3. Let f(x) = 5 + 6x + 12x2 – 2x3 – x4 and let g(x) = f’(x) for -¥ < x < ¥ . At what value of x is g(x) increasing most rapidly?

A. -2

B. –(1/2)

C. 1/2

D. 1

E. 2

 

Solution: The answer is B

 

To ask for the value of x such that g(x) is increasing most rapidly is the same as asking for the value of x that maximizes g’(x). Let h(x) = g’(x) = f"(x). Hence, taking the derivative of f(x) twice yields,

Find the value of x that maximizes h(x) by setting h’(x) = 0 and solving for x, as follows,

Furthermore, notice that h(x) is a parabola that opens downward so that it achieves it’s maximum value at the x coordinate of it’s vertex (x = -b/2a) which is the same value of the x that results in the first derivative of h(x) equaling zero. Also, since h’(x) goes from a positive to a negative value at x = -1/2, then h(x) must attain its maximum value at x = -1/2.