Absolute Extrema

Question

Locate the absolute extrema of function f(x)=2x^3+3x^2-12x on the closed interval [-4, 2].

Answer 1

If we are doing this without the aid of a graphing calculator, we begin by taking the derivative of the given function and we set the derivative equal to 0 so we can determine where the slope is 0 and where the "dips" and "peaks" are for the function on the given interval. So...

f '(x) = 6x² + 6x - 12 ; letting this = 0, we can factor the expression into 6( x+2 )(x - 1 ) so the curve "levels off" at x = -2 and at x = 1. If we use convenient x values on either side of these two critical points, we can verify that a maximum occurs at x = -2 and a minimum occurs at x = 1.

Since we are on a closed interval, we must check the function values at the endpoints, as well as checking the values at our two critical points. For the endpoints, f(-4) = -32 and f( 2) = 4 . For the critical points between the endpoints, f( -2 ) = 20 and f (1) = -7, so the absolute maximum of 20 occurs at x = -2 and the absolute minimum of -32 occurs at x = -4.