# Y p1 x x py y 3 3y 2 3y 1 y 13 y x13

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Unformatted text preview: 0. x2 dx x2 x2 x4 x2 x4 2 2 If f (x) = 0 then (3x + 1) (3x + 2x) = 0. The tangent line is horizontal at x = −1/3, −2/3 (x = 0 is ruled out from the deﬁnition of f ). 8. f (x) = 3 10. (a) x = −2, −1, 1, 3 (d) 9. g (x) = f (x) sin x + 2f (x) cos x − f (x) sin x; g (0) = 2f (0) cos 0 = 2(2)(1) = 4 (a) f (1)g (1) + f (1)g (1) = 3(−2) + 1(−1) = −7 (b) (c) 11. (b) (−∞, −2), (−1, 1), (3, +∞) g (1)f (1) − f (1)g (1) −2(3) − 1(−1) 5 = =− 2 2 g (1) (−2) 4 1 1 3 (d) f (1) = √ 3 = 2 21 2 f (1) (c) (−2, −1), (1, 3) 0 (because f (1)g (1) is constant) The equations of such a line has the form y = mx. The points (x0 , y0 ) which lie on both the line and the parabola and for which the slopes of both curves are equal satisfy y0 = mx0 = x3 − 9x2 − 16x0 , 0 0 so that m = x2 − 9x0 − 16. By diﬀerentiating, the slope is also given by m = 3x2 − 18x0 − 16. 0 0 Equating, we have x2 − 9x0 − 16 = 3x2 − 18x0 − 16, or 2x2 − 9x0 = 0. The root x0 = 0 corresponds 0 0 0 to m = −16, y0 = 0 and the root x0 = 9/2 corresponds to m = −145/4, y0 = −1305/8. So the line y = −16x is tangent to the curve at the point (0, 0), and the line y = −145x/4 is tangent to the curve at the point (9/2, −1305/8). 12. The slope of the line x + 4y = 10 is m1 = −1/4, so we set the negative reciprocal √ d 1 ± 1 + 24 (2x3 − x2 ) = 6x2 − 2x and obtain 6x2 − 2x − 4 = 0 with roots x = = 1, −2/3. 4 = m2 = dx 6 13. 14. d (3x − tan x) = 3 − sec2 x = 1, or sec2 x = 2, The line y − x = 2 has slope m1 = 1 so we set m2 = dx √ sec x = ± 2 so x = nπ ± π/4 where n = 0, ±1, ±2, . . .. f (x) is continuous and diﬀerentiable at any x = 1, so we consider x = 1. (a) (b) lim (x2 − 1) = lim+ k (x − 1) = 0 = f (1), so any value of k gives continuity at x = 1. x→1− x→1 lim f (x) = lim− 2x = 2, and lim+ f (x) = lim+ k = k , so only if k = 2 is f (x) diﬀerentiable at x→1− x = 1. x→1 x→1 x→1 Supplementary Exercises 3 15. 96 The slope of the tangent line is the derivative y = 2x x= 1 (a+b) 2 y = a + b. The slope of the secant is a2 − b2 = a + b, so they are equal. a−b (b, b)2 (a, a)2 x a a+b 2 b 16. To average 60 mi/h one would have to complete the trip in two hours. At 50 mi/h, 100 miles are completed after two hours. Thus time is up, and the speed for the remaining 20 miles would have to be inﬁnite. 17. (a) ∆x = 1.5 − 2 = −0.5; dy = ∆y = −1 −1 ∆x = (−0.5) = 0.5; and (x − 1)2 (2 − 1)2 1 1 − = 2 − 1 = 1. (1.5 − 1) (2 − 1) (b) ∆x = 0 − (−π/4) = π/4; dy = sec2 (−π/4) (π/4) = π/2; and ∆y = tan 0 − tan(−π/4) = 1. (c) ∆x = 3 − 0 = 3; dy = √ ∆y = 18. (a) 19. (a) (b) 20. 21. √ 25 − 32 − √ −0 25 − (0)2 (3) = 0; and 25 − 02 = 4 − 5 = −1. 56 43 − 23 = = 28 4−2 2 (b) (dV /d )| =5 = 3 2 =5 = 3(5)2 = 75 dW dW = 200(t − 15); at t = 5, = −2000; the water is running out at the rate of 2000 gal/min. dt dt W (5) − W (0) 10000 − 22500 = = −2500; the average rate of ﬂow out is 2500 gal/min. 5−0 5 46π 46π π ; let x0 = and x = . Then 180 4 180 π π π x− =1−2 cot 46◦ = cot x = cot − csc2 4 4 4 with a calculator, cot 46◦ = 0.9657. cot 46◦ = cot (a) (a) 1 21 dT =√ √ =√ dL g2 L gL (d) (e) = 0.9651; 51 π π radians and dφ = ±0.5◦ = ±0.5 180 180 radians, h ± dh = 115(1.2349) ± 2.5340 = 142.0135 ± 2.5340, so the height lies between 139.48 m and 144.55 m. If |dh| ≤ 5 then |dφ| ≤ (c) 46π π − 180 4 h = 115 tan φ, dh = 115 sec2 φ dφ; with φ = 51◦ = (b) 22. −x = 25 − x2 5 115 cos2 51 180 π ≈ 0.017 radians, or |dφ| ≤ 0.98◦ . (b) s/m dT > 0 an increase in L gives an increase in T , which is the period. To speed up a clock, Since dL decrease the period; to decrease T , decrease L. √ dT L = − 3/2 < 0; a decrease in g will increase T and the clock runs slower dg g √ √ dT −1 L −3/2 =2 L g = − 3/2 (f ) s3 /m dg 2 g 97 23. Chapter 3 (a) f (x) = 2x, f (1.8) = 3.6 (b) f (x) = (x2 − 4x)/(x − 2)2 , f (3.5) ≈ −0.777778 24. (a) f (x) = 3x2 − 2x, f (2.3) = 11.27 25. f (x) = 2x ln 2; f (2) ≈ 2.772589 (b) f (x) = (1 − x2 )/(x2 + 1)2 , f (−0.5) = 0.48 26. f (x) = xsin x (cos x ln x + sin x/x); f (2) = 0.312141 27. 28. 3(h + 1)2.5 + 580h − 3 1 d 2.5 = 58 + 3x h→0 10h 10 dx vinst = lim 164 ft/s 29. = 58 + x=1 1 (2.5)(3)(1)1.5 = 58.75 ft/s 10 Solve 3x2 − cos x = 0 to get x = ±0.535428. 2500 1 20 0 30. When x4 − x − 1 > 0, f (x) = x4 − 2x − 1; when x4 − x − 1 < 0, f (x) = −x4 + 1, and f is diﬀerentiable in both cases. The roots of x4 − x − 1 = 0 are x1 = −0.724492, x2 = 1.220744. So x4 − x − 1 > 0 on (−∞, x1 ) and (x2 , +∞), and x4 − x − 1 < 0 on (x1 , x2 ). Then lim− f (x) = lim− (4x3 − 2) = 4x3 − 2 and 1 x→x1 -1.5 2 x→x1 x→x+ 1 1.5 x→x1 lim f (x) = lim+ −4x3 = −4x3 which is not equal to 4x3 − 2, so 1 1 f is not diﬀerentiable at x = x1 ; similarly f is not diﬀerentiable at x = x2 . (a) f (x) = 5x...
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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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