### Home > APCALC > Chapter 5 > Lesson 5.5.1 > Problem5-154

5-154.

Without using a graphing calculator, determine the maximum and minimum values of the function $p(x) = (x^2 – 4)^2$ over the interval $[–1, 5]$.

Candidates for global maxima and global minima exist where the derivative equals $0$ AND where the derivative does not exist.... Notice that this is a function that has a closed domain. That means that we already know two candidates for global max and min: the endpoints.

Evaluate the endpoints:
$p\left(−1\right) = 9$
$p\left(5\right)=21^2=441$
These might be the global min and max, respectively. We still need to check the other candidates.

Find additional candidates by setting the derivative equal to zero, and solving for $x.$
$p'\left(x\right)=4x\left(x^2−4\right)$
$0 = 4x\left(x − 2\right)\left(x + 2\right)$
$x = 0, x = 2$ and $x = −2$ are candidates...
but eliminate $x = −2$ because it is not within the given domain.

Evaluate the remaining candidates: $p\left(0\right) = 16, p\left(2\right) = 0$ and compare to the endpoint candidates.

Recall that maxima and minima are $y$-values.
Therefore the global maximum is $441$ and the global minimum is $0$. It should also be noted that $16$ is a local maximum.
Confirm this by sketching $p\left(x\right)$ on your graphing calculator.