< 0 8. + 3x −10 > 0 x2 11. ( ) ( = − 9 f x) x2 = + 3x −10 f x x2 − 9 = x2 + −10 = 3x x2 x2 − + 0 ( 3)( x x − 0 ( + 5)( − 2) x x 9 = 3) 0 = 0 = −3, = 3 are the zeros of . = −5, = x x f 2 are the of . x x f Interval (−∞, − (−5, 2) (2, ∞) 5) Test −6 0 6 Number Value of 8 −10 8 f Conclusion Positive Negative Positive { } < −5 > The solution set is or 2 x x x { } The solution set is − 3 < < 3 or, using x x or, (−3, (−∞, (2, notation, 3) . using interval notation, −5) ∪ ∞ ) . − 1 < 0 x2 12. ( = −1 f x) x2 − > 9. 4 0 x2 x − 1 = x2 ( = − 4 f x) x2 x 0 ( − 1)( + x x − = 4 0 x2 x 1) = 0 − 4) = 0 x( x = −1, = 1 are the zeros of . x x f = 0, = 4 are the zeros of . x x f Interval (−∞, 0) (0, 4) (4, ∞) Test −1 1 5 Number Value of 5 −3 5 f Conclusion Positive Negative Positive { } The solution set is < 0 or > 4 or, x x x { } The solution set is −1 < < 1 or, using x x (−1, (4, ) 1) . ∪ ∞ 0) . 13. + > 12 x2 x + 8x > 0 x2 10. = + + −12 > ( 8x 0 f x) x2 x2 x using ( = + −12 f x) x2 x + 8x = 0 x2 + = + − 12 = 8) 0 x2 x x( x no = = 0 ( + 4)( − 8, 0 are the zeros of . x x x x f – = 3) 0 (−∞, = −4, = 3 are the of x x f
8. x2 + 3x −10 > 0 f ( x ) = x2 + 3x −10 x2 + 3x −10 = 0 ( x + 5)( x − 2) = 0 x = −5, x = 2 are the of f . 11. x2 − 9 < 0 f ( x) = x2 − 9 x2 − 9 = 0 ( x − 3)( x + 3) = 0 x = −3, x = 3 are the zeros of f . Interval (−∞, − 5) (−5, 2) (2, ∞) Test Number −6 0 6 Value of f 8 −10 8 Conclusion Positive Negative Positive The solution set is { x x < −5 or x > 2 } or, The solution set is { x − 3 < x < 3 } or, using using interval notation, (−∞, −5) ∪ (2, ∞ ) . 9. x2 − 4 x > 0 f ( x) = x2 − 4 x x2 − 4 x = 0 x( x − 4) = 0 x = 0, x = 4 are the zeros of f . 12. notation, (−3, 3) . x2 − 1 < 0 f ( x) = x2 −1 x2 − 1 = 0 ( x − 1)( x + 1) = 0 x = −1, x = 1 are the zeros of f . Interval (−∞, 0) (0, 4) (4, ∞) Test Number −1 1 5 Value of f 5 −3 5 Conclusion Positive Negative Positive The solution set is { x x < 0 or x > 4 } or, The solution set is { x −1 < x < 1 } or, using 10. using (−∞, 0) ∪ (4, ∞ ) . x2 + 8x > 0 f ( x) = x2 + 8x x2 + 8x = 0 x( x + 8) = 0 x = – 8, x = 0 are the zeros of f . 13. (−1, 1) . x2 + x > 12 x2 + x −12 > 0 f ( x) = x2 + x −12 x2 + x − 12 = 0 ( x + 4)( x − 3) = 0 x = −4, x = 3 are the of f no
Downloadable-Solution-Manual-for-Precalculus-Concepts-Through-1.docx