What are the absolute extrema of  f(x)= xsqrt(25-x^2) in [-4,5]?

Feb 5, 2016

The absolute minimum is $- \frac{25}{2}$ (at $x = - \sqrt{\frac{25}{2}}$). The absolute maximum is $\frac{25}{2}$ (at $x = \sqrt{\frac{25}{2}}$).

Explanation:

$f \left(- 4\right) = - 12$ and $f \left(5\right) = 0$

$f ' \left(x\right) = \sqrt{25 - {x}^{2}} + \frac{x}{\cancel{2} \sqrt{25 - {x}^{2}}} \cdot - \cancel{2} x$

$= \frac{25 - {x}^{2} - {x}^{2}}{\sqrt{25 - {x}^{2}}} = \frac{25 - 2 {x}^{2}}{\sqrt{25 - {x}^{2}}}$

The critical numbers of $f$ are $x = \pm \sqrt{\frac{25}{2}}$ Both of these are in $\left[- 4 , 5\right]$..

$f \left(- \sqrt{\frac{25}{2}}\right) = - \sqrt{\frac{25}{2}} \sqrt{25 - \frac{25}{2}}$

$= - \sqrt{\frac{25}{2}} \sqrt{\frac{25}{2}} = - \frac{25}{2}$

By symmetry ($f$ is odd), $f \left(\sqrt{\frac{25}{2}}\right) = \frac{25}{2}$

Summary:
$f \left(- 4\right) = - 12$
$f \left(- \sqrt{\frac{25}{2}}\right) = - \frac{25}{2}$
$f \left(\sqrt{\frac{25}{2}}\right) = \frac{25}{2}$
$f \left(5\right) = 0$

The absolute minimum is $- \frac{25}{2}$ (at $x = - \sqrt{\frac{25}{2}}$).
The absolute maximum is $\frac{25}{2}$ (at $x = \sqrt{\frac{25}{2}}$).