 ≤ t ≤ 12. b. Find the position function s(t) for v 1 (t) and v 2 (t). c. How long does it take for the sprinter to finish a 100-meter race using each strategy? Evaluate each integral. 1 26.    3 dx -3 2 12 27.      (-x 2 + 10) dx 27 -1 -1 -3 28.      (-x 2 - 9x - 10) dx 29.      (x 3 + 8x 2 + 21x + 20) dx 28.5 5.33 1 x _ 5x _ 4 3 30.        2 +   4 dx 2.5 31.      (x - 2x - 4x + 8) dx 16.4 -2 -1 -6 -1 ( 5 4 ) -3 695 32. CATAPULT Pumpkins are launched by catapults at The Pumpkin Chunking World Championships in Delaware. A pumpkin launched by a catapult has a velocity of v(t) = -10t + 36 meters per second after t seconds. After 3 seconds, the pumpkin has a height of 68 meters. a. Find the maximum height of the pumpkin. 69.8 m b. Find the pumpkin’s velocity when it hits the ground. Evaluate each integral. about -37 m/s 33–38. See margin. 45. MULTIPLE REPRESENTATIONS In this problem, you will investigate the relationship between the total area and the signed area of a region bound by a curve and the x-axis. a. GEOMETRIC Graph f (x) = x 3 - 6x 2 + 8x, and shade the region bound by f (x) and the x-axis for 0 ≤ x ≤ 4. a, c, e. See Chapter 11 Answer Appendix. 2 b. ANALYTICAL Evaluate    (x 3 - 6x 2 + 8x) dx and 4      (x 3 - 6x 2 + 8x) dx. 2 x 2 33.      (3t 2 + 8t) dt 34.    (10t 4 - 12t 2 + 5) dt x 5 2x 35.      (4t 3 + 10t + 2) dt 6 36.      (-9t 2 + 4t) dt 3 c. VERBAL Make a conjecture for the area found above and below the x-axis. d. ANALYTICAL Calculate the signed area by evaluating 4 -x x 37.    (16t 3 - 15t 2 + 7) dt      (x 3 - 6x 2 + 8x) dx. Then calculate the total area x+3 2 (3t 2 + 6t + 1) dt 38.      0 by evaluating 2x x 0 4; -4 39. SPHERE The volume of a sphere with radius R can be C found by slicing the sphere vertically into circular cross sections and then integrating the areas. ⎜ ⎟ ⎜ 2 ⎟ 4    (x 3 - 6x 2 + 8x) dx +    (x 3 - 6x 2 + 8x) dx . 0 2 0; 8 e. VERBAL Make a conjecture for the difference between the signed area and the total area. H.O.T. Problems Use Higher-Order Thinking Skills r 46. CHALLENGE Evaluate       √ r 2 - x 2   dx, where r is a -r The radius of each cross section is   √ R 2 - x 2  . Thus, the area 2 2  of a cross-section is equal to π(√R - x 2   )  . R Evaluate  (πR 2 - πx 2 )dx to find the volume of a sphere. _  4 πR3 -R constant. (Hint: Find the area between the graph of 1 2 - x 2   and the x-axis.) _ y =   √r   πr 2 2 REASONING Determine whether each statement is always, sometimes, or never true. Explain your reasoning. 47–49. See margin. a b 3 47.      f (x) dx =      f (x) dx a 40. AREA Calculate the area bound by f (x), g (x), and the x-axis on the interval 1 ≤ x ≤ 3. 6 units 2 b b -a 48.      f (x) dx =      f (x) dx a -b b |b| 49.      f (x) dx =      f (x) dx |a| a See Chapter 11 Answer Appendix. 50. PROOF Prove that for constants n and m b b b    (n + m) dx =      n dx +      m dx. a a a n 51. REASONING Describe the values of f (x),  f(x i)∆x, and i=1 b    f (x) dx, where the graph of f (x) lies below the x-axis The integral      0 x k dx gives a reasonable estimate of the n sum of the series  i k . Use the integral to estimate each i=1 sum and then find the actual sum. 20 100 i=1 25 i=1 30 i 3 44,152.52; 44,100 42. i 2 41. 43. i 4 i=1 2,156,407.8; 2,153,645 338,358.38; 338,350 44. i 5 i=1 134,167,641.6; 133,987,425 696 | Lesson 11-6 | The Fundamental Theorem of Calculus See margin. 52. WRITING IN MATH Explain why the constant term C in an antiderivative can be disregarded when evaluating a definite integral. See margin. 53. WRITING IN MATH Write an outline that could be used to describe the steps involved in finding the area of the region between the graph of y = 6x 2 and the x-axis on the interval [0, 2]. See Chapter 11 Answer Appendix. Copyright © McGraw-Hill Education a for a ≤ x ≤ b. n + 0.5 Spiral Review Use limits to approximate the area between the graph of each function and the x-axis given by the definite integral. 54.    2   14x 2 dx 74 _   -2 2 3 6 30 55.      (x + 2) dx 0 Use the Quotient Rule to find the derivative of each function. 56. j(k) = _   4 3 k 8 - 7k 2k + 11k 8k + 55k + 42k + 154k j ′(k ) = ___        4 3 2 11 10 4 3 (2k + 11k )  2n 3 + 4n n + 1 57. g (n) = _   2 58. FASHION The average prices for three designer handbags on an online auction site are shown. a. If a random sample of 35 A-style handbags is selected, find the probability that the mean price is more than AED 138 if the standard deviation of the population is AED 9. 2.4% 2n + 2n + 4 g ′(n) = __   2 2 4 2 (n + 1)  Handbag Style Average Price (AED) A 135 b. If a random sample of 40 C-style handbags is selected, find the probability that the mean price is between AED 150 and AED 155 if the standard deviation of the population is AED 12. 79.7% B 145 C 152 59. BASEBALL The average age of a major league baseball player is normally distributed with a mean of 28 and a standard deviation of 4 years. a. About what percent of major league baseball players are younger than 24? 16% 24 b. If a team has 35 players, about how many are between the ages of 24 and 32? 60. Find two pairs of polar coordinates for the point with the given rectangular coordinates (3, 8), if -2π ≤ θ ≤ 2π. (8.54, 1.21), (-8.54, 4.35) Skills Review for Standardized Tests 61. SAT/ACT In the figure, C and E are midpoints of the edges of the cube. A triangle is to be drawn with R and S as two of the vertices. Which of the following points should be the third vertex of the triangle if it is to have the largest possible perimeter? B 62. The work in joules required to pump all of the water out of a 10-meter by 5-meter by 2-meter swimming pool is 2 490,000x dx. If you evaluate this integral, given by  0 what is the required work? A A F F 980,000 J B B G 985,000 J C C H 990,000 J D D J 995,000 J E E 63. FREE RESPONSE A particle moves along a line so that its position at any time t ≥ 0 is given by the function s(t) = t 2 - 3t + 10 where s is measured in meters and t is measured in seconds. Copyright © McGraw-Hill Education a. Find the displacement of the particle during the first 4 seconds. That is, how far has the particle moved from its original starting position after 4 seconds? 4 m b. Find the average velocity of the particle during the first 4 seconds. 1 m/s s′(t ) = 2t - 3 -1 m/s; 5 m/s c. Write an equation for the instantaneous velocity of the particle at any time t. d. Find the instantaneous velocity of the particle when t = 1 and t = 4. e. At what value for t does s (t) attain a minimum value? 1.5 s f. What does the value of t you found in part d represent in terms of the motion of the particle? See margin. 