# rogawskiet4e_ssm_ch04.pdf - 4 APPLICATIONS OF THE...

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This preview shows page 1 out of 102 pages. Unformatted text preview: 4 APPLICATIONS OF THE DERIVATIVE 4.1 Linear Approximation and Applications Preliminary Questions 1. True or False? The Linear Approximation says that the vertical change in the graph is approximately equal to the vertical change in the tangent line. This statement is true. The linear approximation does say that the vertical change in the graph is approxi­ mately equal to the vertical change in the tangent line. SOLUTION 2. Estimate g(1.2) − g(1) if g' (1) = 4. SOLUTION Using the Linear Approximation, g(1.2) − g(1) ≈ g' (1)(1.2 − 1) = 4(0.2) = 0.8 3. Estimate f (2.1) if f (2) = 1 and f ' (2) = 3. SOLUTION Using the Linearization, f (2.1) ≈ f (2) + f ' (2)(2.1 − 2) = 1 + 3(0.1) = 1.3 4. Complete the following sentence: The Linear Approximation shows that up to a small error, the change in output Δ f . is directly proportional to The Linear Approximation tells us that up to a small error, the change in output Δ f is directly proportional to the change in input Δx when Δx is small. SOLUTION Exercises In Exercises 1 6, use Eq. (1) to estimate Δ f = f (3.02) − f (3). 1. f (x) = x2 Let f (x) = x2 . Then f ' (x) = 2x and Δ f ≈ f ' (3)Δx = 6(0.02) = 0.12. SOLUTION 3. f (x) = x−1 1 Let f (x) = x−1 . Then f ' (x) = −x−2 and Δ f ≈ f ' (3)Δx = − (0.02) = −0.00222. 9 SOLUTION 5. f (x) = SOLUTION √ x + 16 x f (x) = e2x . Then f ' (x) = 2e2x and Let Δ f ≈ f ' (3)Δx = 2e6 (0.02) = 0.04e6 = 16.137 7. The cube root of 27 is 3. How much larger is the cube root of 27.2? Estimate using the Linear Approximation. Let f (x) = x1/3 , a = 27, and Δx = 0.2. Then f ' (x) = Approximation is SOLUTION Δ f ≈ f ' (a)Δx = Thus, √3 27.2 is larger than 1 −2/3 x 3 and f ' (a) = f ' (27) = 1 . 27 The Linear 1 (0.2) = 0.0074074 27 √3 27 by approximately 0.0074074. 1 2 CHAPTER 4 APPLICATIONS OF THE DERIVATIVE In Exercises 9 12, use Eq. (1) to estimate Δ f . Use a calculator to compute both the error and the percentage error. √ 9. f (x) = 1 + x, a = 3, Δx = 0.2 Let f (x) = (1 + x)1/2 , a = 3, and Δ x = 0.2. Then f ' (x) = 12 (1 + x)−1/2 , f ' (a) = f ' (3) = SOLUTION Δ f ≈ f ' (a)Δx = 1 4 and 1 (0.2) = 0.05 4 The actual change is Δ f = f (a + Δx) − f (a) = f (3.2) − f (3) = √ 4.2 − 2 ≈ 0.049390 The error in the Linear Approximation is therefore |0.049390 − 0.05| = 0.000610; in percentage terms, the error is 0.000610 × 100% ≈ 1.24% 0.049390 11. f (x)f =x 1 , 1 + x2 a = 3, Let f (x) = SOLUTION 1 , 1+x2 Δx =Δ 0.5 2x ' ' a = 3, and Δx = 0.5. Then f ' (x) = − (1+x 2 )2 , f (a) = f (3) = −0.06 and Δ f ≈ f ' (a)Δx = −0.06(0.5) = −0.03 The actual change is Δ f = f (a + Δx) − f (a) = f (3.5) − f (3) ≈ −0.0245283 The error in the Linear Approximation is therefore | − 0.0245283 − (−0.03)| = 0.0054717; in percentage terms, the error is    0.0054717  × 100% ≈ 22.31%  −0.0245283  In Exercises 13 20 using Linear Approximation, estimate Δ f for a change in x from x = a to x = b. Use the estimate to approximate f (b), and ﬁnd the error using a calculator. 