# How do use the first derivative test to determine the local extrema y=x(sqrt(8-x^2))?

Aug 13, 2015

$x = \pm 2 , y = \pm 4$

#### Explanation:

A stationary point can be obtained from $f ' \left(x\right) = 0$

$y = x \sqrt{8 - {x}^{2}}$
$y ' = \sqrt{8 - {x}^{2}} - {x}^{2} / \sqrt{8 - {x}^{2}} = \frac{- 2 \left({x}^{2} - 4\right)}{\sqrt{8 - {x}^{2}}}$

$\frac{- 2 \left({x}^{2} - 4\right)}{\sqrt{8 - {x}^{2}}} = 0$
${x}^{2} = 4$
$x = \pm 2 , y = \pm 4$

You can then proceed to find whether these points are local maxima or minima by taking the second derivative and substituting the values of the stationary points, $\alpha$:
$f ' ' \left(\alpha\right) > 0 \Rightarrow$ minima
$f ' ' \left(\alpha\right) < 0 \Rightarrow$ maxima

$y ' ' = - \frac{2 x \left(12 - {x}^{2}\right)}{8 - {x}^{2}} ^ \left(\frac{3}{2}\right)$
$y ' {'}_{x = 2} = - 4 < 0 \Rightarrow$ maxima
$y ' {'}_{x = - 2} = 4 > 0 \Rightarrow$ minima

graph{xsqrt(8-x^2) [-5, 5, -6, 6]}