Find and Classify Extrema of a Function

Question

(No calculator allowed)

Find and classify the relative extrema of the function g(x) = 4x³-3x²-6x+5. Remember to explain in the most concise and precise way possible.

A) Relative maximum at x = -1; relative minimum at x = 2

B) Relative minimum at x = -1/3; relative maximum at 3/2

C) Relative maximum at x = 1; relative minimum at x = -1

D) Relative minimum at x = 1; relative maximum at x = -1/2

Answer 1

Answer 2

Given g(x) = 4x³-3x²-6x+5 (Domain: all real numbers)

Find 1st derivative: g'(x) = 12x² - 6x - 6

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

1. Let g'(x) = 12x² - 6x - 6 = 0.
2. Observe: This is a quadratic equation => Use quadratic formula to find x: x = (-b ±√(b²-4ac))/2a, where a=12, b=-6, c=-6 (coefficients of x², x¹, x⁰)
3. Thus, x = 1 or x = -1/2

Find 2nd derivative: g''(x) = 24x - 6

1. With x = 1, g''(1) = 24(1) - 6 = 18 > 0 (=> think of concave upward) => relative min at x = 1
2. With x = -1/2, g''(-1/2) = 24(-1/2) - 6 = -18 < 0 (=> think of concave downward) => relative max at x = -1/2

Answer: D