Let f(x) (a) The derivative of ƒ(x) is ƒ'(x) = 6x^2+6x-336 = 2x³ + 3x² - 336x. Complete this problem without a graphing calculator. (b) As a comma-separated list, the critical points of fare x = Since f is continuous on the closed interval [–9, 15], ƒ has both an absolute maximum and an absolute minimum on the interval [-9, 15] according to the Extreme Value Theorem. To find the extreme values, we evaluate f at the endpoints and at the critical points. -8,7 (c) As a comma-separated list, the y-values corresponding to the critical points and endpoints are y = -8,7 (d) The minimum value of for I 0 15lic 1510

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