) 25. = −1; x2 ( ) = 3x + 3 g x ( f x { Solution set: −1} . ( ) = 0 a. f x − 1 = x2 0 ( ( −1) + 1) = 0 x x = 1; = −1 x x { Solution set: − 1,1} { } The solution set is < −1 or > 4 x x x . (−∞, (4, or, using interval notation, −1) ∪ ( ) = b. 0 g x ) ∞ . 3x + 3 = ( ) 0 ≥ g. 1 f x = −3 3x −1 > 1 x2 = −1 x − 2 > 0 x2 ( ) ( c. = f x g ( )( ) ) > 2 2 0 x x x − + −1 = + 3x x2 The zeros are = 2 and = 2 . ( ) x – x 3 2 { − − = −1, 3x 4 0 Solution set: 4} . x2 ( ( − 4) + 1) = x x ( ) d. > 0 f x 0 = 4; = −1 x x − 1 > 0 x2 ( ( −1) + 1) > 0 x x = − 2 p x x The zeros are = and = 1 x –1 x { } . <− > The solution set is 2 or 2 or, x x x (−∞, using interval notation, − 2 , ∞ ) ( ) 2 ∪ . 26. ( ) −3x = + 3 g x ( ) = −1; f x x2 { } < −1 > The solution set is or 1 x x x (−∞, (1, or, using interval notation, −1) ∪ ) ∞ . ( ) ≤ e. 0 g x
25. f ( x ) = x2 −1; g ( x ) = 3x + 3 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} . The solution set is { x x < −1 or x > 4 } or, using interval notation, (−∞, −1) ∪ (4, ∞ ) . g. f ( x ) ≥ 1 x2 −1 > 1 x2 − 2 > 0 c. f ( x ) = g ( x ) ( x − 2 )( x + 2 ) > 0 x2 −1 = 3x + 3 The zeros are x = – 2 and x = 2 . ( ) 2 x2 − 3x − 4 = 0 ( x − 4) ( x + 1) = 0 x = 4; x = −1 Solution set: { −1, 4} . d. f ( x ) > 0 x2 − 1 > 0 ( x −1) ( x + 1) > 0 p x = x − 2 The zeros are x = –1 and x = 1 . The solution set is { x x <− 2 or x > 2 } or, using interval notation, (−∞, − 2 ) ∪ ( 2 , ∞ ) . 26. f ( x ) = x2 −1; g ( x ) = −3x + 3 The solution set is { x x < −1 or x > 1 } or, using interval notation, (−∞, −1) ∪ (1, ∞ ) . e. g ( x ) ≤ 0
Downloadable-Solution-Manual-for-Precalculus-Concepts-Through-1.docx