+ ≤ − = 3x 3 0 1 0 x2 ( ( 3x ≤ −3 −1) + 1) = 0 x x ≤ −1 x = 1; = −1 x x { {−1, The solution set is ≤ −1} or, x x Solution set: 1} . (−∞, −1] interval notation, . ( ) b. = 0 g x ( ) ( −3x + 3 = 0 f. > f x g −3x = −3 ) = x 1 x {1} Solution set: . −1 > 3x + 3 x2 − 3x − 4 > 0 x2 ( ) ( ) ( ( c. = f x g x − 4) + 1) > 0 x x −1 = −3x + 3 x2 The zeros are = and = 4 . x x –1 + 3x − 4 = 0 x2 ( ) = − 3x − 4 p x x2 ( ( + − = 4) 1) 0 x x = −4; = 1 x x ( ) = a. 0 {−4, f x Solution set: 1} .
3x + 3 ≤ 0 3x ≤ −3 x ≤ −1 The solution set is { x x ≤ −1} or, interval notation, (−∞, −1] . f. f ( x ) > g ( x ) x2 −1 > 3x + 3 x2 − 3x − 4 > 0 ( x − 4) ( x + 1) > 0 The zeros are x = –1 and x = 4 . p ( x ) = x2 − 3x − 4 a. f ( x ) = 0 x2 − 1 = 0 ( x −1) ( x + 1) = 0 x = 1; x = −1 Solution set: {−1, 1} . b. g ( x ) = 0 −3x + 3 = 0 −3x = −3 x = 1 Solution set: {1} . c. f ( x ) = g ( x ) x2 −1 = −3x + 3 x2 + 3x − 4 = 0 ( x + 4) ( x − 1) = 0 x = −4; x = 1 Solution set: {−4, 1} .
Downloadable-Solution-Manual-for-Precalculus-Concepts-Through-1.docx