Ex x 2 0 so ex 0 impossible or x 2 0 x 2 32 e2x

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Unformatted text preview: 4 (b) (d) 31. f (x) = −3/(x − 1)2 (e) 2 32. f (x) = 2x sin x + x cos x 34. f (x) = 36. f (x) = 37. f (x) = 6x2 + 8x − 17 (3x + 2)2 x2 cos f (x) = −1/x2 √ f (x) = 3x/ 3x2 + 5 33. f (x) = (c) f (x) = −1/2x3/2 (f ) f (x) = 3 cos 3x √ 1 − 2 x sin 2x √ f (x) = 2x 35. -1.5 (1 + x2 ) sec2 x − 2x tan x (1 + x2 )2 √ √ x − 2x3/2 sin x 2x7/2 −2x5 sin x − 2x4 cos x + 4x4 + 6x2 sin x + 6x − 3x cos x − 4x sin x + 4 cos x − 8 √ 2x2 x4 − 3 + 2(2 − cos x)2 Horizon Module 3 98 CHAPTER 3 HORIZON MODULE 1. x1 = l1 cos θ1 , x2 = l2 cos(θ1 + θ2 ), so x = x1 + x2 = l1 cos θ1 + l2 cos(θ1 + θ2 ) (see Figure 3 in text); similarly y1 = l1 sin θ1 + l2 sin(θ1 + θ2 ). 2. Fix θ1 for the moment and let θ2 vary; then the distance r from (x, y ) to the origin (see Figure 3 in text) is at most l1 + l2 and at least l1 − l2 if l1 ≥ l2 and l2 − l1 otherwise. For any fixed θ2 let θ1 vary and the point traces out a circle of radius r. (a) {(x, y ) : 0 ≤ x2 + y 2 ≤ 2l1 } (b) (c) 3. {(x, y ) : l1 − l2 ≤ x2 + y 2 ≤ l1 + l2 } {(x, y ) : l2 − l1 ≤ x2 + y 2 ≤ l1 + l2 } (x, y ) = (l1 cos θ + l2 cos(θ1 + θ2 ), l1 sin θ1 + l2 sin(θ1 + θ2 )) √ = (cos(π/4) + 3 cos(5π/12), sin(π/4) + 3 sin(5π/12)) = √ √ √ 2+3 6 7 2+3 6 , 4 4 4. x = (1) cos 2t + (1) cos(2t + 3t) = cos 2t + cos 5t, y = (1) sin 2t + (1) sin(2t + 3t) = sin 2t + sin 5t y 5. y 2 -2 y 2 2 x 2 -2 2 x x 1 -2 -2 -2 v1 = 1, v2 = 4 v1 = 3, v2 = 5 6. (a) v1 = 4, v2 = 1 x = 2 cos t, y = 2 sin t, a circle of radius 2 7. 2 9 = [3 sin(θ1 + θ2 )]2 + [3 cos(θ1 + θ2 )]2 = [5 − 3 sin θ1 ]2 + [3 − 3 cos θ1 ]2 = 25 − 30 sin θ1 + 9 sin2 θ1 + 9 − 18 cos θ1 + 9 cos2 θ1 = 43 − 30 sin θ1 − 18 cos θ1 , so 15 sin θ1 + 9 cos θ1 = 17 (b) 1 = sin2 θ1 + cos2 θ2 = 17 − 9 cos θ1 15 2 + cos θ1 , or 306 cos2 θ1 − 306 cos θ1 = −64 √ 1 5 17 (c) cos θ1 = 153 ± − 4(153)(32) /306 = ± 2 102 (e) If θ1 = 0.792436 rad, then θ2 = 0.475882 rad ≈ 27.2660◦ ; if θ1 = 1.26832 rad, then θ2 = −0.475882 rad ≈ −27.2660◦ . (153)2 8. 9. dθ2 dθ2 dθ1 dx dθ1 dθ1 = −3 sin θ1 − (3 sin(θ1 + θ2 )) + = −3 (sin θ1 + sin(θ1 + θ2 )) − 3 (sin(θ1 + θ2 )) dt dt dt dt dt dt dy dθ1 dx dy dθ2 dθ2 dθ1 − 3(sin(θ1 + θ2 )) ; similarly =x + 3(cos(θ1 + θ2 )) . Now set = 0, = 1. = −y dt dt dt dt dt dt dt 1 x = 3 cos(π/3) + 3 cos(−π/3) = 6 = 3 and y = 3 sin(π/3) − 3 sin(π/3) = 0; equations (4) 2 dθ1 dθ2 dθ2 = 0, 3 + 3 cos(π/3) = 1 with solution dθ2 /dt = 0, dθ1 /dt = 1/3. become 3 sin(π/3) dt dt dt dθ1 dθ2 dθ1 = 0 and −3 −3 = 1, with solution dθ1 /dt = 0, dθ2 /dt = −1/3. (b) x = −3, y = 3, so −3 dt dt dt (a) CHAPTER 4 Logarithmic and Exponential Functions EXERCISE SET 4.1 1. (a) f (g (x)) = 4(x/4) = x, g (f (x)) = (4x)/4 = x, f and g are inverse functions (b) f (g (x)) = 3(3x − 1) + 1 = 9x − 2 = x so f and g are not inverse functions (c) f (g (x)) = 3 (x3 + 2) − 2 = x, g (f (x)) = (x − 2) + 2 = x, f and g are inverse functions (d) f (g (x)) = (x1/4 )4 = x, g (f (x)) = (x4 )1/4 = |x| = x, f and g are not inverse functions 2. (a) They are inverse functions. 2 -2 2 -2 (b) The graphs are not reflections of each other about the line y = x. 2 -2 2 -2 (c) They are inverse functions provided the domain of g is restricted to [0, +∞) 5 0 5 0 (d) They are inverse functions provided the domain of f (x) is restricted to [0, +∞) 2 0 2 0 3. yes; all outputs (the elements of row two) are distinct (b) 4. (a) no; f (1) = f (6) (a) no; it is easy to conceive of, say, 8 people in line at two different times (b) no; perhaps your weight remains constant for more than a year (c) yes, since the function is increasing, in the sense that the greater the volume, the greater the weight 99 Exercise Set 4.1 100 5. (a) yes (b) yes (c) 6. (a) no (d) no, the horizontal line test fails yes (e) (b) yes, horizontal line test 6 no (f ) no 10 -1 -3 3 3 -2 -10 7. (a) no, the horizontal line test fails (b) no, the horizontal line test fails (c) yes, horizontal line test 9. (a) 8. (a) no, the horizontal line test fails (b) no, the horizontal line test fails (c) yes, horizontal line test f has an inverse because the graph passes the horizontal line test. To compute f −1 (2) start at 2 on the y -axis and go to the curve and then down, so f −1 (2) = 8; similarly, f −1 (−1) = −1 and f −1 (0) = 0. (b) domain of f −1 is [−2, 2], range is [−8, 8] y (c) 8 4 -2 -1 1 2 x -4 -8 10. −∞ < x ≤ −1; −1 ≤ x ≤ 2; and 2 ≤ x < +∞. (a) f (x) = 2x + 8; f < 0 on (−∞, −4) and f > 0 on (−4, +∞); not one-to-one (b) f (x) = 10x4 + 3x2 + 3 ≥ 3 > 0; f (x) is positive for all x, so f is one-to-one (c) f (x) = 2 + cos x ≥ 1 > 0 for all x, so f is one-to-one (a) f (x) = 3x2 + 6x = x(3x + 6) changes sign at x = −2, 0, so f is not one-to-one (b) 12. the horizontal line test fails (b) 11. (a) f (x) = 5x4 + 24x2 + 2 ≥ 2 > 0; f is positive for all x, so f is one-to-one (c) 1 ; f is one-to-one because: (x + 1)2 if x1 < x2 < −1 then f > 0 on [x1 , x2 ], so f (x1 ) = f (x2 ) if −1 < x1 < x2 then f >...
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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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