697 11 Study Guide and Review Chapter Summary KeyConcepts KeyVocabulary Estimating Limits Graphically • • (Lesson 11-1) antiderivative instantaneous rate of change The limit of a function f (x ) as x approaches c exists if and only if both one-sided limits exist and are equal. definite integral instantaneous velocity A limit of f (x ) as x approaches c does not exist if f (x ) approaches a different value from the left of c than from the right of c, increases or decreases without bound from the left and/or right of c, or oscillates back and forth between two values. derivative integration difference quotient lower limit differential equation one-sided limit differential operator regular partition differentiation right Riemann sum direct substitution tangent line indefinite integral two-sided limit indeterminate form upper limit Evaluating Limits Algebraically • • Limits of polynomial and rational functions can often be found by using direct substitution. If you evaluate a limit and reach the indeterminate form _  0 , simplify the 0 expression algebraically by factoring and dividing out a common factor or by rationalizing the numerator or denominator and then dividing out any common factors. Tangent Lines and Velocity • (Lesson 11-2) VocabularyCheck Choose the correct term to complete each sentence. (Lesson 11-3) The instantaneous rate of change of the graph of f (x ) at the point (x, f (x )) is the slope m of the tangent line given by f (x + h ) - f (x ) m =      lim __   . h 1. The slope of a nonlinear graph at a specific point is the and can be represented by the slope of the tangent line to the graph at that point. instantaneous rate of change h→0 2. The process of evaluating an integral is called . integration 3. Limits of polynomial and rational functions can be found by , so long as the denominator of the rational function evaluated at c is not 0. direct substitution 4. The function F (x ) is the • (Lesson 11-4) The derivative of f (x ) = x nis f ′ (x ) given by f ′ (x ) = nx n - 1, where n is a real number. Area Under a Curve and Integration • (Lesson 11-5) The area of a region under the graph of a function is b  f (x ) dx =   limf (x i) ∆x, where a and b are the lower and n→∞ i =1 a b-a   , and x = a + i∆x. upper limits, respectively, ∆x = _ n The Fundamental Theorem of Calculus • • i (Lesson 11-6) The antiderivative of f (x ) = x is F (x ) given by F (x ) = _   n + 1 + C, for some constant C. n x n + 1 If F (x ) is the antiderivative of the continuous function f (x ), then b    f (x ) dx = F (b ) - F (a ). a 698 | Chapter 11 | Study Guide and Review 5. Since it is not possible to determine the limit of a function with 0 in 0 the denominator, it is customary to describe the resulting fraction _   0 as having a(n) . indeterminate form 6. To find the limits of rational functions at infinity, divide the numerator and denominator by the power of x that occurs in the highest function. 7. The process of finding a derivative is called n antiderivative 8. If a function is preceded by a(n) take the derivative of the function. differentiation . d _   , then you are to dx differential operator 9. The velocity or speed achieved at any specific point in time is the . instantaneous velocity 10. The indefinite integral of f (x ) is defined by  f (x ) dx = F (x ) + C . Copyright © McGraw-Hill Education Derivatives of f (x ). Lesson-by-Lesson Review 11-1 Estimating Limits Graphically Estimate each limit using a graph. Support your conjecture using a table of values. 11–12. See margin. 11.lim (2x - 7 ) x→3 12. lim (0.5x 4 + 3x 2 - 5 ) Example 1 x - 4 _ using a graph. Support your conjecture Estimate      lim   2 x→2 x - 2 using a table of values. Analyze Graphically The graph of f (x ) = _   x - 2 x 2 - 4 x→1 Estimate each one-sided or two-sided limit, if it exists. x 2 + x - 6 13.lim+ _ 5 x-2 suggests that as x gets closer to 2, the corresponding function value   x - 2 is 4. approaches 4. Therefore, we can estimate that lim _ x 2 - 4 x→2 x→2  14.lim __   x - 4 x→4 x 2 + x + 20 does not exist Estimate each limit, if it exists. 15.lim __   x→4 x 2 - 8x + 16 ∞ 16.lim __   x - 2 x→2 does not exist 9 x 2 - 7x - 10 Support Numerically Make a table of values, choosing x-values that approach 2 from either side. ✔ x approaches 2 11-2 Evaluating Limits Algebraically Use the properties of limits to evaluate each limit. 17.lim __   x x→5 x 2 + 2x + 10 x→-1 19 Use direct substitution, if possible, to evaluate each limit. If not possible, explain why not. Not possible; when x = 25, the denominator is 0. 20. lim (-3x 3 - 2x 2 + 15) -17 19. lim _   √ 2 Copyright © McGraw-Hill Education x→25   x + 1 x- 5 x→2 Evaluate each limit. 21.    lim __   2 x+2 x→-2 x - 2x - 8 3 _1 - 6 2) 22. lim ( 2 - 4x + x -∞ x→∞ 1.9 1.99 1.999 f (x ) 3.9 3.99 3.999 2 2.001 2.01 2.1 4.001 4.01 4.1 Example 2 Use direct substitution, if possible, to evaluate each limit. If not possible, explain why not. 9 18.    lim (5x 2 - 2x + 12) x x approaches 2 a. lim ( 2x 3- x 2+ 4x + 1) x→2 This is the limit of a polynomial function. Therefore, direct substitution can be used to find the limit. lim ( 2x 3 - x 2 + 4x + 1) = 2( 2 ) 3 - 2 2 + 4( 2 ) + 1 x→2 = 16 - 4 + 8 + 1 or 21 2x - 7 _ b.   lim   x→-4 2 - x 2 This is the limit of a rational function, the denominator of which is nonzero when x = -4. Therefore, direct substitution can be used to find the limit. 2(-4) - 7     lim _   =_   =_   2 - 16 or _ 14 x→-4 2 - x  2 2 - (-4) 2 2x - 7 -8 - 7 15 699 11 Study Guide and Review Continued 11-3 Tangent Lines and Velocity Find the slope of the lines tangent to the graph of each function at the given points. f (x + h ) - f (x ) h →0 24. y = x + 2; (0, 2) and (-1, 3) 0; -2 2 The distance d an object is above the ground t seconds after it is dropped is given by d (t ). Find the instantaneous velocity of the object at the given value for t. 25. y = -x 2 + 3x 26. y = x 3 + 4x m = -2x + 3 h →0 f (2 + h ) - f (2) 2 x=2 2 (2 + h )  - 2   = lim __   h f (2 + h ) = (2 + h ) 2 and f (2) = 2  2 = lim __      h Multiply. h →0 4 + 4h + h 2 - 4 h (4 + h ) = lim _   h Simplify and factor. = lim (4 + h ) Divide by h. h →0 Find the instantaneous velocity if the position of an object in meters is defined as h (t ) for given values of time t given in seconds. 11 m /s 28. h (t ) = -5t 2 - 12t + 130; t = 3.5 Instantaneous Rate of Change Formula = lim __   h h →0 m = 3x 2 + 4 27. h (t ) = 5t + 6t 2; t = 0.5 Find the slope of the line tangent to the graph of y = x 2 at (2, 4). m = lim __   h -1; -1 23. y = 6 - x ; (-1, 7) and (3, 3) Example 3 h →0 = 4 + 0 or 4Sum Property of Limits and Limits of Constant and Identity Functions -47 m /s Therefore, the slope of the line tangent to the graph of y = x 2at (2, 4) is 4. Find an equation for the instantaneous velocity v (t ) if the path of an object is defined as h (t ) for any point in time t. 29. h (t ) = 12t 2 - 5 30. h (t ) = 8 - 2t 2 + 3t v (t ) = -4t + 3 v (t ) = 24t 11-4 Derivatives Evaluate limits to find the derivative of each function. Then evaluate the derivative of each function for the given values of each variable. 31. g (t ) = -t 2 + 5t + 11; t = -4 and 1 32. m ( j ) = 10 j - 3; j = 5 and -3 m ′( j ) = 10; m ′(5) = 10 and m ′(-3) = 10 ′(t ) = -2t + 5; g g ′(-4) = 13 and g ′(1) = 3 Find the derivative of each function. 33. p (v ) = -9v + 14 5 35. t (x ) = -3   √ x 6  33–36. See margin. 34. z (n ) = 4n 2 + 9n 3 _   1 _   36. g (h ) = 4h 4 - 8h 2 + 5 Example 4 x - 5 _  . Find the derivative of h (x ) =   2 x 3 + 2 f (x ) Let f (x ) = x 2 - 5, and g (x ) = x 3 + 2. So, h (x ) = _  g (x ) . Find the derivatives of f (x ) and g (x ) f (x ) = x 2 - 5 Original equation f ′(x ) = 2x Power and Constant Rules 3 g (x ) = x + 2 2 g ′(x ) = 3x Original equation Power and Constant Rules Use the quotient rule to find the derivative of each function. 