( 3)( x x x = + +1)( −1)   π π  110. 2 5 x − − 440 2(110) 220 − = (2) (2)3 2(2) 5 f = − − = −1 π π ( f (3) (3)3 2(3) 5 16 f = − − = The dimensions are 110 yards by 220 yards. The graph passes through (2, which is below the x) –1), π and (3, 16), which is above the Since x-axis, x-axis. = the graph of is continuous, it must cross the 105. Answers will vary. f x-axis somewhere between 2 and 3 to get from one of these x3 106. 10 y = 2 points to the other. 10 y = − 2 3x + 3x 4 2 10 y =± − 3x 111. ( 2 f x) x = − +1 x Since there are values of that give more than one x (−x) (−x)4 2(−x)2 f = − +1 value for (for example, if = 0, then y x 2 x4 x2 = − +1 10 0 y =± − = Since (−x) ( the function is even. 10 ), the equation does not define f f x), = ± as a function of y x. Thus, the graph is symmetric with respect to the y-axis.
= ( x + 3)( x +1)( x −1)  π  π  440 − 2(110) = 220 π π The dimensions are 110 yards by 220 yards. π 105. Answers will vary. 110. f ( x) = x3 − 2 x − 5 f (2) = (2)3 − 2(2) − 5 = −1 f (3) = (3)3 − 2(3) − 5 = 16 The graph passes through (2, –1), which is below the x-axis, and (3, 16), which is above the x-axis. Since the graph of f is continuous, it must cross the x-axis somewhere between 2 and 3 to get from one of these 106. 3x + y 2 = 10 y 2 = 10 − 3x points to the other. 4 2 y =± 10 − 3x 111. f ( x) = x − 2 x +1 Since there are values of x that give more than one value for y (for example, if x = 0, then f (−x) = (−x)4 − 2(−x)2 +1 = x4 − 2 x2 +1 y =± 10 − 0 = ± 10 ), the equation does not define Since f (−x) = f ( x), the function is even. y as a function of x. Thus, the graph is symmetric with respect to the y-axis.
Precalculus-Essentials-5th-Edition-Blitzer-Solutions-Manual.docx