Relative Extrema of a Function

Question

(No calculator allowed)

Find the locations and types of all relative extrema for the function, f(x) = 2x³+4x²-8x-11. Remember to explain in the most concise and precise way possible.

A) f(x) has a relative maximum at x = -1/2

B) f(x) has a relative minimum at x = 2/3

C) f(x) has a relative maximum at -2/3 and a relative minimum at x = 4

D) f(x) has a relative maximum at x = -2 and a relative minimum at x = 2/3

Answer 1

Given f(x) = 2x³+4x²-8x-11 (Domain: all real numbers)

Find 1st derivative: f'(x) = 6x² + 8x - 8

Find critical values (i.e. solve for x when f'(x) = 0):

1. Let f'(x) = 6x² + 8x - 8 = 0.
2. Observe: This is a quadratic equation => Use quadratic formula to find x: x = (-b ±√(b²-4ac))/2a, where a=6, b=8, c=-8 (coefficients of x², x¹, x⁰)
3. Thus, x = 2/3 or x = -2

Find 2nd derivative: f''(x) = 12x + 8

1. With x = 2/3, f''(2/3) = 12(2/3) + 8 = 16 > 0 (=> think of concave upward) => relative min at x = 2/3
2. With x = -2, f''(-2) = 12(-2) + 8 = -16 < 0 (=> think of concave downward) => relative max at x = -2

Answer: D