# A b y c y x y x 1 2 x 1 26 a b y x 27 a f

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Unformatted text preview: wing down because the slope decreases as t increases from t1 to t2 . s 12. t0 t t1 13. It is a straight line with slope equal to the velocity. 14. (a) decreasing (slope of tangent line decreases with increasing time) (b) increasing (slope of tangent line increases with increasing time) (c) increasing (slope of tangent line increases with increasing time) (d) decreasing (slope of tangent line decreases with increasing time) (a) 72◦ F at about 4:30 P.M. (c) decreasing most rapidly at about 9 P.M.; rate of change of temperature is about −7◦ F/h (slope of estimated tangent line to curve at 9 P.M.) 15. (b) about (67 − 43)/6 = 4◦ F/h 16. For V = 10 the slope of the tangent line is about −0.25 atm/L, for V = 25 the slope is about −0.04 atm/L. 17. (a) during the ﬁrst year after birth (b) about 6 cm/year (slope of estimated tangent line at age 5) (c) the growth rate is greatest at about age 14; about 10 cm/year (d) Growth rate (cm/year) 40 30 20 10 t (yrs) 5 10 15 20 Exercise Set 3.2 18. 68 (a) (c) vave = 16(6)2 − 16(0)2 = 96 ft/s 6−0 (b) 16(3)2 − 16(0)2 = 48 ft/s 3−0 vave = (d) The rock will hit the ground when 16t2 = 576, t2 = 36, t = 6 s (only t ≥ 0 is meaningful) vinst = lim 16t2 − 16(6)2 16(t2 − 36) 1 1 = lim t1 →6 t1 →6 t1 − 6 t1 − 6 = lim 16(t1 + 6) = 192 ft/s t1 →6 20. (a) 5(40)3 = 320,000 ft (b) vave = 320,000/40 = 8,000 ft/s (c) 19. 5t3 = 135 when the rocket has gone 135 ft, so t3 = 27, t = 3 s; vave = 135/3 = 45 ft/s. (d) = lim 5(t2 + 40t1 + 1600) = 24, 000 ft/s 1 t1 →40 [3(3)2 + 3] − [3(1)2 + 1] = 13 mi/h 3−1 (a) vave = (b) vinst = lim (a) vave = (b) 21. 5t3 − 5(40)3 5(t3 − 403 ) 1 1 = lim t1 →40 t1 →40 t1 − 40 t1 − 40 vinst = lim vinst = lim (3t2 + t1 ) − 4 (3t1 + 4)(t1 − 1) 1 = lim = lim (3t1 + 4) = 7 mi/h t1 →1 t1 →1 t1 →1 t1 − 1 t1 − 1 6(4)4 − 6(2)4 = 720 ft/min 4−2 6t4 − 6(2)4 6(t4 − 16) 1 1 = lim t1 →2 t1 →2 t1 − 2 t1 − 2 6(t2 + 4)(t2 − 4) 1 1 = lim 6(t2 + 4)(t1 + 2) = 192 ft/min 1 t1 →2 t1 →2 t1 − 2 = lim EXERCISE SET 3.2 1. f (1) = 2, f (3) = 0, f (5) = −2, f (6) = −1/2 2. f (4) < f (0) < f (2) < 0 < f (−3) 3. (b) 4. f (2) = m = 3 f (−1) = m = 5. (c) the same, f (2) = 3 4−3 =1 0 − (−1) 6. y y 1 0 x x 7. y − (−1) = 5(x − 3), y = 5x − 16 9. f (x + h) − f (x) 3(x2 + 2xh + h2 ) − 