− 3x −10 = x2 0 ( − 5)( + 2) x x 7. − 3x −10 < 0 x2 = 0 = − − ( 3x 10 f x) x2 = 5, =− 2 are the zeros of . x x f {x 3. a. graph of is above the axis. Thus, < f x- x −2 > or 2} or, using interval notation, x ( > 0 f x) (−∞, (2, ) when the { } −2) ∪ ∞ . The solution set is − 2 < < 5 or, using x x b. ( ≤ 0 when the graph of is below or f x) f {x (−2, interval notation, 5) . intersects the Thus, − 2 ≤ ≤ 2} x-axis. x [−2, or, using interval notation, 2] .
7. x2 − 3x −10 < 0 f ( x) = x2 − 3x − 10 x2 − 3x −10 = 0 ( x − 5)( x + 2) = 0 x = 5, x =− 2 are the zeros of f . 3. a. f ( x) > 0 when the graph of f is above the x- axis. Thus, {x x < −2 or x > 2} or, using interval notation, (−∞, −2) ∪ (2, ∞ ) . The solution set is { x − 2 < x < 5 } or, using b. f ( x) ≤ 0 when the graph of f is below or intersects the x-axis. Thus, {x − 2 ≤ x ≤ 2} or, using interval notation, [−2, 2] . interval notation, (−2, 5) .
Downloadable-Solution-Manual-for-Precalculus-Concepts-Through-1.docx