13. f (x) = SOLUTION • • • SOLUTION • x, a = 25, b = 26 √ Let f (x) = x, a = 25, and Δ x = 1. Then f ' (x) = 12 x−1/2 and f ' (a) = f ' (25) = a = 100, b = 101 = Let f (x) = √1 x , a = 100, and Δ x = 1. Then f ' (x) = √1 , x d (x−1/2 ) dx 1 = − 21 x−3/2 and f ' (a) = − 21 ( 1000 ) = −0.0005. The Linear Approximation is Δ f ≈ f ' (a)Δx = −0.0005(1) = −0.0005. Using the Linear Approximation, f (b) ≈ f (a) + Δ f = • 1 . 10 The Linear Approximation is Δ f ≈ f ' (a)Δx = 101 (1) = 0.1. Using the Linear Approximation, f (b) ≈ f (a) + Δ f = 5 + 0.1 = 5.1. √ Using a calculator, the error in this estimate is | 26 − 5.1| ≈ |5.0990195 − 5.1| = 0.0009805. 15. f (x) = • √ 1 + (−0.0005) = 0.0995 10 Using a calculator, the error in this estimate is    1  − 0.0995 ≈ |0.0995037 − 0.0995| = 3.7 × 10−6  √  101 17. f (x) = x1/3 , a = 8, b = 9 SOLUTION • • Let f (x) = x1/3 , a = 8, and Δ x = 1. Then f ' (x) = 3 1 x−2/3 and f ' (a) = f ' (8) = 1 . 12 The Linear Approximation is Δ f ≈ f ' (a)Δ x = 121 (1) = 0.083333. Using the Linear Approximation, f (b) ≈ f (a) + Δ f = 2 + 0.083333 = 2.083333. √3 Using a calculator, the error in this estimate is | 9 − 2.083333| ≈ |2.080084 − 2.083333| = 0.003249. S E C T I O N 4.1 Linear Approximation and Applications 3 19. f (x) = e x , a = 0, b = −0.1 SOLUTION • • • Let f (x) = e x , a = 0, and Δx = −0.1. Then f ' (x) = e x and f ' (a) = f ' (0) = 1. The Linear Approximation is Δ f ≈ f ' (a)Δx = 1(−0.1) = −0.1. Using the Linear Approximation, f (b) ≈ f (a) + Δ f = 1 + (−0.1) = 0.9. Using a calculator, the error in this estimate is |e−0.1 − 0.9| ≈ |0.904837 − 0.9| = 0.004837. In Exercises 21 28, ﬁnd the linearization at x = a and then use it to approximate f (b). 21. f (x) = x4 , SOLUTION a = 1, b = 0.96 Let f (x) = x . Then f ' (x) = 4x3 , f (a) = 1, and f ' (a) = 4. The linearization at a = 1 is 4 L(x) = f ' (a)(x − a) + f (a) = 4(x − 1) + 1 = 4x − 3 and f (b) = f (0.96) ≈ L(0.96) = 4(0.96) − 3 = 0.84 23. f (x) = sin2 x, SOLUTION a = π4 , b= 1.1π 4 2 1 ' ' Let x f (x) = sin x. Then f (x) = 2 sin x cos x = sin 2x, f (a) = 2 , and f (a) = 1. The linearization at a = π 4 is ( π) 1 π 1 + = x− + L(x) = f ' (a)(x − a) + f (a) = 1 x − 4 2 4 2 and ( f (b) = f ) ( ) 1.1π 1.1π 1.1π π 1 0.1π 1 ≈L = − + = + ≈ 0.578540 4 4 4 4 2 4 2 25. f (x) = (1 + x)−1/2 , a = 0, b = 0.08 (x) Let f (x) = (1 + x)−1/2=. Th. en f ' (x) = − 1 (1 + x)−3/2 , f (a) = 1, and f ' (a) = − 1 . The linearization at a = 0 is SOLUTION 2 2 x 1 L(x) = f ' (a)(x − a) + f (a) = − x + 1 2 and 1 f (b) = f (0.08) ≈ L(0.08) = − (0.08) + 1 = 0.96 2 √ 27. f (x) = e x , SOLUTION a = 1, 1/ b = 0.85 √ Let f (x) = e x . Then f ' (x) = 1 1 √ √ e x , f (a) = e, and f ' (a) = e 2 2 x so the linearization of f (x) at a = 1 is L(x) = f ' (a)(x − a) + f (a) = 1 1 e(x − 1) + e = e(x + 1) 2 2 and f (b) = f (0.85) ≈ L(0.85) = 1 e(0.85 + 1) = 0.925e ≈ 2.514411 2 [Eq. (3)]. In Exercises 29 x32, ln xestimate = Δy using b = differentials 02 f (x) = 29. y = cos x, SOLUTION a = π6 , dx = 0.014 Let f (x) = cos x. Then f ' (x) = − sin x and Δy ≈ dy = f ' (a)dx = − sin 31. y = 10 − x2 , 2 + x2 a = 1, dx = 0.010 (π) 6 (0.014) = −0.007 4 CHAPTER 4 APPLICATIONS OF THE DERIVATIVE SOLUTION Let f (x) = 10 − x2 . Then 2 + x2 f ' (x) = 24x (2 + x2 )(−2x) − (10 − x2 )(2x) =− (2 + x2 )2 (2 + x2 )2 and Δy ≈ dy = f ' (a)dx = − 24 (0.01) = −0.026667 9 33. Estimate f (4.03) for f (x) as in Figure 9. 1 y y 5 f (x) (10, 4) Tangent line (4, 2) x FIGURE 9 SOLUTION Using the Linear Approximation, f (4.03) ≈ f (4) + f ' (4)(0.03). From the ﬁgure, we ﬁnd that f (4) = 2 and f ' (4) = 4−2 1 = 10 − 4 3 Thus, 1 f (4.03) ≈ 2 + (0.03) = 2.01 3 √ √ √ √ 2.1 − 2 or 9.1 − 9? Explain using the Linear Approximation. √ 1 SOLUTION thethe Li Linear ApApproximation at x = a gives ne t qu Let-s f (x) = x, eand Δ x = 0.1. t Then is a f ' (x) = 2 x−1/2 and 35. Which is larger: Δf = √ a + 0.1 − √ 1 0.05 a ≈ f ' (a)(0.1) = a−1/2 (0.1) = √ 2 a We see that Δ f decreases as a increases. In particular √ √ 0.05 2.1 − 2 ≈ √ 2 is larger than √ √ 0.05 9.1 − 9 ≈ 3 37. Box ofﬁce revenue at a multiplex cinema in Paris is R(p) = 3600p − 10p3 euros per showing when the ticket price is p euros. Calculate R(p) for p = 9 and use the Linear Approximation to estimate ΔR if p is raised or lowered by 0.5 euro. Let R(p) = 3600p − 10p3 . Then R(9) = 3600(9) − 10(9)3 = 25110 euros. Moreover, R' (p) = 3600 − 30p2 , so by the Linear Approximation, SOLUTION ΔR ≈ R' (9)Δp = 1170Δp If p is raised by 0.5 euros, then ΔR ≈ 585 euros; on the other hand, if p is lowered by 0.5 euros, then ΔR ≈ −585 euros. 39. A thin silver wire has length L = 18 cm when the temperature is T = 30◦ C. Estimate ΔL when T decreases to 25◦ C if the coefﬁcient of thermal expansion is k = 1.9 × 10−5◦ C−1 (see Example 3). SOLUTION We have dL = kL = (1.9 × 10−5 )(18) = 3.42 × 10−4 cm/◦ C dT The change in temperature is ΔT = −5◦ C, so by the Linear Approximation, the change in length is approximately ΔL ≈ 3.42 × 10−4 ΔT = (3.42 × 10−4 )(−5) = −0.00171 cm At T = 25◦ C, the length of the wire is approximately 17.99829 cm. 41. The atmospheric pressure at altitude h (kilometers) for 11 ≤ h ≤ 25 is approximately P(h) = 128e−0.157h kilopascals Linear Approximation and Applications S E C T I O N 4.1 5 (a) Estimate ΔP at h = 20 when Δh = 0.5. (b) Compute the actual change, and compute the percentage error in the Linear Approximation. SOLUTION (a) Let P(h) = 128e−0.157h . Then P' (h) = −20.096e−0.157h . Using the Linear Approximation, ΔP ≈ P' (h)Δh = P' (20)(0.5) = −0.434906 