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# Ism et chapter_3

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### Ism et chapter_3

1. 1. −1/2 1 f (x0 ) = f (0) = (2 · 0 + 9) = 3 So, L(x) = f (x0 ) + f (x0 ) (x − x0 ) 1 = 3 + (x − 0) 3Chapter 3 =3+ x 1 3 (b) Using √ the approximation L(x) to esti- √ mate 8.8, we get 8.8 = f (−0.1) ≈Applications of 1 L(−0.1) = 3 + (−0.1) = 3 − 0.033 = 3 2.967Diﬀerentiation 4. (a) f (x) = 2 , x0 = 1 x f (x0 ) = f (1) = 2 2 f (x) = − 2 and so f (1) = −2 x3.1 Linear Approximations The linear approximation is L(x) = 2 + (−2) (x − 1) and Newtons Method √ (b) Using the approximation L(x) to estimate 1. (a) f (x) = x, x0 = 1 2 2 √ , we get = f (0.99) ≈ L(0.99) = f (x0 ) = f (1) = 1 = 1 0.99 0.99 1 2 + (−2)(0.99 − 1) = 2.02 f (x) = x−1/2 2 1 5. (a) f (x) = sin 3x, x0 = 0 f (x0 ) = f (1) = f (x0 ) = sin(3 · 0) = 0 2 So, f (x) = 3 cos 3x L(x) = f (x0 ) + f (x0 ) (x − x0 ) f (x0 ) = f (0) = 3 cos(3 · 0) = 3 1 So, = 1 + (x − 1) L(x) = f (x0 ) + f (x0 ) (x − x0 ) 2 1 1 = 0 + 3 (x − 0) = + x 2 2 = 3x (b) √ Using the approximation L(x) to estimate √ (b) Using the approximation L(x) to esti- 1.2, we get 1.2 = f (1.2) ≈ L(1.2) = 1 1 mate sin(0.3), we get sin(0.3) = f (0.1) ≈ + (1.2) = 1.1 L(0.1) = 3(0.1) = 0.3 2 2 2. (a) f (x0 ) = f (0) = 1 and 6. (a) f (x) = sin x, x0 = π 1 −2/3 f (x0 ) = sin π = 0 f (x) = (x + 1) 3 f (x) = cos x 1 So, f (0) = f (x0 ) = f (π) = cos π = −1 3 The linear approximation is, The Linear approximation is, 1 1 L(x) = f (x0 ) + f (x0 ) (x − x0 ) L(x) = 1 + (x − 0) = 1 + x 3 3 = 0 + (−1) (x − π) = π − x (b) Using the approximation L(x) to estimate √ √ 3 3 (b) Using the approximation L(x) to esti- 1.2, we get 1.2 = f (0.2) ≈ L(0.2) = mate sin(3.0), we get sin(3.0) = f (3.0) ≈ 1 1 + (0.2) = 1.066 L(3.0) = π − 3.0 3 √ √ 4 3. (a) f (x) = 2x + 9, √0 = 0 x 7. (a) f (x) = √ 16 + x, x0 = 0 4 f (x0 ) = f (0) = 2 · 0+9 = 3 f (0) = 16 + 0 = 2 1 1 f (x) = (2x + 9) −1/2 ·2 f (x) = (16 + x)−3/4 2 4 1 1 = (2x + 9) −1/2 f (0) = (16 + 0)−3/4 = 4 32 150
2. 2. 3.1. LINEAR APPROXIMATIONS AND NEWTONS METHOD 151 L(x) = f (0) + f (0)(x − 0) 36 L(72) = 120 + (72 − 80) 1 20 =2+ x = 120 + 1.8(−8) 32 1 = 105.6 cans = 2 + (0.04) = 2.00125 32 168 − 120 1 (b) L(x) = f (100) + (x − 100) (b) L(0.08) = 2 + (0.08) = 2.0025 100 − 80 32 48 1 L(94) = 168 − (94 − 100) (c) L(0.16) = 2 + (0.16) = 2.005 20 32 = 168 − 2.4(−6) 8. (a) f (x) = sin x, x0 = 0 = 182.4 cans f (0) = 0 f (x) = cos x 142 − 128 f (0) = cos 0 = 1 11. (a) L(x) = f (200) + (x − 200) 220 − 200 L(x) = f (0) + f (0) (x − 0) 14 L(208) = 128 + (208 − 200) =0+1·x 20 L(0.1) = 0.1 = 128 + 0.7(8) = 133.6 (b) f (x) = sin x, x0 = π 142 − 136 √ 3 π 3 (b) L(x) = f (240) + (x − 240) f = 220 − 240 3 2 6 π π 1 L(232) = 136 − (232 − 240) f = cos = 20 3 3 2 = 136 − 0.3(−8) = 138.4 π π π L(x) = f +f x− √ 3 3 3 3 1 π 14 − 8 L(1) = + 1− ≈ 0.842 12. (a) L(x) = f (10) + (x − 10) 2 2 3 10 − 5 2π 6 (c) f (x) = sin x, x0 = L(8) = 14 + (−2) = 11.6 √ 3 5 2π 3 14 − 8 f = (b) L(x) = f (10) + (x − 10) 3 2 10 − 5 2π 2π 1 6 f = cos =− L(12) = 14 + (2) = 16.4 3 3 2 5 2π 2π 2π L(x) = f +f x− 13. f (x) = x3 + 3x2 − 1 = 0, x0 = 1 3 3 3 √ f (x) = 3x2 + 6x 3 1 2π = − x− 2√ 2 3 f (x0 ) 9 3 1 9 2π (a) x1 = x0 − L = − − ≈ 0.788 f (x0 ) 4 2 2 4 3 13 + 3 · 12 − 1 18 − 14 =1− 9. (a) L(x) = f (20) + (x − 20) 3 · 12 + 6 · 1 20 − 30 3 2 4 =1− = L(24) ≈ 18 − (24 − 20) 9 3 10 f (x1 ) = 18 − 0.4(4) x2 = x1 − f (x1 ) = 16.4 games 2 3 2 2 2 3 +3 3 −1 14 − 12 = − 3 2 2 2 (b) L(x) = f (40) + (x − 40) 3 3 +6 3 30 − 40 2 79 f (36) ≈ 12 − (36 − 40) = ≈ 0.5486 10 144 = 12 − 0.2(−4) (b) 0.53209 = 12.8 games 120 − 84 14. f (x) = x3 + 4x2 − x − 1, x0 = −110. (a) L(x) = f (80) + (x − 80) f (x) = 3x2 + 8x − 1 80 − 60
3. 3. 152 CHAPTER 3. APPLICATIONS OF DIFFERENTIATION f (x0 ) 30 (a) x1 = x0 − f (x0 ) 3 1 = −1 − =− 20 −6 2 f (x1 ) y x2 = x1 − f (x1 ) 10 1 0.375 =− − = −0.4117647 2 −4.25 0 −5.0 −2.5 0.0 2.5 5.0 (b) The root is x ≈ −0.4064206546. x −10 Start with x0 = −5 to ﬁnd the root near −5: 15. f (x) = x4 − 3x2 + 1 = 0, x0 = 1 x1 = −4.718750, x2 = −4.686202, f (x) = 4x3 − 6x x3 = −4.6857796, x4 = −4.6857795 f (x0 ) (a) x1 = x0 − 18. From the graph, we see two roots: f (x0 ) 14 − 3 · 12 + 1 1 =1− = 15 4 · 13 − 6 · 1 2 10 f (x1 ) x2 = x1 − f (x1 ) 5 -1 0 1 2 3 4 1 4 1 2 0 1 2 −3 2 +1 = − 2 1 3 1 -5 4 2 −6 2 -10 5 = 8 -15 -20 (b) 0.61803 16. f (x) = x4 − 3x2 + 1, x0 = −1 f (xi ) Use xi+1 = xi − with f (x) = 4x3 − 6x f (xi ) f (x) = x4 − 4x3 + x2 − 1, and f (x) = 4x3 − 12x2 + 2x f (x0 ) Start with x0 = 4 to ﬁnd the root below 4: (a) x1 = x0 − x1 = 3.791666667, x2 = 3.753630030, x3 = f (x0 ) −1 1 3.7524339, x4 = 3.752432297 = −1 − =− Start with x = −1 to ﬁnd the root just above 2 2 f (x1 ) −1: x2 = x1 − x1 = −0.7222222222, f (x1 ) x2 = −0.5810217936, 1 0.3125 x3 = −0.5416512863, =− − = −0.625 2 2.5 x4 = −0.5387668233, x5 = −0.5387519962 (b) The root is x ≈ −0.6180339887. f (xi ) f (xi ) 17. Use xi+1 = xi − with 19. Use xi+1 = xi − with f (xi ) f (xi ) f (x) = x3 + 4x2 − 3x + 1, and f (x) = x5 + 3x3 + x − 1, and f (x) = 3x2 + 8x − 3 f (x) = 5x4 + 9x2 + 1