# What are the absolute extrema of f(x) =x^4 − 8x^2 − 12 in[-3,-1]?

May 30, 2016

$- 3$ (occurring at $x = - 3$) and $- 28$ (occurring at $x = - 2$)

#### Explanation:

Absolute extrema of a closed interval occur at the endpoints of the interval or at $f ' \left(x\right) = 0$.

That means we'll have to set the derivative equal to $0$ and see what $x$-values that gets us, and we'll have to use $x = - 3$ and $x = - 1$ (because these are the endpoints).

So, starting with taking the derivative:
$f \left(x\right) = {x}^{4} - 8 {x}^{2} - 12$
$f ' \left(x\right) = 4 {x}^{3} - 16 x$

Setting it equal to $0$ and solving:
$0 = 4 {x}^{3} - 16 x$
$0 = {x}^{3} - 4 x$
$0 = x \left({x}^{2} - 4\right)$
$x = 0$ and ${x}^{2} - 4 = 0$
Thus the solutions are $0 , 2 ,$ and $- 2$.

We immediately get rid of $0$ and $2$ because they are not on the interval $\left[- 3 , - 1\right]$, leaving only $x = - 3 , - 2 ,$ and $- 1$ as the possible places where extrema can occur.

Finally, we evaluate these one by one to see what the absolute min and max are:
$f \left(- 3\right) = - 3$
$f \left(- 2\right) = - 28$
$f \left(- 1\right) = - 19$

Therefore $- 3$ is the absolute maximum and $- 28$ is the absolute minimum on the interval $\left[- 3 , - 1\right]$.