So the tangent line is horizontal at x 1 115 2 2 2 3x

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Unformatted text preview: part (a), (1.02)3 ≈ 1 + 3(0.02) = 1.06. From part (b), (1.02)3 ≈ 1 + 3(0.02) = 1.06. (a) f (x) ≈ f (2) + f (2)(x − 2) = 1/2 + (−1/22 )(x − 2) = (1/2) − (1/4)(x − 2) (b) f (2 + ∆x) ≈ f (2) + f (2)∆x = 1/2 − (1/4)∆x (c) From part (a), 1/2.05 ≈ 0.5 − 0.25(0.05) = 0.4875, and from part (b), 1/2.05 ≈ 0.5 − 0.25(0.05) = 0.4875. 12. 13. 14. 15. (a) √ (b) dy = 5 sec2 x dx x 1−x− √ 2 1−x 2 − 3x dx = √ dx 2 1−x (b) f (1 + ∆x) ≈ f (1) + f (1)∆x = 1 + 3∆x √ f (x) ≈ f (x0 ) + f √ 0 )(x − x0 ) = 1 + (1/(2 1)(x − 0) = 1 + (1/2)x, so with x0 = 0 and (x x = −0.1, we have 0.9 = f (−0.1) ≈ 1 + (1/2)(−0.1) = 1 − 0.05 = 0.95. With x = 0.1 we have √ 1.1 = f (0.1) ≈ 1 + (1/2)(0.1) = 1.05. (b) y ∆y dy ∆y dy –0.1 16. (a) (b) 0.1 x f (x) ≈ f (x0 ) + f (x0 )(x − x0 ) = 1/2 − 1/(2 · 43/2 ) (x − 4) = 1/2 − (x − 4)/16, so with x0 = 4 √ and x = 3.9 we have 1/ 3.9 = f (3.9) ≈ 0.5 − (−0.1)/16 = 0.50625. If x0 = 4 and x = 4.1 then √ 1/ 4.1 = f (4.1) ≈ 0.5 − (0.1)/16 = 0.49375 y ∆y dy dy ∆y x 3.9 4 4.1 17. f (x) = (1 + x)15 and x0 = 0. Thus (1 + x)15 ≈ f (x0 ) + f (x0 )(x − x0 ) = 1 + 15(1)14 (x − 0) = 1 + 15x. 18. f (x) = √ 1 1 1 and x0 = 0, so √ ≈ f (x0 ) + f (x0 )(x − x0 ) = 1 + (x − 0) = 1 + x/2 2(1 − 0)3/2 1−x 1−x 91 Chapter 3 −1 1 ≈1+ (x − 0) = 1 − x 1+x (1 + 0)2 19. tan x ≈ tan(0) + sec2 (0)(x − 0) = x 21. x4 ≈ (1)4 + 4(1)3 (x − 1). Set ∆x = x − 1; then x = ∆ + 1 and (1 + ∆x)4 = 1 + 4∆x. 22. 23. √ x≈ √ 20. √ 1 1 + √ (x − 1), and x = 1 + ∆x, so 1 + ∆x ≈ 1 + ∆x/2 21 1 1 11 1 1 = − = − ∆x (x − 1), and 2 + x = 3 + ∆x, so 2+x 2 + 1 (2 + 1)2 3 + ∆x 39 24. (4 + x)3 = (4 + 1)3 + 3(4 + 1)2 (x − 1) so, with 4 + x = 5 + ∆x we get (5 + ∆x)3 = 125 + 75∆x 25. The local linear approximation sin x ≈ x gives sin 1◦ = sin(π/180) ≈ π/180 = 0.0174533 and a calculator gives sin 1◦ = 0.0174524. The relative error | sin(π/180) − (π/180)|/(sin π/180) = 0.000051 is very small, so for such a small value of x the approximation is very good. √ (b) Use x0 = 45◦ (this assumes you know, or can approximate, 2/2). (a) (c) 26. π 44π 44π 45π π radians, and 45◦ = = radians. With x = and x0 = we obtain 180 180 4 4 √ 180 √ 2 2 −π π π 44π π 44π ≈ sin + cos − = + = 0.694765. With a sin 44◦ = sin 180 4 4 180 4 2 2 180 calculator, sin 44◦ = 0.694658. 