Hi! The absolute extrema of a continuous function f(x) on a closed interval [a,b] either occur at the endpoints (x = a or x = b) or at the critical numbers of f(x) inside the open interval (a,b).
In this case, the endpoints are x = -1 and x = 6.
Let's find the critical numbers by solving f'(x) = 0.
f'(x) = 0
12x3 - 72x2 + 60x = 0
12x(x2 - 6x + 5) = 0
12x(x - 1)(x - 5) = 0
The critical numbers of f(x) are thus x = 0, 1, and 5, all of which are in the open interval (-1,6). We can determine the absolute max of f(x) on [-1,6] by plugging the critical numbers and the endpoints into f(x).
f(-1) = 57
f(0) = 0
f(1) = 9
f(5) = -375
f(6) = -216
The absolute max is f(-1) = 57, and the absolute min is f(5) = -375.
To determine local max and min values, use the Second Derivative Test. Local extrema only occur at critical numbers, so we only need to test x = 0, 1, and 5.
f''(x) = 36x2 - 144x + 60
f''(0) = 60 >0 ⇒ (0,0) is a local minimum
f''(1) = -48 < 0 ⇒ (1,9) is a local maximum
f''(5) = 240 > 0 ⇒ (5,-375) is a local minimum