37. f (m ) = _ 5 - 3m 5 + 2m -25   f ′(m) = _ (5 + 2m) 2 m (q ) = 2q 4 - q 2 + 9 38. __      q 2 - 12 4q 5 - 96q 3 + 6q m ′(q ) = __      (q 2 - 12) 2 700 | Chapter 11 | Study Guide and Review f ′(x )g (x ) - f (x )g ′(x ) h ′(x ) = __      2 [ g (x )]  Quotient Rule 2x (x 3 + 2) - (x 2 - 5) 3x 2 = ___         Substitution (x 3 + 2) 2 = __     3 2 -x 4 + 15x 2 + 4x (x + 2)  Simplify. Copyright © McGraw-Hill Education Use f (x ), f ′(x ), g (x ), and g ′(x ) to find the derivative of h (x ). 11-5 Area Under a Curve and Integration Example 5 Approximate the area of the shaded region for each function using right endpoints and 5 rectangles. Use limits to find the area of the region between the graph of y = 2 39. x 2 and the x -axis on the interval [0, 2], or      2 x 2 dx. 2 40. 0 First, find ∆x and x i.   n ∆x = _ b-a 11.28 units 2 5.16 units 2 Use limits to find the area between the graph of each function and the x-axis given by the definite integral. 2 4_   units 2 2 41.    2x 2 dx 3 42.      (2x 3 - 1) dx 0 2 43.      (x 2 + x ) dx 0 4 44.    (3x 2 - x ) dx 1 n 2 i=1 0 ( )( ) ( ) n 4i 4 =   lim _   _ 2 n→∞ 2 Simplify. ( n (n + 1)(2n + 1) 4 4 1 37 _   units 2 6 n→∞ n n ) 2 ( 8(2n 2 + 3n + 1) ) =   lim __    2 2 4_   units 2 n→∞ 3 3n ⎡8 ⎤ =   lim ⎢_   · 2   + _   + _   2 ⎥ n→∞ 3 55 _   units 2 ⎣ 1 2 ( 3 n )⎦ 1 n Find all antiderivatives for each function. 5 G (n ) = _   n 2 - 2n + C 2 9   q 2 - 2q + C R (q ) = -q 3 + _ 46. r (q ) = -3q + 9q - 2 2 _4 3 2 47. m (t ) = 6t - 12t + 2t - 11 M (t ) =   t - 4t + t 2 - 11t + C 48. p (h ) = 7h 6 + 4h 5 - 12h 3 - 4 2 P (h ) = h 7 + _   h 6 - 3h 4 - 4h + C 3 2 3 Evaluate each integral. 49.  8x dx 8 _   x 3 + C 3 50.  (2x 2 - 4) dx 3 4 2 3 5 52.    (-x + 4x - 2x + 5x ) dx 1 _4 a. f (x ) = 5 x Rewrite with a negative exponent. f (x ) = 4x -5 F (x ) = _   -5 + 1 + C 4x -5 + 1 1 = -1x -4 + C or - _ 4 + C x f (x ) = x 2 - 7 3 51.      (2x 2 - 4 + 5x 3 + 3x 4 ) dx Factor and divide each term by n  2. Constant Multiple of a Power Simplify. b. f (x ) = x 2- 7 2 _   x 3 - 4x + C 5 n (n + 1)(2n + 1) __ i=1 1 16 =_   3 or 5 _    Limit theorems and 3 simplify. 45. g (n ) = 5n - 2 2 n Multiply and divide by n. Example 6 3 _ _ 2i 2 2 2i 2  2x 2dx =   lim 2 _   n _   n x i=   n and ∆x =   n n→∞ Find all antiderivatives for each function. 2 _2 a = 0 and ∆x =   n =lim _ _ 2 · __       i 2 =      6 11-6 The Fundamental Theorem of Calculus Copyright © McGraw-Hill Education b = 2 and a = 0 n i =1 n 3 1 Formula for ∆x 2-0 2 ∆x = _   n or _   n 2 2i xi= 0 + i _ n or _   n 2466.53 units 2 2 3294 units  Original equation = x 2 - 7x 0 F (x ) = _   2 + 1 - _   0 + 1 + C x 2 + 1 7x 0 + 1 =_   3 x 3 - 7x + C 1 Rewrite the function so each term has a power of x. Antiderivative Rule Simplify. 701 11 Study Guide and Review Continued Applications and Problem Solving 53. STAMPS Suppose the value v of a stamp in dirhams after t years 450 can be represented as v (t ) = __   . 5 + 25(0.4) t a–b. See margin. a. Complete the following table. Years 0 (Lesson 11-1) 1 2 57. ARCHERY An archer shoots an arrow with a velocity of 11 meters per second towards the target. Suppose the height s of the arrow in meters t seconds after it is shot is defined as s (t ) = -5t 2 + 11t + 0.5. (Lesson 11-3) 3 Value b. Graph the function for 0 ≤ t ≤ 10. c. Use the graph to estimate  lim v (t ), if it exists. 90 t→∞ d. Explain the relationship between the limit of the function and the value of the stamp. Sample answer: The stamp will peak in value at about AED 90. 54. ANIMALS A wildlife conservation’s population P in hundreds after 120 t years can be estimated by P (t ) = __ . 1 + 24e -0.25t (Lesson 11-1) a. Use a graphing calculator to graph the function for 0 ≤ t ≤ 50. See margin. b. Estimate   lim __ , if it exists. -0.25t 120 t→∞ 1 + 24e 120 c. Interpret your results from part b. Over time, the animal population will approach a maximum of 120,000. a. Find an equation for the instantaneous velocity v (t ) of the arrow. v (t ) = -10t + 11 b. How fast is the arrow traveling 0.5 second after it is shot? 6 m/s c. For what value of t will the arrow reach its maximum height? ≈1.09 s d. What is the maximum height of the arrow? ≈6.5 m 58. DESIGN The owner of a ski resort is designing a new logo to put on his employees’ uniforms. The design is in the shape of the region shown in the figure. If this part of the design is to be sewn on to the uniforms, how much material is required if x is given in centimeters? (Lesson 11-6) 12.15 cm2 55. COLLECTORS The value of Faris’ coin collection has been increasing every year over the last five years. Following this trend, the value v of   4t + 19 . the coins after t years can be modeled by v (t ) = _ 800t - 21 (Lesson 11-2) a. Find   lim v (t ). t→∞ 200 b. What does the limit of the function imply about the value of Faris’ coin collection? Do you agree? See margin. c. After 10 years, a coin dealer offers Faris AED 300 for his collection. Should Faris sell his collection? Explain your answer. 56. ROCKETS A rocket is launched into the sky with an upward velocity of 50 meters per second. Suppose the height d of the rocket in meters t seconds after it is launched is modeled by d (t ) = -5t 2 + 50t + 2.7. (Lesson 11-3) a. Find an equation for the instantaneous velocity v (t ) of the rocket. v (t ) = -10t + 50 b. How fast is the rocket traveling 1.5 seconds after it is launched? 34 m /s c. For what value of t will the rocket reach its maximum height? ≈4.69 s d. What is the maximum height that the rocket will reach? ≈108 m 702 | Chapter 11 | Study Guide and Review 59. FROGS A frog can jump with a velocity modeled by v (t ) = -10t + 8, where t is given in seconds and velocity is given in meters per second. (Lesson 11-6) a. Find the position function s (t ) for the frog’s jump. Assume that for t = 0, s (t ) = 0. s (t ) = -5t 2 + 8t b. How long will the frog stay in the air when it jumps? 1.63 s 60. BIRDS A cardinal in a tree 6 meters above the ground drops some food. The instantaneous velocity of his food can be defined as v (t ) = -10t, where t is given in seconds and velocity is measured in meters per second. (Lesson 11-6) a. s (t ) = -5t 2 + 6 a. Find the position function s (t ) of the dropped food. b. Find how long it will take for the food to hit the ground. 1.12 s Copyright © McGraw-Hill Education Yes; the offer is more than double the projected value. 11 Practice Test Find the derivative of each function. Estimate each one-sided or two-sided limit, if it exists. x 2 - 16 2. lim _   x - 4 8 -6 1.  lim+ √ x + 4  - 8 x→0  f ′ (x ) = -3 4 _   21. b (c ) = 4c 2 - 8c 3 + 5c 5 See margin. 20. f (x ) = -3x - 7 x→4 1 _   4 _   Estimate each limit, if it exists. 3.lim _   x→7 x - 7 6 4.   lim x 3 + 5x 2 - 2x + 21 ∞ x→∞ C (x ) = __   . x 100x + 7105 a. Determine the limit of the function as x approaches infinity. b. Interpret the results from part a. See margin. 