3 74 dy dx 3x 2 x 27 3x 2 x dy dx 28 5x6

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Unformatted text preview: f (x) = lim ∆x→0 ∆x[(x + ∆x)2/3 + (x + ∆x)1/3 x1/3 + x2/3 ] f (x) = lim = lim ∆x→0 21. 1 1 = 2/3 (x + ∆x)2/3 + (x + ∆x)1/3 x1/3 + x2/3 3x f (t + h) − f (t) [4(t + h)2 + (t + h)] − [4t2 + t] = lim h→0 h→0 h h f (t) = lim 4t2 + 8th + 4h2 + t + h − 4t2 − t h→0 h = lim 8th + 4h2 + h = lim (8t + 4h + 1) = 8t + 1 h→0 h→0 h = lim 22. 4 4 4 π (r + h)3 − πr3 π (r3 + 3r2 h + 3rh2 + h3 − r3 ) dV 3 3 3 = lim = lim h→0 h→0 dr h h 4 = lim π (3r2 + 3rh + h2 ) = 4πr2 h→0 3 23. (a) D (b) F (c) B (d) C (e) A (f ) E 24. Any function of the form f (x) = x + k has slope 1, and thus the derivative must be equal to 1 everywhere. y 3 1 -2 x -2 25. (a) (b) y (c) y x y x 1 2 x -1 26. (a) (b) y x 27. (a) f (x) = x2 and a = 3 (c) y x (b) f (x) = √ y x x and a = 1 71 28. 29. 30. Chapter 3 (a) (b) f (x) = x7 and a = 1 f (x) = cos x and a = π [4(x + h)2 + 1] − [4x2 + 1] 4x2 + 8xh + 4h2 + 1 − 4x2 − 1 dy = lim = lim = lim (8x + 4h) = 8x h→0 h→0 dx h→0 h h dy = 8(1) = 8 dx x=1 dy = lim dx h→0 5 +1 − x+h h = limh→0 dy dx x=−2 5 +1 x 5x − 5(x + h) 5 5 − x(x + h) = lim x + h x = lim h→0 h→0 h h 5x − 5x − 5h −5 5 = lim =− 2 h→0 x(x + h) hx(x + h) x =− 5 5 =− (−2)2 4 31. y = −2x + 1 32. 1.5 5 -2 2 0 2.5 0 -3 33. (b) h (f (1 + h) − f (1))/h 0.5 0.1 0.01 0.001 0.0001 0.00001 1.6569 1.4355 1.3911 1.3868 1.3863 1.3863 34. (b) h (f (1 + h) − f (1))/h 0.5 0.1 0.01 0.001 0.0001 0.00001 0.50489 0.67060 0.70356 0.70675 0.70707 0.70710 35. (a) dollars/ft (b) (c) As you go deeper the price per foot may increase dramatically, so f (x) is roughly the price per additional foot. If each additional foot costs extra money (this is to be expected) then f (x) remains positive. (d) From the approximation 1000 = f (300) ≈ so the extra foot will cost around $1000. 36. f (301) − f (300) we see that f (301) ≈ f (300) + 1000, 301 − 300 (a) gallons/dollar (b) The increase in the amount of paint that would be sold for one extra dollar. (c) It should be negative since an increase in the price of paint would decrease the amount of paint sold. f (11) − f (10) we see that f (11) ≈ f (10) − 100, so an increase of one From −100 = f (10) ≈ 11 − 10 dollar would decrease the amount of paint sold by around 100 gallons. (d) 37. (a) F ≈ 200 lb, dF/dθ ≈ 60 lb/rad (b) µ = (dF/dθ)/F ≈ 60/200 = 0.3 38. (a) dN/dt ≈ 34 million/year; in 1950 the world population was increasing at the rate of about 