− ≥ 5 0 x2 ( ) g. ≥ f x 1 { } The solution set is < 1 or > 1 or, x x x − ≥ 4 x2 (−∞, (1, ) ∪ ∞ using interval notation, 1) . 1 2 ( − x e. ) g 5 ≥ ( ) x )( 5 + 0 ≤ 0 x The zeros are − + 1 ≤ 0 x = 5 and = 5 . x x – − ≤ 1 0 x2 (1 ) (1 ) − + ≤ 0 x x ( ( ) = p x The zeros are = and = 1 . x –1 x )( − 5 + x x ) 5 (−∞, (− ) ( Interval 5 5, 5 5, − ∞ ) ) Test 0 −3 3 Number −5 of 4 4 p Conclusion Positive Negative Positive { The solution set is ≤− 5 or ≥ x x x { } (−∞, } The solution set is ≤ −1 or ≥ 1 x x x using interval notation, − ⎤ ⎡ 5 or, [1, (−∞, or, using interval notation, −1] ∪ ) ∞ 5, ) ∞ . ( ) ( ) . > f. f x g x ) 30. − = 2 + 1; x2 x − 2 + 1 = x2 x − + 1 ( x2 ( ) f x = − + 1 g x x2 ( ) {−1, a. = f x Solution set: 1} . 0 ( ) ( ) = c. f x g x − 2 + 1 = 0 x2 x − 2 + 1 = − + 1 ( ( x2 x x2 −1) − 1) = x x − = 2x2 2 0 x 0 ( 2x − 1) = 0 x = 1 x = 0, = 1 x x {1} Solution set: . {0, Solution set: 1} . ( ) b. = g x ( ) d. > 0 f x 0 − + = 2 1 0 x2 x − + 1 = 0 x2 ( ( −1) − 1) = 0 x x − = 1 0 x2 (1 ) (1 ) The zero is = 1 . x − + = x x 0 2x2 − 2 = 0 x = 1; = −1 x x ( − 2x 1) = 0 x
g. f ( x ) ≥ 1 x2 − 4 ≥ 1 x2 − 5 ≥ 0 The solution set is { x x < 1 or x > 1 } or, using interval notation, (−∞, 1) ∪ (1, ∞ ) . ( x − 5 )( x + 5 ) ≥ 0 e. g ( x ) ≤ 0 2 The zeros are x = – 5 and x = 5 . − x + 1 ≤ 0 1 − x2 ≤ 0 (1 − x ) (1 + x ) ≤ 0 p ( x ) = ( x − 5 )( x + 5 ) The zeros are x = –1 and x = 1 . Interval (−∞, − 5 ) (− 5, 5 ) ( 5, ∞ ) Test Number −3 0 3 of p 4 −5 4 Conclusion Positive Negative Positive The solution set is { x x ≤− 5 or x ≥ using interval notation, (−∞, − ⎤ ⎡ 5 } or, 5, ∞ ) . The solution set is { x x ≤ −1 or x ≥ 1 } or, using interval notation, (−∞, −1] ∪ [1, ∞ ) . f. f ( x ) > g ( x ) 30. f ( x ) = x2 − 2 x + 1; g ( x ) = − x2 + 1 x2 − 2 x + 1 = − x2 + 1 a. f ( x ) = 0 x2 − 2 x + 1 = 0 ( x −1) ( x − 1) = 0 x = 1 Solution set: {1} . b. g ( x ) = 0 − x2 + 1 = 0 1 − x2 = 0 (1 − x ) (1 + x ) = 0 x = 1; x = −1 Solution set: {−1, 1} . c. f ( x ) = g ( x ) x2 − 2 x + 1 = − x2 + 1 2x2 − 2 x = 0 2x ( x − 1) = 0 x = 0, x = 1 Solution set: {0, 1} . d. f ( x ) > 0 x2 − 2 x + 1 = 0 ( x −1) ( x − 1) = 0 The zero is x = 1 . 2x2 − 2 x = 0 2x ( x − 1) = 0
Downloadable-Solution-Manual-for-Precalculus-Concepts-Through-1.docx