100 7. lim (2x 3 - 12x + 3) x→9 1353 8. SCHOOL The number of students S absent with the flu after t days 2000t 2 + 4    . at a school can be modeled by S (t ) = __ 1 + 50t 2 a. How many students are initially sick? 4 t 2 25. FOOTBALL The marginal cost c of producing x footballs is represented by c (x ) = 15 - 0.005x. a. Determine the function that represents the actual cost 2 function. C (x ) = 15x - 0.0025x Use limits to find the area between the graph of each function and the x-axis given by the definite integral. 4 26.    (x 2 - 3x + 4 ) dx 1 8 27.    10x 4 dx 3 b. How many students will eventually be sick? 1 b. Determine the cost of increasing the daily production from 1500 footballs to 2000 footballs. AED 3125 Use direct substitution, if possible, to evaluate each limit. If not possible, explain why not. -25 -_ w ′( y ) = 4y 3 + 3y 2 23. g (x ) = (x 2 - 4)(2x - 5) g ′(x ) = 6x 2 - 10x - 8 t 2 - 1 t 3 + 4t 2 + t _ 24. h (t ) = __   h ′ (t ) =   t 2 5. ELECTRONICS The average cost C in dirhams of x number of personal digital assistants can be modeled by x 2 _  1 1  _ 22. w ( y ) = 3y 3 + 6y 2 does not exist 6.lim  __ x→5   √ x - 4  - 2 2 _   1 2 65,050 units 2 5 40 10 _   units 2 28.    (7 - 2x + 4x 2 ) dx 2 156 units 2 Evaluate each limit. 9.  lim (x 2 - 7x + 2) 10.   lim (2x 3 - 8x 2 - 5) ∞ x→∞ x→∞ 2x 3 - x - 1 11.  lim __ 4 0 3 x→∞ √ 25 + x - 4 12.   lim __ 0 x→∞ -x + 7x + 4 x 1 1 _   -_   13. MULTIPLE CHOICE Evaluate lim _   . x x+3 x→0 A -_ B 0 ∞ 3 A 1 C _   9 1 9 D does not exist Find the slope of the lines tangent to the graph of each function at the given points. 14. y = x 2 + 2x - 8; (-5, 7) and (-2, -8) 15. y = _   3 + 2; (-1, -2) and 2, _   2 Copyright © McGraw-Hill Education -12; - _ ( 5) 4 x 16. y = (2x + 1) 2; (-3, 25) and (0, 1) -8; -2 -20; 4 3 4 Find all antiderivatives for each function. 4 3 2 + 8a + C 29. d (a ) = 4a 3 + 9a 2 - 2a + 8 D(a) = a + 3a - a 30. w (z ) = _   4 z 4 + _   6 z 2 - _   5 3 1 2   z 5 + _   z 3 - _   5 z + C W (z ) = _ 3 20 1 18 2 Evaluate each integral. 31.  (5x 3 - 6x 2 + 4x - 3) dx 4 32.    (x 2 + 4x - 2) dx 1 5 _   x 4 - 2x 3 + 2x 2 - 3x + C 4 45 33. AREA Calculate the area bound by f (x ) and g (x ) on the interval 2 ≤ x ≤ 4. H Find an equation for the instantaneous velocity v (t ) if the path of an object is defined as h (t ) for any point in time t. 17. h (t ) = 9t + 3t 2 v (t ) = 9 + 6t 18. h (t ) = 10t 2 - 7t 3 v (t ) = 20t - 21t 2 19. h (t ) = 3t 3 - 2 + 4t v (t ) = 9t 2 + 4 5 F 17 _  12 1 G 17 _  3 1 H 15 _     3 J 16 703 11 Connect to AP Calculus The Chain Rule Objective Differentiate composite functions using the Chain Rule. Learning to differentiate polynomials by using the Power Rule led to the evaluation of definite integrals. This allowed for the calculation of the area found between a curve and the x-axis. However, our integration work was limited to basic polynomial functions. So, how can we calculate the areas found between the x-axis and the curves defined by composite functions such as h (x ) = (x 4 - 3) 2 or t (x ) =   √x 3 - 4 ? Just as in Lesson 11-4, You must learn how to differentiate these functions before you can integrate. Begin with h (x ). You can use the Product Rule to develop the derivative. h (x ) = (x 4 − 3) 2 Original equation = (x 4 - 3)(x 4 - 3) 3 4 Rewrite without the power. 4 3 h ′(x ) = 4x (x - 3) + (x - 3)4x 3 4 = 2(x - 3)4x Product Rule Simplify. Although the derivative of h(x) can be simplified further, leave it as shown in order to derive a rule for composite functions. Activity 1 Derivative of a Composite Function Find the derivative of k(x) = (x 4 - 3) 3. Step 1 Rewrite k(x) to include the factor (x 4 - 3) 2. k(x) = (x 4 - 3)(x 4 − 3) 2 Step 2 Let m(x) = (x 4 - 3) and n(x) = (x 4 - 3) 2. Calculate the derivative of each function. m'(x) = 4x 3 Step 3 n'(x) = 2(x 4 − 3)4x 3 Power Rule Derived above Use the Product Rule to find k'(x). k′(x) = m'(x)n(x) + m(x)n'(x) 2. Sample answer: The power of h(x) is 2, and the power of k(x) is 3. Similar to the Power Rule, the exponent moves down and becomes a factor of the derivative. 3. S ample answers: 4(x 4 − 3) 34 x 3 or 16x 3 ( x 4 − 3) 3 704 | Chapter 11 4 = (x - 3) 24x 3 4 = 3(x - 4 + 2(x - 3) 24x 3 3) 24x 3 Substitution Simplify. Add. Therefore, k′(x) = 3(x 4 - 3) 24x 3 . CHECK You can use a graphing calculator to evaluate the derivative of a function at a point. Graph k (x), and from the CALC menu, select 6:dy/dx. After the screen returns to the graphing window, press 1 and then . The value of k'(1) = 48. Now verify the answer found in Step 3 by substituting x = 1 into k′(x). k'(1) = 3(1 4 − 3) 24(1) 3or 48 ✔ Analyze the Results 1. Make a conjecture as to why h'(x) and k'(x) have 4x 3 as a factor. 2. Make a conjecture as to why h'(x) has 2 as a factor and k'(x) has 3 as a factor. 3. Without rewriting p(x) as a product, find the derivative of p(x) = (x 4 - 3) 4. Copyright © McGraw-Hill Education 1. Sample answer: 4x 3 is the derivative of the expression within the parentheses. = 4x 3 (x 4 - 3) 2 + (x 4 - 3)2(x 4 - 3)4x 3 Notice that for the derivatives of h(x) and k(x), each expression was multiplied by its exponent, 1 was subtracted from its exponent, and then each expression was multiplied by the derivative of the expression inside the parentheses. Multiply by exponent. 4 h(x) = (x - Multiply by exponent. Subtract 1 from exponent. 3) 2 4 k(x) = (x - Multiply by derivative. 3) 3 Subtract 1 from exponent. Multiply by derivative. h'(x) = 2(x 4 - 3) 2 - 14 x  3 k'(x) = 3(x 4 − 3) 3 - 14 x 3 = 2(x 4 - 3)4x 3 = 3(x 4 - 3) 24x 3 Recall that we can decompose a function. For example, h(x) = (x 4 - 3) 2can be written as h(x) = f [g(x)], where f(x) = x 2 and g(x) = (x 4 - 3). This can be used to develop the Chain Rule. KeyConcept Function The derivative of a composite function f [ g (x )] is the derivative of the outer function f evaluated at the inner function g multiplied by the derivative of the inner function g. Symbols If g is differentiable at x and f is differentiable at g (x ), then f [ g (x )] = f ' [ g (x )]g '(x ).  _ dx d Activity 2 Derivative of a Composite Function Find the derivative of t(x) =   √ x 3 - 4 . WatchOut! Step 1 Decompose t(x) into two functions f and g. Substitution To help with the Chain Rule, g (x ) can be represented by u. For example, if y = f [g (x)], then y = f (u ). The derivative of y becomes y ' = f '(u ) · u '. Copyright © McGraw-Hill Education Words Step 2 Find f '(x) and g'(x).  _ 1   √ 100 + 8x   25 + 2x 9. f '(x ) = __ 2  √ 25 + 2x or __ 25 + 2x m 2 + 4 1 1 -_ f '(x) = _   2 x 2 g'(x) = 3x 2 Power Rule Step 3 Substitute into the Chain Rule. Chain Rule t'(x) = f ' [g(x)] g'(x) 1 - _ _1 =   (x 3 - 4)  2 2 3x 2 _1 -  _ 1 f '(x ) =   x 2 , g (x ) = x 3 - 4, and g '(x ) = 3x 2 2 =_   2  √ x 3 - 4  Simplify. 3 3x 2   √x - 4  =_   3 Simplify 3x 2 2(x − 4) Analyze the Results 4. Can t'(x) be found using the same method that was used earlier to find h'(x)? Explain. Model and Apply m√ m 2 + 4  __ g(x) = x 3 - 4 Original equation f (x) =   √xor x 2 4. No; sample answer: t (x ) cannot be rewritten as the product of expressions that are raised to the first power. 5. s '(t ) = 8t (t 2 − 1) 3 6. b '(x ) = -30(1 − 5x ) 5 7. c '(r ) = 3(3r − 2r 2)  2 · (3 − 4r ) 8. h '(x ) = 3(x 3 + x − 1) 2 · (3x 2 + 1) 10. g'(m) =   g(x) = x 3 - 4 f (x) =   √ x Find the derivative of each function. 