advertisement

MATH 147 Quiz Key #7 3/31/2016 (1) Consider the function f ( x ) = 13 x3 + 21 x2 − 6x + 2. (a) Determine whether f is increasing or decreasing. (b) Determine where f is concave up or down and the x-coordinates of all inflection points. Solution: (a) f 0 ( x ) = x2 + x − 6. Therefore, f 0 ( x ) = ( x + 3) ( x − 2) = 0. Therefore, the critical values are at x = −3 and x = 2. We now make a sign chart as follows: x+3 x−2 f 0 (x) (−∞, −3) (−3, 2) (2, ∞) − + + − − + + − + Therefore, f is increasing on (−∞, −3) ∪ (2, ∞) and decreasing on (−3, 2) . (b) f 00 ( x ) = 2x + 1. Hence, the only possible point of inflection is at x = − 21 . We make a sign chart: − 12 , ∞ −∞, − 21 f 00 ( x ) − + Therefore, f is concave up on − 21 , ∞ and concave down on −∞, − 21 , and it has a point of inflection at x = − 12 . 1 2 (2) Consider the function f ( x ) = e− x . (a) Determine whether f is increasing or decreasing. (b) Determine where f is concave up or down and the x-coordinates of all inflection points. Solution: (a) f 0 (x) = 2 2 d h 2i d h − x2 i = e− x · e − x = −2xe− x . dx dx To find the critical values of f , we need to set the derivative equal to zero. However, since no exponential function is ever zero, f 0 ( x ) = 0 only when x = 0. Now, we make a sign chart: −2x 2 e− x 0 f (x) (−∞, 0) (0, ∞) + − + + + − Hence, f is increasing on (−∞, 0) and decreasing on (0, ∞) . (b) i 2 2 d h d d h − x2 i f 00 ( x ) = −2xe− x = e . [−2x ] · e− x − 2x · dx dx dx 2 2 2 ∴ f 00 ( x ) = −2e− x − 2x (−2x ) e− x = −2e− x 1 − 2x2 . Now to solve f 00 ( x ) = 0, we only need to find solutions of 1 − 2x2 = 0. 1 − 2x2 = 0 2x2 = 1. 1 x2 = . 2 1 x = ±√ . 2 Hence, the values at which inflection points can occur are these two values. We now set up a sign chart as follows: √1 , ∞ −∞, − √1 − √1 , √1 2 −2e− x 1 − 2x2 f 00 ( x ) 2 2 − − + 2 2 − − + − − + ∪ √1 , ∞ and concave down on Therefore, f is concave up on −∞, − √1 2 2 − √1 , √1 , as well as inflection points at x = ± √1 . 2 2 2 2