( ( ( ) ( ) 0 = + 1) − 3) b. = x x f x g x − + 4 = −2 + 1 x2 x 73. a and d. 0 = − 2x − 3 x2 + 1 = 0 x − = 3 0 x or = 3 x = −1 x (3) (3) = −2 + 1 = −6 + 1 = g (1) (−1)2 c. = − + 4 = −1 + f −5 = 4 3 (1) (−1) = −2 + 1 = 2 + 1 g = 3 (3) (3)2 = − + 4 = −9 + 4 = f ( ) ( ) b. = f x g x −5 2 2 Thus, the graphs of and intersect at the f g − + = + − 5x 3x 4 x x (−1, (3, points 3) and −5) . 0 = 2 − 2 − 4 x 2 x − − 0 = 2 x2 x and 72. a d. ( ( = + − 0 1) 2) x x + = 1 0 or x − = 2 0 x = −1 x = 2 x (−1) (−1)2 (−1) c. = − + 5 = −1 − 5 f = −6 (−1) (−1)2 (−1) = + 3 − 4 = 1 − 3 − g 4 = −6 (2) (2)2 (2) = − + 5 = −4 + 10 = 6 f (2) = + − = + − = 22 3(2) 4 4 6 4 6 g Thus, the graphs of and intersect at the f g (−1, (2, points −6) and 6) . 0 = + 2x − 8 x2 ( ( ( ) ( ) 0 = + 4) − 2) = x x b. f x g x − + 9 = 2 + 1 x2 x 74. a and d. + 4 = 0 or x − 2 = 0 x = −4 x = 2 x (2) (2)2 = − + 9 = −4 + 9 f (−4) (−4)2 c. = − + 9 = −16 + 9 = f = 5 −7 (−4) (−4) = 2 + 1 = −8 + 1 g = −7
b. f ( x ) = g ( x ) − x2 + 4 = −2 x + 1 0 = x2 − 2x − 3 0 = ( x + 1) ( x − 3) 73. a and d. x + 1 = 0 or x = −1 x − 3 = 0 x = 3 c. f (1) = − (−1)2 + 4 = −1 + 4 = 3 g (1) = −2 (−1) + 1 = 2 + 1 = 3 f (3) = − (3)2 + 4 = −9 + 4 = −5 g (3) = −2 (3) + 1 = −6 + 1 = −5 b. f ( x ) = g ( x ) 2 2 Thus, the graphs of f and g intersect at the points (−1, 3) and (3, −5) . − x + 5x = x 0 = 2 x + 3x − 4 2 − 2 x − 4 72. a and d. 0 = x2 − x − 2 0 = ( x + 1) ( x − 2) x + 1 = 0 or x = −1 x − 2 = 0 x = 2 c. f (−1) = − (−1)2 + 5 (−1) = −1 − 5 = −6 g (−1) = (−1)2 + 3 (−1) − 4 = 1 − 3 − 4 = −6 f (2) = − (2)2 + 5 (2) = −4 + 10 = 6 g (2) = 22 + 3(2) − 4 = 4 + 6 − 4 = 6 Thus, the graphs of f and g intersect at the points (−1, −6) and (2, 6) . b. f ( x ) = g ( x ) − x2 + 9 = 2 x + 1 0 = x2 + 2x − 8 0 = ( x + 4) ( x − 2) 74. a and d. x + 4 = 0 or x = −4 x − 2 = 0 x = 2 c. f (−4) = − (−4)2 + 9 = −16 + 9 = −7 g (−4) = 2 (−4) + 1 = −8 + 1 = −7 f (2) = − (2)2 + 9 = −4 + 9 = 5
Downloadable-Solution-Manual-for-Precalculus-Concepts-Through-1.docx