3x2 = lim = lim 3(2x + h) = 6x; f (3) = 3(3)2 = 27, h→0 h→0 h→0 h h f (3) = 18 so y − 27 = 18(x − 3), y = 18x − 27 f (x) = lim 8. y − 3 = −4(x + 2), y = −4x − 5 69 10. Chapter 3 f (x + h) − f (x) (x + h)2 − (x + h) − (x2 − x) = lim = lim (2x + h − 1) = 2x − 1; h→0 h→0 h→0 h h f (2) = 22 − 2 = 2, f (2) = 3 so y − 2 = 3(x − 2), y = 3x − 4 f (x) = lim f (x + h) − f (x) (x + h)3 − x3 = lim = 3x2 ; f (0) = 03 = 0, h→0 h→0 h h f (0) = 0 so y − 0 = (0)(x − 0), y = 0 11. f (x) = lim 12. 13. 14. 15. f (x + h) − f (x) 2(x + h)3 + 1 − (2x3 + 1) = lim = lim (6x2 + 6xh + 2h2 ) = 6x2 ; h→0 h→0 h→0 h h f (−1) = 2(−1)3 + 1 = −1, f (−1) = 6 so y + 1 = 6(x + 1), y = 6x + 5 f (x) = lim √ √ f (x + h) − f (x) x+1+h− x+1 = lim f (x) = lim h→0 h→0 h h √ √ √ √ 1 x+1+h− x+1 x+1+h+ x+1 1 √ √ √ = lim = lim √ =√ ; h→0 h 2 x+1 x + 1 + h + x + 1 h→0 x + 1 + h + x + 1 √ 1 1 5 1 f (8) = 8 + 1 = 3, f (8) = so y − 3 = (x − 8), y = x + 6 6 6 3 f (x + h) − f (x) (x + h)4 − x4 = lim = lim (4x3 + 6x2 h + 4xh2 + h3 ) = 4x3 ; h→0 h→0 h→0 h h f (−2) = (−2)4 = 16, f (−2) = −32 so y − 16 = −32(x + 2), y = −32x − 48 f (x) = lim x − (x + ∆x) 1 1 − x(x + ∆x) f (x) = lim x + ∆x x = lim ∆x→0 ∆x→0 ∆x ∆x = lim ∆x→0 16. −∆x 1 1 = lim − =− 2 x∆x(x + ∆x) ∆x→0 x(x + ∆x) x 1 1 x2 − (x + ∆x)2 −2 (x + ∆x)2 x x2 (x + ∆x)2 = lim f (x) = lim ∆x→0 ∆x→0 ∆x ∆x 2 x2 − x2 − 2x∆x − ∆x2 −2x∆x − ∆x2 −2x − ∆x = lim 2 = lim 2 =− 3 2 ∆x(x + ∆x)2 ∆x→0 ∆x→0 x ∆x(x + ∆x)2 ∆x→0 x (x + ∆x)2 x x = lim 17. [a(x + ∆x)2 + b] − [ax2 + b] ax2 + 2ax∆x + a(∆x)2 + b − ax2 − b = lim ∆x→0 ∆x→0 ∆x ∆x f (x) = lim 2ax∆x + a(∆x)2 = lim (2ax + a∆x) = 2ax ∆x→0 ∆x→0 ∆x = lim 18. 1 1 (x + 1) − (x + ∆x + 1) − (x + ∆x) + 1 x + 1 (x + 1)(x + ∆x + 1) f (x) = lim = lim ∆x→0 ∆x→0 ∆x ∆x = lim x + 1 − x − ∆x − 1 −∆x = lim ∆x(x + 1)(x + ∆x + 1) ∆x→0 ∆x(x + 1)(x + ∆x + 1) = lim −1 1 =− (x + 1)(x + ∆x + 1) (x + 1)2 ∆x→0 ∆x→0 √ 19. f (x) = lim ∆x→0 = lim ∆x→0 1 1 √ √ −√ x − x + ∆x x x + ∆x = lim √√ ∆x→0 ∆x x x + ∆x ∆x 1 x − (x + ∆x) −1 √ √ = lim √ √ = − 3/2 √√ √ √ 2x ∆x x x + ∆x( x + x + ∆x) ∆x→0 x x + ∆x( x + x + ∆x) Exercise Set 3.2 20. 70 (x + ∆x)1/3 − x1/3 , but a3 − b3 = (a − b)(a2 + ab + b2 ) so with a = (x + ∆x)1/3 ∆x→0 ∆x and b = x1/3 (x + ∆x) − x...
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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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