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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online