# A 25 f 20 if v 48 then 60 wci 16t 55 thus t

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Unformatted text preview: 1 so g (x) = 3(x − 2) − (x + 1) = 2x − 7; x < −1 7 − 2x, 5 − 4x, −1 ≤ x < 2 g (x) = 2x − 7, x≥2 (a) V = (8 − 2x)(15 − 2x)x (b) −∞ < x < +∞, −∞ < V < +∞ (c) 0 < x < 4 (d) minimum value at x = 0 or at x = 4; maximum value somewhere in between (can be approximated by zooming with graphing calculator) (a) x = 3000 tan θ (b) θ = nπ + π/2 for n an integer, −∞ < n < ∞ (c) 0 ≤ θ < π/2, 0 ≤ x < +∞ 16. (d) 3000 ft 6000 6 0 0 17. (i) x = 1, −2 causes division by zero (ii) g (x) = x + 1, all x 18. (i) x = 0 causes division by zero (ii) g (x) = 19. (a) 25◦ F 20. If v = 48 then −60 = WCI = 1.6T − 55; thus T = (−60 + 55)/1.6 ≈ −3◦ F. 21. √ If v = 8 then −10 = WCI = 91.4 + (91.4 − T )(0.0203(8) − 0.304 8 − 0.474); thus √ T = 91.4 + (10 + 91.4)/(0.0203(8) − 0.304 8 − 0.474) and T = 5◦ F √ x + 1 for x ≥ 0 (b) 2◦ F (c) −15◦ F 22. The WCI is given by three formulae, but the ﬁrst and third don’t √ work with the data. Hence √ −15 = WCI = 91.4 + (91.4 − 20)(0.0203v − 0.304 v − 0.474); set x = v so that v = x2 and obtain 0.0203x2 − 0.304x − 0.474 + (15 + 91.4)/(91.4 − 20) = 0. Use the quadratic formula to ﬁnd the two roots. Square them to get v and discard the spurious solution, leaving v ≈ 25. 23. Let t denote time in minutes after 9:23 AM. Then D(t) = 1000 − 20t ft. EXERCISE SET 1.3 1. (e) seems best, though only (a) is bad. y 2. (e) seems best, though only (a) is bad and (b) is not good. y 0.5 0.5 x -1 1 -1 1 -0.5 -0.5 x Exercise Set 1.3 3. 6 (b) and (c) are good; (a) is very bad. 4. (b) and (c) are good; (a) is very bad. y y 15 -11 14 -12 13 -13 12 -14 x -2 5. -1 0 1 -2 2 [−3, 3] × [0, 5] 6. -1 0 1 2 x [−4, 2] × [0, 3] y y 2 4 1 2 -3 7. -2 -3 -1 1 2 -2 -1 1 x x 3 (a) window too narrow, too short (c) good window, good spacing (b) window wide enough, but too short (d) window too narrow, too short y -5 -100 5 10 15 20 x -200 -300 -400 -500 (e) window too narrow, too short 8. (a) window too narrow (c) good window, good tick spacing (b) window too short (d) window too narrow, too short y 50 -16 -12 -8 -4 4 x -50 -100 -150 -200 -250 (e) shows one local minimum only, window too narrow, too short 7 Chapter 1 9. [−5, 14] × [−60, 40] 10. [6, 12] × [−100, 100] y y 100 40 20 50 -5 5 x 10 8 -20 10 x 12 -50 -40 -100 -60 11. [−0.1, 0.1] × [−3, 3] 12. [−1000, 1000] × [−13, 13] y y 3 10 2 5 1 x -0.1 x 1000 -1000 0.1 -1 -5 -2 -10 -3 13. [−250, 1050] × [−1500000, 600000] 14. [−3, 20] × [−3500, 3000] y -1000 y 1000 x 1000 -500000 5 10 x 15 -1000 -2000 15. [−2, 2] × [−20, 20] 16. [1.6, 2] × [0, 2] y y 20 2 10 1.5 x -2 -1 1 1 2 0.5 -10 x -20 17. 1.6 depends on graphing utility 18. 1.7 1.8 2 depends on graphing utility y y 6 6 4 4 2 -4 1.9 -2 2 2 -2 4 x -4 -2 2 -2 -4 -4 -6 -6 4 x Exercise Set 1.3 19. 8 (a) f (x) = √ √ (b) f (x) = − 16 − x2 16 − x2 y y 4 3 2 1 -4 -4 -2 2 4 x 4 y (d) 4 4 3 2 -4 2 -1 -2 -3 -4 x y (c) -2 -2 2 4 2 x 1 -2 1 -4 2 3 4 (e) No; the vertical line test fails. 20. √ (b) y = ± x2 + 1 (a) y = ±3 1 − x2 /4 y 4 2 -4 -2 2 4 x -2 -4 21. y (a) y (b) (c) y 1 1 x -1 x x -1 (d) 1 1 -1 2 y (e) y 1 (f ) y 1 1 π 2π x −π 1 π —1 x x -1 1 -1 y 22. 1 x -1 1 -1 23. The portions of the graph of y = f (x) which lie below the x-axis are reﬂected over the x-axis to give the graph of y = |f (x)|. 9 Chapter 1 24. Erase the portion of the graph of y = f (x) which lies in the left-half plane and replace it with the reﬂection over the y -axis of the portion in the right-half plane (symmetry over the y -axis) and you obtain the graph of y = f (|x|). 25. (a) for example, let a = 1.1 (b) y y 3 2 1 µ 1 26. They are identical. 27. 2 3 x y y 15 10 x 5 x -1 28. 1 2 3 This graph is very complex. We show three views, small (near the origin), medium and large: (a) (b) y (c) y y 10 x 2 -1 x -50 -1 1000 x 40 29. (a) (b) y y 1 1.5 0.5 1 x -3 -2 -1 1 2 3 -0.5 0.5 -1 -3 (c) 1 -1 -2 2 3 x y (d) y 1.5 1.5 1 0.5 -1 1 1 2 3 x 0.5 x -2 -1 1 Exercise Set 1.3 10 30. y 1 x 2 31. (a) stretches or shrinks the graph in the y -direction; ﬂips it if c changes sign (b) As c increases, the parabola moves down and to the left. If c increases, up and right. y y c=2 4 8 2 -1 c=1 4 x 1 c=2 6 c=1 2 c = -1.5 -2 c = -1.5 -2 -1 1 x 2 (c) The graph rises or falls in the y -direction with changes in c. y c = +2 8 c = .5 6 c = -1 4 2 -2 32. -1 1 2 x y (a) 2 x 1 2 —2 (b) x-intercepts at x = 0, a, b. Assume a < b and let a approach b. The two branches of the curve come together. If a moves past b then a and b switch roles. y y y 3 3 2 2 2 1 1 3 x -1 -2 -3 1 2 1 1 3 -1 a = 0.5 b = 1.5 -2 -3 2 3 x 1 -1 a=1 b = 1.5 -2 -3 a = 1.5 b = 1.6 2 3 x 11 33. Chapter 1 The curve oscillates between the lines y = x and y = −x with increasing rapidity as |x| increases. 34. The curve oscillates between the lines y = +1 and y = −1, inﬁnitely many times in any neighborhood of x = 0. y y 30...
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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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