kilopascals (b) The actual change in pressure is P(20.5) − P(20) = −0.418274 kilopascals The percentage error in the Linear Approximation is −0.434906 − (−0.418274) × 100% ≈ 3.98% −0.418274 43. Newton’s Law of Gravitation shows that if a person weighs w pounds on the surface of the earth, then his or her weight at distance x from the center of the earth is W(x) = wR2 x2 (for x ≥ R) where R = 3960 miles is the radius of the earth (Figure 10). (a) Show that the weight lost at altitude h miles above the earth’s surface is approximately ΔW ≈ −(0.0005w)h. Hint: Use the Linear Approximation with dx = h. (b) Estimate the weight lost by a 200-lb football player ﬂying in a jet at an altitude of 7 miles. h 960 3 FIGURE 10 The distance to the center of the earth is 3960 + h miles. SOLUTION (a) Using the Linear Approximation ΔW ≈ W ' (R)Δx = − 2wR2 2wh h=− ≈ −0.0005wh R R3 (b) Substitute w = 200 and h = 7 into the result from part (a) to obtain ΔW ≈ −0.0005(200)(7) = −0.7 pounds 45. (a) (b) (c) A stone tossed vertically into the air with initial velocity v cm/s reaches a maximum height of h = v2 /1960 cm. Estimate Δh if v = 700 cm/s and Δv = 1 cm/s. Estimate Δh if v = 1000 cm/s and Δv = 1 cm/s. In general, does a 1-cm/s increase in v lead to a greater change in h at low or high initial velocities? Explain. SOLUTION A stone tossed vertically with initial velocity v cm/s attains a maximum height of h(v) = v2 /1960 cm. Thus, h' (v) = v/980. 1 (a) If v = 700 and Δv = 1, then Δh ≈ h' (v)Δv = 980 (700)(1) ≈ 0.71 cm. 1 ' (b) If v = 1000 and Δv = 1, then Δh ≈ h (v)Δv = 980 (1000)(1) = 1.02 cm. (c) A 1 centimeter per second increase in initial velocity v increases the maximum height by approximately v/980 cm. Accordingly, there is a bigger effect at higher velocities. In Exercises 47 and 48, use the following fact derived from Newton’s Laws: An object released at an angle θ with initial velocity acc v ft/s travels a horizontal distance s= 1 2 v sin 2θ ft (Figure 11) 32 6 CHAPTER 4 APPLICATIONS OF THE DERIVATIVE y s x FIGURE 11 Trajectory of an object released at an angle θ. 47. A player located 18.1 ft from the basket launches a successful jump shot from a height of 10 ft (level with the rim of the basket), at an angle θ = 34◦ and initial velocity v = 25 ft/s. (a) Show that Δs ≈ 0.255Δθ ft for a small change of Δθ. (b) Is it likely that the shot would have been successful if the angle had been off by 2◦ ? (c) Estimate Δs if θ = 34◦ , v = 25 ft/s, and Δv = 2. Using Newton’s laws and the given initial velocity of v = 25 ft/s, the shot travels s = sin 2t ft, where t is in radians. (a) If θ = 34◦ (i.e., t = 17 π), then 90 ( ) ( ) 625 17 625 17 π cos π Δt = cos π Δθ · ≈ 0.255Δθ Δs ≈ s' (t)Δt = 16 45 16 45 180 SOLUTION 625 32 1 2 v 32 sin 2t = (b) If Δθ = 2◦ , this gives Δs ≈ 0.51 ft, in which case the shot would not have been successful, having