# 4 18 x axis because x 5y 2 9 gives x 5y 2 9 x axis

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Unformatted text preview: none y x x −1 −1 0 0 1 −1 2 −5 3 1 (b) y x f (x) (b) −3 1 −2 5 −1 −1 0 0 1 1 2 −5 y x x 67. (a) even (b) odd 68. neither; odd; even 69. (a) (c) f (−x) = (−x)2 = x2 = f (x), even f (−x) = | − x| = |x| = f (x), even (e) f (−x) = (f ) f (−x) = 2 = f (x), even (c) odd (d) neither (b) f (−x) = (−x)3 = −x3 = −f (x), odd (d) f (−x) = −x + 1, neither x3 + x (−x)3 − (−x) =− = −f (x), odd 1 + (−x)2 1 + x2 3 -1 Exercise Set 1.4 18 x-axis, because x = 5(−y )2 + 9 gives x = 5y 2 + 9 x-axis, y -axis, and origin, because x2 − 2(−y )2 = 3, (−x)2 − 2y 2 = 3, and (−x)2 − 2(−y )2 = 3 all give x2 − 2y 2 = 3 (c) origin, because (−x)(−y ) = 5 gives xy = 5 (a) y -axis, because (−x)4 = 2y 3 + y gives x4 = 2y 3 + y (b) origin, because (−y ) = (c) 71. (a) (b) 70. x-axis, y -axis, and origin because (−y )2 = |x| − 5, y 2 = | − x| − 5, and (−y )2 = | − x| − 5 all give y 2 = |x| − 5 72. (−x) x gives y = 2 3 + (−x) 3 + x2 73. 3 -4 4 2 -3 3 –3 74. (a) -2 Whether we replace x with −x, y with −y , or both, we obtain the same equation, so by Theorem 1.4.3 the graph is symmetric about the x-axis, the y -axis and the origin. (b) y = (1 − x2/3 )3/2 (c) For quadrant II, the same; for III and IV use y = −(1 − x2/3 )3/2 . (For graphing it may be helpful to use the tricks that precede Exercise 29 in Section 1.3.) 75. 76. y y 5 2 4 3 2 1 -1 77. 1 2 3 4 x x 2 (a) (b) y y 2 1 O C c o x O C c o x 19 78. Chapter 1 (a) (b) y y 1 1 x -1 -1 1 1 23 -1 (c) x (d) y y 1 3 −π π/2 x -1 1 x -1 1 79. Yes, e.g. f (x) = xk and g (x) = xn where k and n are integers. 80. If x ≥ 0 then |x| = x and f (x) = g (x). If x &lt; 0 then f (x) = |x|p/q if p is even and f (x) = −|x|p/q if p is odd; in both cases f (x) agrees with g (x). EXERCISE SET 1.5 1. 2. 3 3 − (8/3) 1 0 − (8/3) 2 3−0 =− , =− , = 0−2 2 0−6 18 2−6 3 (b) Yes; the ﬁrst and third slopes above are negative reciprocals of each other. (a) −1 − 3 3−3 −1 − 3 −1 − (−1) = 0, = 2, = 0, =2 −3 − 5 5−7 7 − (−1) −3 − (−1) (b) Yes; there are two pairs of equal slopes, so two pairs of parallel lines. (a) 3. III &lt; II &lt; IV &lt; I 4. III &lt; IV &lt; I &lt; II 5. (a) −5 − (−1) 1 − (−1) 1 − (−5) = 2, = 2, = 2. Since the slopes connecting all pairs of points are 1 − (−2) −2 − 0 1−0 equal, they lie on a line. (b) 4−2 2−5 4−5 1 = −1, = 3, = . Since the slopes connecting the pairs of points are not −2 − 0 0−1 −2 − 1 3 equal, the points do not lie on a line. 6. The slope, m = −2, is obtained from (a) If x = 9 then y = 1. 7. The slope, m = 3, is equal to (a) If x = 5 then y = 14. y−5 , and thus y − 5 = −2(x − 7). x−7 (b) If y = 12 then x = 7/2. y−2 , and thus y − 2 = 3(x − 1). x−1 (b) If y = −2 then x = −1/3. Exercise Set 1.5 20 y−0 1 y−5 = or y = x/2. Also = 2 or y = 2x − 9. Solve simultaneously x−0 2 x−7 to obtain x = 6, y = 3. 8. (a) Compute the slopes: 9. (a) 5−0 2−0 and the second is . Since they are negatives of each other we get 1−x 4−x 2(4 − x) = −5(1 − x) or 7x = 13, x = 13/7. The ﬁrst slope is 10. (a) 27◦ (b) 135◦ (c) 63◦ (d) 91◦ 11. (a) 153◦ (b) 45◦ (c) 117◦ (d) 89◦ 12. √ (a) m = tan φ = − 3/3, so φ = 150◦ 13. (a) m = tan φ = 14. y = 0 and x = 0 respectively (b) m = tan φ = 4, so φ = 76◦ √ 3, so φ = 60◦ (b) m = tan φ = −2, so φ = 117◦ 15. y = −2x + 4 6 -1 1 0 16. y = 5x − 3 17. 2 Parallel means the lines have equal slopes, so y = 4x + 7. 12 –1 2 –8 -1 1 0 18. The slope of both lines is −3/2, so y − 2 = (−3/2)(x − (−1)), or 3x + 2y = 1 4 -2 1 –4 19. 1 The negative reciprocal of 5 is −1/5, so y = − x + 6. 5 12 -9 9 0 21 20. Chapter 1 The slope of x − 4y = 7 is 1/4 whose negative reciprocal is −4, so y − (−4) = −4(x − 3) or y + 4x = 8. 9 0 18 –3 21. 4 − (−7) y − (−7) = , or y = 11x − 18 x−1 2−1 7 0 4 -9 22. 6−1 y−1 = , or y = −5x − 9 x − (−2) −3 − (−2) 15 –4 0 –5 23. (a) m1 = m2 = 4, parallel (c) m1 = m2 = 5/3, parallel (b) m1 = 2 = −1/m2 , perpendicular (d) (e) 24. If A = 0 and B = 0 then m1 = −A/B = −1/m2 , perpendicular; if A = 0 or B = 0 (not both) then one line is horizontal, the other vertical, so perpendicular. neither (a) m1 = m2 = −5, parallel m1 = −4/5 = −1/m2 , perpendicular (c) (b) m1 = 2 = −1/m2 , perpendicular (d) If B = 0, m1 = m2 = −A/B ; if B = 0 both are vertical, so parallel (e) neither 25. (a) m = (0 − (−3))/(2 − 0)) = 3/2 so y = 3x/2 − 3 (b) m = (−3 − 0)/(4 − 0) = −3/4 so y = −3x/4 26. (a) m = (0 − 2)/(2 − 0)) = −1 so y = −x + 2 (b) m = (2 − 0)/(3 − 0) = 2/3 so y = 2x/3 27. 5 − (−4) = 9/10 ft/s. 10 − 0 (a) The velocity is the slope, which is (b) x = −4 (c) The line has slope 9/10 and passes through (0, −4), so has equation x = 9t/10 − 4; at t = 2, x = −2.2. (d) t = 80/9 Exercise Set 1.5 22 5−1 = 2 m/s 4−2 (b) x − 1 = 2(t − 2) or x = 2t − 3 28. (a) v =...
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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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