( ) > d. 0 f x − > 1 0 x2 ( ( −1) + 1) > 0 x x The are = and = 1 . x x –1 { } The solution set is <− 2 or > 2 x x x or, using interval notation, { } (−∞, ) ) The solution set is < −1 or > 1 x x x − ∞ 2 2 , . ( or, ∪ (−∞, (1, interval notation, −1) ∪ ∞ ( ) ( ) 27. ) . ( ) ( ) e. ≤ g x = a. 0 f x = − + f x x2 0 1; = 4 + 1 g x x −3x + 3 ≤ 0 − + 1 = x2 0 2 −3x ≤ −3 1 − = 0 x = 1; = −1 x x ≥ 1 x {−1, { Solution set: 1} . The solution set is ≥ 1} or, using x x ( ) [1, ) = ∞ b. 0 interval notation, . g x 4 + 1 = 0 x ( ) ( ) f. > f x g x 4 = −1 x 1 −1 > −3x + 3 x2 =− + − > 3x 4 0 x2 (1 ) (1 ) − + = 0 x x ( ( + 4) − 1) > 0 x x 4 The zeros are = and = 1 . x x ⎧ 1 ⎫ –4 x ( ) = + 3x p x x2 Solution set: . ⎨− ⎬ ⎩ 4 ⎭ − 4 { The solution set is < −4 or > 1 x x x } (−∞, or, using interval notation, (1, ) −4) ∪ ∞ . ( ) ≥ g. 1 f x
d. f ( x ) > 0 x2 − 1 > 0 ( x −1) ( x + 1) > 0 The are x = –1 and x = 1 . { } The solution set is x x <− 2 or x > 2 or, using interval notation, The solution set is { x x < −1 or x > 1 } or, (−∞, − 2 ) ∪ ( 2 , ∞ ) . interval notation, (−∞, −1) ∪ (1, ∞ ) . 27. ( ) ( ) e. g ( x ) ≤ 0 f x = − x2 + 1; a. f ( x ) = 0 g x = 4 x + 1 −3x + 3 ≤ 0 − x2 + 1 = 0 2 −3x ≤ −3 1 − x = 0 x ≥ 1 The solution set is { x x ≥ 1} or, using interval notation, [1, ∞ ) . f. f ( x ) > g ( x ) x2 −1 > −3x + 3 x2 + 3x − 4 > 0 (1 − x ) (1 + x ) = 0 x = 1; x = −1 Solution set: {−1, 1} . b. g ( x ) = 0 4 x + 1 = 0 4 x = −1 1 =− ( x + 4) ( x − 1) > 0 The zeros are x = –4 and x = 1 . x 4 ⎧ 1 ⎫ p ( x ) = x2 + 3x − 4 Solution set: ⎨− ⎬ . ⎩ 4 ⎭ The solution set is { x x < −4 or x > 1 } or, using interval notation, (−∞, −4) ∪ (1, ∞ ) . g. f ( x ) ≥ 1
Downloadable-Solution-Manual-for-Precalculus-Concepts-Through-1.docx