34 million per year. (b) dN/dt 34 ≈ ≈ 0.014 = 1.4 %/year N 2490 Exercise Set 3.2 72 41. (a) T ≈ 120◦ F, dT /dt ≈ −4.5◦ F/min (b) 39. k = (dT /dt)/(T − T0 ) ≈ (−4.5)/(120 − 75) = −0.1 lim f (x) = lim √ 3 x = 0 = f (0), so f is continuous at x = 0. √ 3 f (0 + h) − f (0) h−0 1 = lim = lim 2/3 = +∞, so lim h→0 h→0 h→0 h h h f (0) does not exist. x→0 y x→0 2 -2 42. lim f (x) = lim(x − 2)2/3 = 0 = f (2) so f is continuous at x = 2. x→2 x→2 x 2 y f (2 + h) − f (2) h −0 1 = lim = lim 1/3 which does not exist h→0 h→0 h h h so f (2) does not exist. 2/3 5 lim h→0 x 2 43. lim f (x) = lim+ f (x) = f (1), so f is continuous at x = 1. x→1− x→1 y f (1 + h) − f (1) [(1 + h) + 1] − 2 = lim− = lim− (2 + h) = 2; h→0 h→0 h→0 h h f (1 + h) − f (1) 2(1 + h) − 2 = lim+ = lim+ 2 = 2, so f (1) = 2. lim h→0+ h→0 h→0 h h 2 5 lim− -3 44. lim f (x) = lim+ f (x) = f (1) so f is continuous at x = 1. x→1− x y x→1 f (1 + h) − f (1) [(1 + h)2 + 2] − 3 = lim− = lim− (2 + h) = 2; h→0 h→0 h→0 h h f (1 + h) − f (1) [(1 + h) + 2] − 3 = lim+ = lim+ 1 = 1, so f (1) lim h→0+ h→0 h→0 h h does not exist. 5 lim− -3 45. 3 3 x f is continuous at x = 1 because it is differentiable there, thus lim f (1 + h) = f (1) and so f (1) = 0 h→0 f (1 + h) f (1 + h) − f (1) f (1 + h) because lim exists; f (1) = lim = lim = 5. h→0 h→0 h→0 h h h f (x + h) − f (x) , but (with h f (h) + 5xh y = h) f (x + h) = f (x) + f (h) + 5xh so f (x + h) − f (x) = f (h) + 5xh and f (x) = lim = h→0 h f (h) lim ( + 5x) = 3 + 5x. h→0 h 46. Let x = y = 0 to get f (0) = f (0) + f (0) + 0 so f (0) = 0. f (x) = lim 47. f (x) = limh→0 h→0 f (x + h) − f (x) f (x)f (h) − f (x) f (x)[f (h) − 1] f (h) − f (0) = lim = lim = f (x) lim h→0 h→0 h→0 h h h h = f (x)f (0) = f (x) 73 Chapter 3 EXERCISE SET 3.3 1. 28x6 2. 5. 0 6. 9. 3ax2 + 2bx + c 13. f (x) = (3x2 + 6) 3. 24x7 + 2 4. 2x3 √ 1 7. − (7x6 + 2) 3 8. 2 x 5 1 a 2 2x + √ 11. 24x−9 + 1/ x 1 b −3x−4 − 7x−8 15. 10. −36x11 1 d 2x − dx 4 = 18x2 − 3 x + 12 2 14. + 2x − 1 4 5 12. −42x−7 − √ 2x 1 1 √−2 2x x 1 d (3x2 + 6) = (3x2 + 6)(2) + 2x − dx 4 (6x) d d (7 + x5 ) + (7 + x5 ) (2 − x − 3x3 ) = (2 − x − 3x3 )(5x4 ) + (7 + x5 )(−1 − 9x2 ) dx dx = −24x7 − 6x5 + 10x4 − 63x2 − 7 16. f (x) = (2 − x − 3x3 ) d d (2x−3 + x−4 ) + (2x−3 + x−4 ) (x3 + 7x2 − 8) dx dx = (x3 + 7x2 − 8)(−6x−4 − 4x−5 ) + (2x−3 + x−4 )(3x2 + 14x) = −15x−2 − 14x−3 + 48x−4 + 32x−5 17. f (x) = (x3 + 7x2 ...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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