been off half a foot. π, the shot travels (c) Using Newton’s laws and the ﬁxed angle of θ = 34◦ = 17 90 s= 1 2 17 v sin π 32 45 With v = 25 ft/s and Δv = 2 ft/s, we ﬁnd Δs ≈ s' (v)Δv = 1 17π (25) sin · 2 = 2.897 ft 16 45 49. The radius of a spherical ball is measured at r = 25 cm. Estimate the maximum error in the volume and surface area if r is accurate to within 0.5 cm. The volume and surface area of the sphere are given by V = 43 πr3 and S = 4πr2 , respectively. If r = 25 and Δr = ±0.5, then SOLUTION ΔV ≈ V ' (25)Δr = 4π(25)2 (0.5) ≈ 3927 cm3 and ΔS ≈ S ' (25)Δr = 8π(25)(0.5) ≈ 314.2 cm2 51. The volume (in liters) and pressure P (in atmospheres) of a certain gas satisfy PV = 24. A measurement yields V = 4 = computation. with a possible error of ±0.3 L. Compute P and estimate the maximum error in this Given PV = 24 and V = 4, it follows that P = 6 atmospheres. Solving PV = 24 for P yields P = 24V −1 . Thus, P = −24V −2 and SOLUTION ' ΔP ≈ P' (4)ΔV = −24(4)−2 (±0.3) = ±0.45 atmospheres 53. Approximate f (2) if the linearization of f (x) at a = 2 is L(x) = 2x + 4? f (2) ≈ L(2) = 2(2) + 4 = 8. √ √ 55. Estimate 16.2 using the linearization L(x) of f (x) = x at a = 16. Plot f and L on the same set of axes and determine whether the estimate is too large or too small. SOLUTION SOLUTION Let f (x) = x1/2 , a = 16, and Δx = 0.2. Then f ' (x) = 21 x−1/2 and f ' (a) = f ' (16) = 81 . The linearization to f (x) is L(x) = f ' (a)(x − a) + f (a) = 1 1 (x − 16) + 4 = x + 2 8 8 √ Thus, we have 16.2 ≈ L(16.2) = 4.025. Graphs of f (x) and L(x) are shown below. Because the graph of L(x) lies above the graph of f (x), we expect that the estimate from the Linear Approximation is too large. Linear Approximation and Applications S E C T I O N 4.1 7 y 5 4 3 2 1 L(x) 0 5 f (x) 10 15 20 25 x In Exercises 57 65, approximate using linearization and use a calculator to compute the percentage error. 1 57. √ 17 SOLUTION 1 Let f (x) = x−1/2 , a = 16, and Δx = 1. Then f ' (x) = − 21 x−3/2 , f ' (a) = f ' (16) = − 128 and the linearization to f (x) is L(x) = f ' (a)(x − a) + f (a) = − Thus, we have √1 17 ≈ L(17) ≈ 0.24219. The percentage error in this estimate is     59. 1 1 1 3 (x − 16) + = − x+ 128 4 128 8 √1 17  − 0.24219   × 100% ≈ 0.14%  √1 17 1 (10.03)2 SOLUTION Let f (x) = x−2 , a = 10 and Δx = 0.03. Then f ' (x) = −2x−3 , f ' (a) = f ' (10) = −0.002 and the linearization to f (x) is L(x) = f ' (a)(x − a) + f (a) = −0.002(x − 10) + 0.01 = −0.002x + 0.03 Thus, we have 1 ≈ L(10.03) = −0.002(10.03) + 0.03 = 0.00994 (10.03)2 The percentage error in this estimate is  1   (10.03)2 − 0.00994    × 100% ≈ 0.0027% 1   (10.03)2 61. (64.1)1/3 SOLUTION Let f (x) = x1/3 , a = 64, and Δx = 0.1. Then f ' (x) = 31 x−2/3 , f ' (a) = f ' (64) = f (x) is 1 1 8 (x − 64) + 4 = x+ 48 48 3 L(x) = f ' (a)(x − a) + f (a) = Thus, we have (64.1)1/3 ≈ L(64.1) ≈ 4.002083. The