5. s(t) = (t 2 - 1) 4 6. b(x) = (1 - 5x) 6 8. h(x) = (x 3 + x - 1) 3 9. f(x) =   √ 100 + 8x 7. c(r) = (3r - 2r 2 ) 3 10. g(m) =   √ m 2 + 4  705 Student Handbook Symbols, Formulas, and Key Concepts Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EM-2 Arithmetic Operations and Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EM-3 Algebraic Formulas and Key Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EM-3 Geometric Formulas and Key Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .EM-5 Trigonometric Functions and Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EM-6 Parent Functions and Function Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EM-7 Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EM-7 Statistics Formulas and Key Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EM-8 Statistical Tables Table A: The Standard Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EM-9 Table B: Student's t-Distribution (the t Distribution) . . . . . . . . . . . . . . . . . . . . . . . . . EM-11 Table C: The Chi-Square Distribution (X2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EM-12 Chapter Sourced From, EM. Lesson Summaries, from Integrated Math IV © 2012 & Appendix, from Bluman Elementary Statistics;9e Appendix © 2015 & End Matter/ Glossary, from Integrated Math IV © 2012 Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EM-1 Copyright © McGraw-Hill Education EM-2 | Symbols, Formulas, and Key Concepts Symbols ⌀ empty set is not equal to ∼p negation of p, not p ≈ is approximately equal to p⋀q conjunction of p and q ∼ is similar to p∨q disjunction of p and q >, ≥ is greater than, is greater than or equal to p⟶q conditional statement, if p then q <, ≤ is less than, is less than or equal to p⟷q biconditional statement, p if and only if q -a opposite or additive inverse of a |a| absolute value of a Geometry ∠ angle   √ a principal square root of a △ triangle a:b ratio of a to b ° degree (x, y) ordered pair π pi (x, y, z ) ordered triple  angles the imaginary unit m∠A degree measure of ∠A b = √ b nth root of b line containing points A and B 핈 rational numbers 핀 irrational numbers AB  ̶̶ AB AB  핑 integers AB ray with endpoint A containing B ̶̶ measure of AB , distance between 핎 whole numbers points A and B 핅 natural numbers ǁ is parallel to ∞ infinity ∦ is not parallel to -∞ negative infinity ⊥ is perpendicular to [] endpoint included △ triangle () endpoints not included  parallelogram log bx logarithm base b of x n-gon polygon with n sides log x common logarithm of x vector a ln x natural logarithm of x ω omega, angular speed a AB  ⎜ AB ⎟ α alpha, angle measure A′ the image of preimage A β beta, angle measure → is mapped onto γ gamma, angle measure ⊙A circle with center A theta, angle measure ⁀ AB minor arc with endpoints A and B lambda, wavelength ⁀ ABC major arc with endpoints A and C ϕ phi, angle measure ⁀ mAB degree measure of arc AB a vector a |a| magnitude of vector a Algebra ≠ i _n 1 θ Copyright © McGraw-Hill Education λ n Sets and Logic ∈ is an element of segment with endpoints A and B vector from A to B magnitude of the vector from A to B Trigonometry sin x sine of x cos x cosine of x tan x tangent of x ⊂ is a subset of sin-1 x Arcsin x ∩ intersection cos-1 x Arccos x ∪ union tan-1 x Arctan x EM-1 Symbols Functions f(x) Probability and Statistics f(x) = { f of x, the value of f at x P(a) probability of a piecewise-defined function permutation of n objects taken r at a time f(x) = ⎜ x⎟ absolute value function P(n, r) or n P r C(n, r) or n C r f(x) = 〚x〛 function of greatest integer not greater P (A ) probability of A than x P (A|B ) the probability of A given that B has f(x, y ) f of x and y, a function with two variables, already occurred x and y n! Factorial of n (n being a natural number) [ f ◦ g](x) f of g of x, the composition Σ sigma (uppercase), summation of functions f and g µ mu, population mean f -1(x) inverse of f(x) σ sigma (lowercase), population Calculus   lim   x→c m sec combination of n objects taken r at a time standard deviation 2 limit as x approaches c σ population variance slope of a secant line s sample standard deviation f ′(x ) derivative of f (x ) sample variance ∆ delta, change s 2 k indefinite integral  b      definite integral F(x ) antiderivative of f (x ) a  summation from n = 1 to k ̶ x x-bar, sample mean H 0 H a null hypothesis n=1 alternative hypothesis Measures Metric Customary Length 1 mile (mi) = 1760 yards (yd) 1 mile = 5280 feet (ft) 1 yard = 3 feet 1 foot = 12 inches (in) 1 yard = 36 inches 1 kilometer (km) = 1000 meters (m) 1 meter = 100 centimeters (cm) 1 centimeter = 10 millimeters (mm) Volume and Capacity 1 liter (L) = 1000 milliliters (mL) 1 kiloliter (kL) = 1000 liters Weight and Mass 1 kilogram (kg) = 1000 grams (g) 1 gram = 1000 milligrams (mg) 1 metric ton (t) = 1000 kilograms EM-2 | Symbols, Formulas, and Key Concepts 1 ton (T) = 2000 pounds (lb) 1 pound = 16 ounces (oz) Copyright © McGraw-Hill Education 1 gallon (gal) = 4 quarts (qt) 1 gallon = 128 fluid ounces (fl oz) 1 quart = 2 pints (pt) 1 pint = 2 cups (c) 1 cup = 8 fluid ounces Arithmetic Operations and Relations Identity For any number a, a + 0 = 0 + a = a and a · 1 = 1 · a = a. Substitution (=) If a = b, then a may be replaced by b. Reflexive (=) a=a Symmetric (=) If a = b, then b = a. Transitive (=) If a = b and b = c, then a = c. Commutative For any numbers a and b, a + b = b + a and a · b = b · a. Associative For any numbers a, b, and c, (a + b) + c = a + (b + c) and (a · b) · c = a · (b · c). Distributive For any numbers a, b, and c, a(b + c) = ab + ac and a(b - c) = ab - ac. Additive Inverse For any number a, there is exactly one number -a such that a + (-a) = 0. Multiplicative Inverse For any number _   b , where a, b ≠ 0, there is exactly one number _   a such that _   b · _   a = 1. Multiplicative (0) For any number a, a · 0 = 0 · a = 0. Addition (=) For any numbers a, b, and c, if a = b, then a + c = b + c. Subtraction (=) Multiplication and Division (=) Addition (>)* For any numbers a, b, and c, if a = b, then a - c = b - c. For any numbers a, b, and c, with c ≠ 0, if a = b, then ac = bc and _   c = _   c . Subtraction (>)* For any numbers a, b, and c, if a > b, then a - c > b - c. Multiplication and Division (>)* Zero Product a b a a b b For any numbers a, b, and c, if a > b, then a + c > b + c. For any numbers a, b, and c, a b 1. if a > b and c > 0, then ac > bc and _   c > _   c . 