44◦ = tan x ≈ tan 0 + sec2 0(x − 0) = x, so tan 2◦ = tan(2π/180) ≈ 2π/180 = 0.034907, and with a calculator tan 2◦ = 0.034921 √ (b) use x0 = π/3 because we know tan 60◦ = tan(π/3) = 3 (a) (c) 60π π 61π π 61π π and x = we have tan 61◦ = tan ≈ tan + sec2 with x0 = = 3 180 180 180 3 3 √ π ◦ = 1.8019, and with a calculator tan 61 = 1.8040 3+4 180 61π π − 180 3 27. f (x) = x4 , f (x) = 4x3 , x0 = 3, ∆x = 0.02; (3.02)4 ≈ 34 + (108)(0.02) = 81 + 2.16 = 83.16 28. f (x) = x3 , f (x) = 3x2 , x0 = 2, ∆x = −0.03; (1.97)3 ≈ 23 + (12)(−0.03) = 8 − 0.36 = 7.64 29. f (x) = 30. f (x) = 31. f (x) = 32. f (x) = 33. f (x) = sin x, f (x) = cos x, x0 = 0, ∆x = 0.1; sin 0.1 ≈ sin 0 + (cos 0)(0.1) = 0.1 √ √ √ 1 1 1 = 8.0625 x, f (x) = √ , x0 = 64, ∆x = 1; 65 ≈ 64 + (1) = 8 + 16 16 2x √ √ √ 1 1 x, f (x) = √ , x0 = 25, ∆x = −1; 24 ≈ 25 + (−1) = 5 − 0.1 = 4.9 10 2x √ √ √ 1 1 x, f (x) = √ , x0 = 81, ∆x = −0.1; 80.9 ≈ 81 + (−0.1) ≈ 8.9944 18 2x √ √ √ 1 1 x, f (x) = √ , x0 = 36, ∆x = 0.03; 36.03 ≈ 36 + (0.03) = 6 + 0.0025 = 6.0025 12 2x 34. f (x) = tan x, f (x) = sec2 x, x0 = 0, ∆x = 0.2; tan 0.2 ≈ tan 0 + (sec2 0)(0.2) = 0.2 35. f (x) = cos x, f (x) = − sin x, x0 = π/6, ∆x = π/180; √ π π 1 3 ◦ ◦ = − ≈ 0.8573 cos 31 ≈ cos 30 + − 2 180 2 360 = Exercise Set 3.6 92 Let f (x) = (1 + x)k and x0 = 0. Then (1 + x)k ≈ 1k + k (1)k−1 (x − 0) = 1 + kx. Set k = 37 and x = 0.001 to obtain (1.001)37 ≈ 1.037. With a calculator (1.001)37 = 1.03767. (c) 37. (a) (b) 36. The approximation is (1.1)37 ≈ 1 + 37(0.1) = 4.7, and the calculator value is 34.004. The error is due to the relative largeness of f (1)∆x = 37(0.1) = 3.7. √ f (x) = x + 3 and x0 = 0, so √ √ √ 1 1 x + 3 ≈ 3 + √ (x − 0) = 3 + √ x, and 23 23 √ 1 3 + √ x < 0.1 if |x| < 1.692. f (x) − 23 0 -2 | 38. 1 1 1 1 1 1 (x − 0) = + x, so √ ≈√ + 3 54 9−x 9−x 9 2(9 − 0)3/2 1 1 + x < 0.1 if |x| < 5.5114 and f (x) − 3 54 f (x) = √ 2 -0.1 f (x) – ( 3 + 2 13 x)| 0.06 -6 6 | f (x) – ( 39. tan x ≈ tan 0 + (sec2 0)(x − 0) = x, and | tan x − x| < 0.1 if |x| < 0.6316 0 1 3 + 1 54 )| x 0.06 -0.8 0.8 0 | f (x) – x | 40. 1 1 −5(2) ≈ + (x − 0) = 1 − 10x, and 5 5 (1 + 2x) (1 + 2 · 0) (1 + 2 · 0)6 |f (x) − (1 − 10x)| < 0.1 if |x| < 0.0372 0.06 -0.8 0.8 0 | f (x) – x | 41. 3 3 dx, x = 2, dx = 0.03; ∆y ≈ dy = (0.03) = 0.0225 dy = √ 4 2 3x − 2 42. dy = √ x dx, x = 1, dx = −0.03; ∆y ≈ dy = (1/3)(−0.03) = −0.01 +8 x2 93 Chapter 3 43. dy = 44. dy = 45. (a) (b) 1 − x2 dx, x = 2, dx = −0.04; ∆y ≈ dy...
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