# How do you locate the absolute extrema of the function f(x)=x^3-12x on the closed interval [0,4]?

Jul 1, 2015

The absolute extrema of a function, $f$, that is continuous on a closed interval must occur at either a critical number for $f$ in the interval, or at an endpoint of the interval.

#### Explanation:

So consider:

$f \left(x\right) = {x}^{3} - 12 x$

$f ' \left(x\right) = 3 {x}^{2} - 12$.

$f '$ is polynomial, so every critical number for $f$ is a zero for $f '$.

The zeros of $f '$ are: $3 \left({x}^{2} - 4\right) = 0$, $x = - 2 , 2$. But $- 2$ is not in the interval $\left[0 , 4\right]$.
So the only critical number in the interval is $2$.

The minimum and maximum must occur at one of the values

$x =$ $0$ or $2$ or $4$.

To finish, evaluate $f$ at each of these values.

$f \left(0\right) = 0$

$f \left(2\right) = 8 - 24 = - 16$

$f \left(4\right) = 64 - 48 = 16$

The minimum is $- 16$ (it occurs at $2$)
The maximum is $16$ (it occurs at $4$).

To understand what we've done it may help to see the graph:

graph{y=(x^3-12x)*sqrt(4-(x-2)^2)/(sqrt(4-(x-2)^2)) [-52.2, 40.28, -23.87, 22.4]}