2. if a > b and c < 0, then ac < bc and _   c < _   c . a b For any real numbers a and b, if ab = 0, then a = 0, b = 0, or both a and b equal 0. * These properties are also true for <, ≥, and ≤. Algebraic Formulas and Key Concepts Matrices Adding ⎡ a   b ⎤ + ⎡ e   f ⎤ = ⎡a + e b + f⎤ ⎣ c d ⎦ ⎣ g h ⎦ ⎣ c + g d + h⎦ Multiplying by a Scalar ⎤ k ⎡⎣  ac   bd ⎤⎦= ⎡⎣ka   kb kc kd ⎦ Subtracting ⎡ a   b ⎤ - ⎡ e   f ⎤ = ⎡a - e b - f⎤ ⎣ c d ⎦ ⎣ g h ⎦ ⎣ c - g d - h⎦ Multiplying ⎡ a   b ⎤ · ⎡ e   f ⎤ = ⎡ae + bg af + bh ⎤ ⎣ c d ⎦ ⎣ g h ⎦ ⎣ ce + dg cf + dh ⎦ Copyright © McGraw-Hill Education Polynomials Quadratic Formula -b ±√ b 2 - 4ac x =   __    , a ≠ 0 2a Square of a Difference (a - b)2 = (a - b)(a - b) = a 2 - 2ab + b 2 Square of a Sum (a + b)2 = (a + b)(a + b) = a 2 + 2ab + b 2 Product of Sum and Difference (a + b)(a - b) = (a - b)(a + b) = a 2 - b 2 Logarithms Product Property log xab = log xa + log xb Power Property log bm p = p log bm Quotient Property a log x_ b = log xa - log xb, b ≠ 0 Change of Base loga n =  _ log a log bn b EM-3 Algebraic Formulas and Key Concepts Exponential and Logarithmic Functions Compound Interest   n ) A = P(1 + _ Exponential Growth or Decay N=N 0(1 + r)t Continuous Compound Interest A = Pe rt Continuous Exponential Growth or Decay N=N oe kt Product Property log bxy = log bx + log by Power Property log bx p = p log bx Quotient Property log b_   y = log bx - log by Change of Base loga x log bx =   _ log b Logistic Growth r nt x a c f(t ) = _   1 + ae -bt Sequences and Series n th term, Arithmetic a n= a 1+ (n - 1)d Sum of Arithmetic Series S n= n _ or S n= _   2 [2a 1 + (n - 1)d ] 2 Sum of Infinite Geometric Series S=_   1 - r , ⎜r⎟< 1 Power Series  a n x n = a 0 + a 1x + a 2 x 2 + a 3 x 3 + … ( a + a ) 1 n 2 a 1 ∞ n=0 n th term, Geometric a n= a 1r n - 1 Sum of Geometric Series S n= _ or S n= _ , r ≠ 1 1-r Euler’s Formula e iθ = cos θ + i sin θ Exponential Series x n x 2 x 3 … e x=  _   = 1 + x + _   2 ! + _   3! + a 1- a 1r n a 1- a nr 1-r ∞ n=0 n! Binomial Theorem (a + b) n= nC 0 a nb 0 + n C 1 a n - 1b 1 + n C 2 a n - 2b 2 + … + n C ra n - rb r+ … + n C na 0 b n Cosine and Sine Power Series ∞ n 2n (-1)  x x 2 x 4 _ x 6 … cos x =  _   = 1 - _   2! +  _   + 4! 6! n=0 (2n )! n 2n + 1 ∞ (-1)  x x 3 _ x 5 _ x 7 … sin x =   _ = x -  _ +     + 5! 7! 3! n=0 (2n + 1)! Vectors Addition in Plane a + b = 〈a 1 + b 1, a 2 + b 2〉 Addition in Space a + b = 〈a 1 + b 1, a 2 + b 2, a 3 + b 3〉 Subtraction in Plane a - b = 〈a 1 - b 1, a 2 - b 2〉 Subtraction in Space a - b = a + (-b) = 〈a 1 - b 1, a 2 - b 2, a 3 - b 3〉 Scalar Multiplication in Plane ka = 〈ka 1, ka 2〉 Scalar Multiplication in Space k a = 〈ka 1, ka 2, ka 3〉 Dot Product in Plane a · b = a 1b 1 + a 2 b 2 Dot Product in Space a · b = a 1b 1 + a 2 b 2 + a 3b 3 Angle Between Two Vectors cos θ =   _ Projection of u onto v u·v projvu =   _ 2 v a·b |a||b| Triple Scalar Product Equations of a Line on a Coordinate Plane Slope-intercept form of a line Point-slope form of a line y = mx + b y - y 1 = m(x - x 1) EM-4 | Symbols, Formulas, and Key Concepts ⎜ ⎟ t 1 t 2 t 3 t · (u × v) = u 1  u 2   u 3 v 1 v 2 v 3 Copyright © McGraw-Hill Education Magnitude of a Vector |v| =   √ (   x 2 - x 1) 2 + (y 2 - y 1) 2  ( |v| ) Algebraic Formulas and Key Concepts Conic Sections (x - h) 2 = 4p(y - k) or (y - k) 2 = 4p(x - h) Parabola (x - h)2 (y - k) 2 a b Circle _ = 1 or   _ 2 +   2 Ellipse 2 Hyperbola 2 (x - h ) (y - k)    _ +   _ = 1 2 2 b Rotation of Conics x 2 + y 2 = r 2or (x - h) 2 + (y - k)2 = r 2 (x - h )2 (y - k)2 a b _ = 1 or  _ 2 -   2 2 2 a 2 (y - k)  (x - h)    _ -   _ = 1 2 a b x ′ = x cos θ + y sin θ and y ′ = y cos θ - x sin θ Parametric Equations Horizontal Distance y = tv 0 sin θ − _   2 gt 2 + h 0 1 Vertical Position x = tv 0 cos θ Complex Numbers Product Formula z 1z 2 = r 1r 2[cos (θ 1 + θ 2) + i sin (θ 1 + θ 2)] Distinct Roots Formula r p cos   _ + i sin  _ p p 1 _ ( θ + 2nπ ) θ + 2nπ Quotient Formula z 1 r 1 _ =_   [cos (θ - θ ) + i sin (θ - θ )] De Moivre’s Theorem z n = [r(cos θ + i sin θ)] nor r n (cos nθ + i sin nθ) z 2 r 2 1 2 1 2 Geometric Formulas and Key Concepts Coordinate Geometry Slope y 2 - y 1 m =   _ x 2 - x 1, x 2 ≠ x 1 Distance on a coordinate plane d =   √ (   x 2 - x 1) 2 + ( y 2 - y 1) 2  Midpoint on a coordinate plane ,   _ M =   _ 2 2 Midpoint in space ( x + x y + y ) x + x _ y + y _ z + z M = (   _ ,   ,   2 2 2 ) Distance in space d =   √ (    x 2 - x 1) 2 + ( y 2 - y 1) 2 + ( z 2 - z 1) 2  1 2 1 2 1 2 1 2 1 2 Distance on a number line d = ∣a - b∣ Arc length ℓ=_   360 · 2πr Midpoint on a number line M =   _ 2 Pythagorean Theorem a 2 + b 2 = c 2 x a+b Perimeter and Circumference Square P = 4s Rectangle P = 2ℓ + 2w Circle C = 2πr or C = πd Lateral Surface Area Prism L = Ph Pyramid L=_   2 Pℓ Cylinder L = 2πrh Cone L = πrℓ 1 Copyright © McGraw-Hill Education Total Surface Area Prism Pyramid S = Ph + 2B Cone S=_   2 Pℓ + B 1 Sphere S = πrℓ + πr 2 S = 4πr 2 Cylinder Cube S = 2πrh + 2πr 2 S = 6s 2 Volume Prism Pyramid V = Bh Cone V=_   3 Bh Rectangular prism 1 Sphere V=_   3 πr 2 h 1 V=_   3 πr 3 4 Cylinder Cube V = πr 2 h 3 V = s V = ℓwh EM-5 Trigonometric Functions and Identities Trigonometric Functions adj opp Trigonometric Functions sin θ =  _ hyp cos θ = _   hyp hyp _ csc θ =  _ opp =   1 sin θ hyp opp sin θ cos θ adj cos θ sin θ tan θ =  _ =  _ adj sec θ =  _ =_   adj cot θ = _   opp =  _ 1 cos θ Law of Cosines a 2 = b 2 + c 2 - 2bc cos A b 2 = a 2 + c 2 - 2ac cos B c 2 = a 2 + b 2 - 2ab cos C Law of Sines  _ =  _ =  _ b a c  Heron’s Formula Area =   √s(s    - a)(s - b)(s - c)  Linear Speed v=_   t Angular Speed sin A sin B sin C s ω=_   t θ Trigonometric Identities Reciprocal   sin θ = _ 1 csc θ 1 csc θ = _   sin θ cos θ = _   1 sec θ 1 sec θ = _   cos θ tan θ = _   Pythagorean sin 2 θ + cos 2 θ = 1 tan 2 θ + 1 = sec 2 θ cot 2 θ + 1 = csc 2 θ sin θ = cos _   2 - θ (π ) π cos θ = sin ( _   2 - θ ) tan θ = cot _   2 - θ (π ) π cot θ = tan ( _   2 - θ ) sec θ = csc _   2 - θ sin (-θ) = -sin θ cos (-θ) = cos θ tan (-θ) = -tan θ csc (-θ) = -csc θ sec (-θ) = sec θ cot (-θ) = -cot θ Cofunction Odd-Even Sum & Difference Double-Angle 1 cot θ 1 cot θ = _   tan θ (π ) π csc θ = sec ( _   2 - θ ) cos (α + β) = cos α cos β - sin α sin β cos (α - β) = cos α cos β + sin α sin β sin (α + β) = sin α cos β + cos α sin β sin (α - β) = sin α cos β - cos α sin β tan α + tan β tan (α + β) =   __    1 - tan α tan β cos 2θ = cos 2 θ - sin 2 θ sin 2θ = 2 sin θ cos θ tan α - tan β 1 + tan α tan β tan (α - β) =   __    cos 2θ = 2 cos 2 θ - 1 cos 2θ = 1 - 2 sin 2 θ tan 2θ =   _ 2 2 tan θ 1 - tan  θ Power-Reducing sin 2 θ =   _ cos 2 θ =   _ 2 2 Half-Angle  1 - cos θ θ sin  _ = ±   _   2 2 √  1 - cos θ θ  tan  _ = ±√  _ 2 1 + cos θ √ tan  _ =   _ 2 θ sin α sin β = _   2 [cos (α - β) - cos (α + β)] cos α cos β = _   2 [cos (α - β) + cos (α + β)] 1 (α + β) (α - β) α+β α-β sin α - sin β = 2 cos (  _ sin (   _ 2 ) 2 ) sin α + sin β = 2 sin   _ cos   _ 2 2 EM-6 | Symbols, Formulas, and Key Concepts tan  _ =  _ 2 1 - cos θ sin θ θ sin θ 1 + cos θ sin α cos β = _   2 [sin (α + β) + sin (α - β)] 1 cos α sin β = _   2 [sin (α + β) - sin (α - β)] 1 (α + β) (α - β) α+β α-β cos α - cos β = -2 sin (   _ sin (   _ 2 ) 2 ) cos α + cos β = 2 cos   _ cos   _ 2 2 Copyright © McGraw-Hill Education Sum-to-Product 1 - cos 2θ 1 + cos 2θ  1 + cos θ θ cos  _ = ±   _ 2 2 1 Product-to-Sum tan 2 θ =   _ 1 + cos 2θ 1 - cos 2θ Parent Functions and Function Operations Parent Functions Linear Functions Absolute Value Functions y Quadratic Functions y y y = |x | y = x2 y= x O x O x O EM-056A-888482 Exponential and Logarithmic Functions EM-057A-888482 Square Root Functions y= EM-058A-888482 Reciprocal and Rational Functions y y b x, b > y y = x, x ≥ 0 1 x y= 1 ,x≠0 x (0, 0) O O x O x x y = logb x, b > 1 Function Operations EM-060A-888482 EM-059A-888482 EM-061A-888482 Addition (f + g)(x) = f (x) + g(x) Multiplication (f · g)(x) = f(x) · g(x) Subtraction (f - g)(x) = f (x) - g(x) Division _ g (x) = _   g(x) , g(x) ≠ 0 (f) f(x) Calculus Limits Sum   x→c Scalar Multiple   x→c Quotient nth Root   lim [f(x ) + g(x)] =   limf(x ) +   limg(x ) Difference   x→c   lim [k f(x )] = k limf(x ) Product   x→c   x→c   x→c   x→c limf(x ) f(x )   x→c   lim _ = _ , if  limg (x) ≠ 0   x→c g(x )   limg(x )   x→c   x→c n n    lim √ f (x )  = √   limf (x ) , if  limf(x) > 0   x→c when n is even   x→c   x→c Power Velocity   lim [f(x) - g(x)] =limf(x) -limg (x )   x→c   x→c   lim [f(x ) · g(x )] =limf(x ) ·   limg (x )   x→c ⎤ ⎡   lim [f(x) n ] = ⎢ limf(x )⎥⎦ ⎣  x→c   x→c n   x→c Average f(b) - f(a) v avg =   _ b-a Instantaneous f(t + h ) - f(t ) v(t ) = lim   __ h h→0 Copyright © McGraw-Hill Education Derivatives Power Rule If f(x) = x n , f ′(x) = n x n - 1 Sum or Difference If f(x) = g (x) ± h(x), then f ′(x ) = g′(x ) ± h′(x) Product Rule d _ [f(x)g(x)] = f ′(x )g (x ) + f(x )g′(x ) Quotient Rule f ′(x )g (x ) - f (x )g′(x ) d ⎡ f(x ) ⎤ _ d x ⎢  _⎥ =   __    ⎣ g(x ) ⎦ [g (x )]2 dx Integrals Indefinite Integral ∫f (x ) d x = F (x ) + C Fundamental Theorem of Calculus b      f (x ) d x = F(b) - F(a) a EM-7 Statistics Formulas and Key Concepts X-µ z-Values z =   _ σ Binomial Probability P(X ) = n C xp xq n - x= _   (n - x )! x ! p xq n - x n! Confidence Interval, Normal CI = x   ̶ ± E or x  ̶ ± z · Distribution Correlation Coefficient ̶ σ _   √ n ̶ ( x - x  ) ( y- y  ) _ r=_   n - 1  _ s x   s y 1 i i z-Value of a Sample Mean Maximum Error of Estimate ̶   - µ x z =   _ σ ̶ x   σ E=z·σ x  ̶ or z · _   √  n Confidence Interval, t-Distribution s CI = x   ̶ ± t · _   √ t-Test for the Correlation Coefficient n-2  t = r   _2  , degrees of freedom: n - 2  n √ 1 - r Copyright © McGraw-Hill Education EM-8 | Symbols, Formulas, and Key Concepts TABLE A The Standard Normal Distribution Cumulative Standard Normal Distribution z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 -3.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0002 -3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 .0003 -3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005 -3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 .0007 .0007 -3.0 .0013 .0013 .0013 .0012 .0012 .0011 .0011 .0011 .0010 .0010 -2.9 .0019 .0018 .0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014 -2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019 -2.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026 -2.6 .0047 .0045 .0044 .0043 .0041 .0040 .0039 .0038 .0037 .0036 -2.5 .0062 .0060 .0059 .0057 .0055 .0054 .0052 .0051 .0049 .0048 -2.4 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064 -2.3 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084 -2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110 -2.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143 -2.0 .0228 .0222 .0217 .0212 .0207 .0202 .0197 .0192 .0188 .0183 -1.9 .0287 .0281 .0274 .0268 .0262 .0256 .0250 .0244 .0239 .0233 -1.8 .0359 .0351 .0344 .0336 .0329 .0322 .0314 .0307 .0301 .0294 -1.7 .0446 .0436 .0427 .0418 .0409 .0401 .0392 .0384 .0375 .0367 -1.6 .0548 .0537 .0526 .0516 .0505 .0495 .0485 .0475 .0465 .0455 -1.5 .0668 .0655 .0643 .0630 .0618 .0606 .0594 .0582 .0571 .0559 -1.4 .0808 .0793 .0778 .0764 .0749 .0735 .0721 .0708 .0694 .0681 -1.3 .0968 .0951 .0934 .0918 .0901 .0885 .0869 .0853 .0838 .0823 -1.2 .1151 .1131 .1112 .1093 .1075 .1056 .1038 .1020 .1003 .0985 -1.1 .1357 .1335 .1314 .1292 .1271 .1251 .1230 .1210 .1190 .1170 -1.0 .1587 .1562 .1539 .1515 .1492 .1469 .1446 .1423 .1401 .1379 -0.9 .1841 .1814 .1788 .1762 .1736 .1711 .1685 .1660 .1635 .1611 -0.8 .2119 .2090 .2061 .2033 .2005 .1977 .1949 .1922 .1894 .1867 -0.7 .2420 .2389 .2358 .2327 .2296 .2266 .2236 .2206 .2177 .2148 -0.6 .2743 .2709 .2676 .2643 .2611 .2578 .2546 .2514 .2483 .2451 -0.5 .3085 .3050 .3015 .2981 .2946 .2912 .2877 .2843 .2810 .2776 -0.4 .3446 .3409 .3372 .3336 .3300 .3264 .3228 .3192 .3156 .3121 -0.3 .3821 .3783 .3745 .3707 .3669 .3632 .3594 .3557 .3520 .3483 -0.2 .4207 .4168 .4129 .4090 .4052 .4013 .3974 .3936 .3897 .3859 -0.1 .4602 .4562 .4522 .4483 .4443 .4404 .4364 .4325 .4286 .4247 -0.0 .5000 .4960 .4920 .4880 .4840 .4801 .4761 .4721 .4681 .4641 Copyright © McGraw-Hill Education For z values less than -3.49, use 0.0001. Area z 0 EM-9 TABLE A (continued ) Cumulative Standard Normal Distribution z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 0.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389 1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621 1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830 1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015 1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177 1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319 1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441 1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545 1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633 1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706 1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767 2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817 2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857 2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890 2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916 2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936 2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952 2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964 2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974 2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981 2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986 3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990 3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993 3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .9995 3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997 3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998 For z values greater than 3.49, use 0.9999. 0 EM-10 | Normal Distribution table z Copyright © McGraw-Hill Education Area TABLE B Copyright © McGraw-Hill Education Degrees of Freedom Student's t-Distribution (the t Distribution) Confidence intervals One tail, α 80% 0.10 90% 0.05 95% 0.025 Two tails, α 0.20 0.10 0.05 0.02 0.01 1 3.078 2 1.886 3 1.638 4 1.533 5 1.476 6 1.440 7 1.415 8 1.397 9 1.383 10 1.372 11 1.363 12 1.356 13 1.350 14 1.345 15 1.341 16 1.337 17 1.333 18 1.330 19 1.328 20 1.325 21 1.323 22 1.321 23 1.319 24 1.318 25 1.316 26 1.315 27 1.314 28 1.313 29 1.311 30 1.310 32 1.309 34 1.307 36 1.306 38 1.304 40 1.303 45 1.301 50 1.299 55 1.297 60 1.296 65 1.295 70 1.294 75 1.293 80 1.292 90 1.291 100 1.290 500 1.283 1000 1.282 (z) ∞ 1.282 6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.860 1.833 1.812 1.796 1.782 1.771 1.761 1.753 1.746 1.740 1.734 1.729 1.725 1.721 1.717 1.714 1.711 1.708 1.706 1.703 1.701 1.699 1.697 1.694 1.691 1.688 1.686 1.684 1.679 1.676 1.673 1.671 1.669 1.667 1.665 1.664 1.662 1.660 1.648 1.646 1.645 12.706 4.303 3.182 2.776 2.571 2.447 2.365 2.306 2.262 2.228 2.201 2.179 2.160 2.145 2.131 2.120 2.110 2.101 2.093 2.086 2.080 2.074 2.069 2.064 2.060 2.056 2.052 2.048 2.045 2.042 