2 2 ( ) ( ) b. = f x g x 0 = 2 − 6 (2) (2) x = 2 + 1 = 4 + 1 = 5 2 g x Thus, the graphs of and intersect at the f g + − 6 x − + 7 − 6 = x x x (−4, ) (2, points −7 and 5) . ( = − 0 2 3) x x 2 = 0 or x − 3 = 0 x = 0 x = 3 x
b. f ( x ) = g ( x ) 2 2 g (2) = 2 (2) + 1 = 4 + 1 = 5 Thus, the graphs of f and g intersect at the − x + 7 x − 6 = x 0 = 2 x + x − 6 2 − 6 x points (−4, −7 ) and (2, 5) . 0 = 2 x ( x − 3) 2 x = 0 or x = 0 x − 3 = 0 x = 3
Downloadable-Solution-Manual-for-Precalculus-Concepts-Through-1.docx