percentage error in this estimate is    (64.1)1/3 − 4.002083    × 100% ≈ 0.000019% (64.1)1/3 63. cos−1 (0.52) SOLUTION Let f (x) = cos−1 x and a = 0.5. Then √ 2 3 f (x) = − √ , f (a) = f (0) = − 3 1 − x2 1 ' ' ' and the linearization to f (x) is √ 2 3 π (x − 0.5) + L(x) = f (a)(x − a) + f (a) = − 3 3 ' Thus, we have cos−1 (0.52) ≈ L(0.02) = 1.024104. The percentage error in this estimate is  −1   cos (0.52) − 1.024104    × 100% ≈ 0.015% cos−1 (0.52) 1 48 and the linearization to 8 CHAPTER 4 APPLICATIONS OF THE DERIVATIVE 65. e−0.012 SOLUTION Let f (x) = e x and a = 0. Then f ' (x) = e x , f ' (a) = f ' (0) = 1 and the linearization to f (x) is L(x) = f ' (a)(x − a) + f (a) = 1(x − 0) + 1 = x + 1 Thus, we have e−0.012 ≈ L(−0.012) = 1 − 0.012 = 0.988. The percentage error in this estimate is  −0.012   e − 0.988    × 100% ≈ 0.0073% e−0.012 √ √ 67. Show that the Linear Approximation to f (x) = x at3 x2 = 9 yields the estimate 9 + h − 3 ≈ 16 h. Set K = 0.01 and '' show that | f (x)| ≤ K for x ≥ 9. Then verify numerically that the error E satisﬁes Eq. (5) for h = 10−n , for 1 ≤ n ≤ 4. √ SOLUTION Let f (x) = x. Then f (9) = 3, f ' (x) = 12 x−1/2 and f ' (9) = 16 . Therefore, by the Linear Approximation, f (9 + h) − f (9) = √ 9+h−3≈ 1 h 6 Moreover, f '' (x) = − 41 x−3/2 , so | f '' (x)| = 41 x−3/2 . Because this is a decreasing function, it follows that for x ≥ 9, K = max | f '' (x)| ≤ | f '' (9)| = 1 < 0.01 108 From the following table, we see that for h = 10−n , 1 ≤ n ≤ 4, E ≤ 12 Kh2 . √ 1 E = | 9 + h − 3 − 61 h| h Kh2 2 10−1 10−2 10−3 10−4 4.604 × 10−5 4.627 × 10−7 4.629 × 10−9 4.627 × 10−11 5.00 × 10−5 5.00 × 10−7 5.00 × 10−9 5.00 × 10−11 Linear Ap and Challeng to f (x) =es tan Further Insights 4 69. Compute dy/dx at the point P = (2, 1) on the curve y3 + 3xy = 7 and show that the linearization at P is L(x) = − 31 x + 35 . Use L(x) to estimate the y-coordinate of the point on the curve where x = 2.1. SOLUTION Differentiating both sides of the equation y3 + 3xy = 7 with respect to x yields 3y2 dy dy + 3x + 3y = 0 dx dx so dy y =− 2 dx y +x Thus,  dy  1 1  =− 2 =− dx (2,1) 1 +2 3 and the linearization at P = (2, 1) is 1 1 5 L(x) = 1 − (x − 2) = − x + 3 3 3 Finally, when x = 2.1, we estimate that the y-coordinate of the point on the curve is 1 5 y ≈ L(2.1) = − (2.1) + = 0.967 3 3 71. Apply the method of Exercise 69 to P = (−1, 2) on y4 + 7xy = 2 to estimate the solution of y4 − 7.7y = 2 near y = 2. SOLUTION Differentiating both sides of the equation y4 + 7xy = 2 with respect to x yields 4y3 dy dy + 7x + 7y = 0 dx dx so dy 7y =− 3 dx 4y + 7x S E C T I O N 4.2 Extreme Values 9 Thus,  dy  7(2) 14  =− =− 4(2)3 + 7(−1) 25 dx (−1,2) and the linearization at P = (−1, 2) is L(x) = 2 − 14 14 36 (x + 1) = − x + 25 25 25 Finally, the equation y4 − 7.7y = 2 corresponds to x = −1.1, so we estimate the solution of this equation near y = 2 is y ≈ L(−1.1) = − 14 36 (−1.1) + = 2.056 25 25 73. Let Δ f = f (5 + h) − f (5), where f (x) = x2 . Verify directly that E = |Δ f − f ' (5)h| satisﬁes (5) with K = 2. SOLUTION Let f (x) = x2 . Then Δ f = f (5 + h) − f (5) = (5 + h)2 − 52 = h2 + 10h and E = |Δ f − f ' (5)h| = |h2 + 10h − 10h| = h2 ≤ 2h2 2 4.2 Extreme Values Preliminary Questions 1. What is the deﬁnition of a critical point? A critical point is a value of the independent variable x in the domain of a function f at which either f ' (x) = 0 or f (x) does not exist. SOLUTION ' In Questions 2 and 3, which is the correct conclusion, (a) or (b)? 2. If f is not continuous on [0, 1], then (a) f has no extreme values on [0, 1]. (b) f might not have any extreme values on [0, 1]. The correct response is (b): f might not have any extreme values on [0, 1]. Although [0, 1] is closed, because f is not continuous, the function is not guaranteed to have any extreme values on [0, 1]. SOLUTION 3. If f is continuous but has no critical points in [0, 1], then (a) f has no min or max on [0, 1]. (b) Either f (0) or f (1) is the minimum value on [0, 1]. The correct response is (b): either f (0) or f (1) is the minimum value on [0, 1]. Remember that extreme values occur either at critical points or endpoints. If a continuous function on a closed interval has no critical points, the extreme values must occur at the endpoints. SOLUTION 4. For each statement, indicate whether it is true or false. If false, correct the statement or explain why it is false. (a) If f ' (c) = 0, then f (c) is either a local minimum or a local maximum. (b) If f (c) is the absolute maximum of f on an interval I, then f ' (c) = 0. (c) If f is differentiable and f (c) is a local minimum of f , then f ' (c) = 0. (d) If there is one local minimum of f on an interval I, then it is the absolute minimum on I. SOLUTION (a) This statement is false. The condition f ' (c) = 0 is not sufﬁcient to guarantee that f (c) is either a local minimum or a local maximum. For example, consider the function f (x) = x3 . Then f ' (0) = 0, but f (0) is neither a local minimum nor a local maximum. (b) This statement is false. If f (c) is the absolute maximum of f on an interval I, it could be that f ' (c) does not exist or that f ' (c) is any real number if c happens to be an endpoint of the interval I. For example, consider the function f (x) = x2 on the interval [0, 2]. The absolute maximum value of f on [0, 2] is f (2) = 4, but f ' (2) = 4 * 0. (c) This statement is true. (d) This statement is false. If there is one local minimum of f on an interval I, then the absolute minimum might occur at an endpoint of the interval. For example, consider the function f (x) = x3 − x2 on the interval [−1, 1]. The function f has only one local minimum on this interval, at x = 2/3, but the absolute m...
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