2.037 2.032 2.028 2.024 2.021 2.014 2.009 2.004 2.000 1.997 1.994 1.992 1.990 1.987 1.984 1.965 1.962 1.960 31.821 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 2.583 2.567 2.552 2.539 2.528 2.518 2.508 2.500 2.492 2.485 2.479 2.473 2.467 2.462 2.457 2.449 2.441 2.434 2.429 2.423 2.412 2.403 2.396 2.390 2.385 2.381 2.377 2.374 2.368 2.364 2.334 2.330 2.326 63.657 9.925 5.841 4.604 4.032 3.707 3.499 3.355 3.250 3.169 3.106 3.055 3.012 2.977 2.947 2.921 2.898 2.878 2.861 2.845 2.831 2.819 2.807 2.797 2.787 2.779 2.771 2.763 2.756 2.750 2.738 2.728 2.719 2.712 2.704 2.690 2.678 2.668 2.660 2.654 2.648 2.643 2.639 2.632 2.626 2.586 2.581 2.576 One tail 98% 0.01 99% 0.005 Two tails Area α t Area α 2 -t Area α 2 +t EM-11 TABLE C Degrees of Freedom The Chi-Square Distribution (X2) α 0.995 0.99 0.975 0.95 0.90 0.10 1 — — 0.001 0.004 0.016 2.706 0.05 3.841 0.025 2 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 10.597 3 0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.838 4 0.207 0.297 0.484 0.711 1.064 7.779 9.488 11.143 13.277 14.860 5 0.412 0.554 0.831 1.145 1.610 9.236 11.071 12.833 15.086 16.750 6 0.676 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.812 18.548 7 0.989 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.475 20.278 8 1.344 1.646 2.180 2.733 3.490 13.362 15.507 17.535 20.090 21.955 9 1.735 2.088 2.700 3.325 4.168 14.684 16.919 19.023 21.666 23.589 10 2.156 2.558 3.247 3.940 4.865 15.987 18.307 20.483 23.209 25.188 11 2.603 3.053 3.816 4.575 5.578 17.275 19.675 21.920 24.725 26.757 12 3.074 3.571 4.404 5.226 6.304 18.549 21.026 23.337 26.217 28.299 13 3.565 4.107 5.009 5.892 7.042 19.812 22.362 24.736 27.688 29.819 14 4.075 4.660 5.629 6.571 7.790 21.064 23.685 26.119 29.141 31.319 15 4.601 5.229 6.262 7.261 8.547 22.307 24.996 27.488 30.578 32.801 16 5.142 5.812 6.908 7.962 9.312 23.542 26.296 28.845 32.000 34.267 17 5.697 6.408 7.564 8.672 10.085 24.769 27.587 30.191 33.409 35.718 18 6.265 7.015 8.231 9.390 10.865 25.989 28.869 31.526 34.805 37.156 19 6.844 7.633 8.907 10.117 11.651 27.204 30.144 32.852 36.191 38.582 20 7.434 8.260 9.591 10.851 12.443 28.412 31.410 34.170 37.566 39.997 21 8.034 8.897 10.283 11.591 13.240 29.615 32.671 35.479 38.932 41.401 22 8.643 9.542 10.982 12.338 14.042 30.813 33.924 36.781 40.289 42.796 23 9.262 10.196 11.689 13.091 14.848 32.007 35.172 38.076 41.638 44.181 24 9.886 10.856 12.401 13.848 15.659 33.196 36.415 39.364 42.980 45.559 25 10.520 11.524 13.120 14.611 16.473 34.382 37.652 40.646 44.314 46.928 26 11.160 12.198 13.844 15.379 17.292 35.563 38.885 41.923 45.642 48.290 27 11.808 12.879 14.573 16.151 18.114 36.741 40.113 43.194 46.963 49.645 28 12.461 13.565 15.308 16.928 18.939 37.916 41.337 44.461 48.278 50.993 29 13.121 14.257 16.047 17.708 19.768 39.087 42.557 45.722 49.588 52.336 5.024 0.01 6.635 0.005 7.879 13.787 14.954 16.791 18.493 20.599 40.256 43.773 46.979 50.892 53.672 40 20.707 22.164 24.433 26.509 29.051 51.805 55.758 59.342 63.691 66.766 50 27.991 29.707 32.357 34.764 37.689 63.167 67.505 71.420 76.154 79.490 60 35.534 37.485 40.482 43.188 46.459 74.397 79.082 83.298 88.379 91.952 70 43.275 45.442 48.758 51.739 55.329 85.527 90.531 95.023 100.425 104.215 80 51.172 53.540 57.153 60.391 64.278 96.578 101.879 106.629 112.329 116.321 90 59.196 61.754 65.647 69.126 73.291 107.565 113.145 118.136 124.116 128.299 100 67.328 70.065 74.222 77.929 82.358 118.498 124.342 129.561 135.807 140.169 Area α χ2 EM-12 | Normal Distribution table Copyright © McGraw-Hill Education 30 Hybrid education in the Emirati school Within the strategic dimension of the Ministry of Education’s development plans and its endeavor to diversify education channels and overcome all the challenges that may prevent it, and to ensure continuity in all circumstances, the Ministry has implemented a hybrid education plan for all students at all levels of education. Prekindergarten/ kindergarten Study Plan Cycle 1 Cycle 2 Cycle 3 School learning Direct e-learning Self Channels for obtaining a textbook: Electronic units Doctorate Degree / Applied Sciences (3-5 years) Master’s Degree / Applied Sciences (2 years) Education System in the United Arab Emirates Postgraduate Diploma / Applied Sciences (1-2 Years) Bachelor’s Degree / Applied Sciences (3-4 years) Diploma / Postgraduate Diploma (2-3 years) Joining Work Bridging Program Vocational Qualification Vocational Education and Development Centre Continuous Education Home General & Schooling Academic Stream Applied Stream Literacy Bachelor’s Degree Technical & Vocational Streams Specialised Academies Technical General Academic Streams Technical Advanced General Applied Technology Secondary School Program Advanced Elite Top 10 percent of students in national test scores 6 International Classifications Grades 1-4 Ages 6-9 (4 years) 12 3 International Classifications Cycle 1 Basic Education Cycle 2 Elite Program (Grades 5-8) Top 10 percent of students in national test scores Ages 10-14 (4 years) Early Childhood Kindergarten Ages 4-6 (not compulsory) Nursery/ Care Ages 0-4 (not compulsory) International Classifications 12 Cycle 3 / Secondary Stage Grades 9-12 Ages 14-18 (4 years) Grades 5-8 Ages 10-14 (4 years) UAE National Qualifications Framework Level 10 18 International Classifications The Ministry coordinates with national higher education institutions to admit students in various majors in line with the needs of the labour market and future human development plans. Higher Education institutions also determine the number of students that can be admitted according to their capabilities, mission and goals. They also set the conditions for students’ admission to various programmes according to the stream they graduated from, the levels of their performance in the secondary stage, and their results from the Emirates Standard Assessment Test. Integration and coordination between General and Higher Education systems allow for the approval and calculation of school study courses within university studies according to the school stream and university specialisation, which reduces the duration of university studies. International Classifications Advanced University Enrollment (3 years) (3-4 years) Applied 18 Levels 7-8 (4-5 years) Bachelor’s Degree / Applied Sciences Levels 5-6 Postgraduate Diploma (1 year) Levels 1-4 Postgraduate Diploma (1-2 year) International Classifications 22 (2 years) UAE National Qualifications Framework Level 9 (2 years) International Classifications Master’s Degree UAE National Qualifications Framework Dual Degree Bachelor’s / Master’s Theoretical International Classifications Master’s Degree UAE National Qualifications Framework 24 (3-5 years) UAE National Qualifications Framework Doctorate Degree Starting Age All ri right ghtss rreserved. eserved. No part of this bage page or may thebe book r may be repr eproduced, oduced, st stor ored ed in a rretieval etieval sy syst stem em or tr transmitt ansmitted ed in any form or b byy any means w without ithout prior permission in writing of the publisher publisher..

Derivatives and Integration: Limits, Rates of Change, Calculus

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