# How do use the first derivative test to determine the local extrema f(x) = xsqrt(25-x^2)?

Oct 14, 2015

The extrema of your functions are $\setminus \pm \frac{5}{\sqrt{2}}$.

#### Explanation:

You compute the first derivative and find its roots. The derivative is computed as follows:

$D \left(x \cdot \sqrt{25 - {x}^{2}}\right) = D \left(x\right) \cdot \sqrt{25 - {x}^{2}} + x \cdot D \left(\sqrt{25 - {x}^{2}}\right)$.

$D \left(x\right)$ is obviously one, while

$D \left(\sqrt{25 - {x}^{2}}\right) = D {\left(25 - {x}^{2}\right)}^{\frac{1}{2}} = \frac{1}{2} {\left(25 - {x}^{2}\right)}^{- \frac{1}{2}} \cdot D \left(25 - {x}^{2}\right)$

Since $D \left(25 - {x}^{2}\right) = - 2 x$, the whole expression becomes

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

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

$= \frac{\left(25 - {x}^{2}\right) - {x}^{2}}{\sqrt{25 - {x}^{2}}} = \frac{25 - 2 {x}^{2}}{\sqrt{25 - {x}^{2}}}$.

This function equals zero if and only if its numerator equals zero, so

$f ' \left(x\right) = 0 \setminus \iff 2 {x}^{2} = 25 \setminus \iff {x}^{2} = \frac{25}{2} \setminus \iff x = \setminus \pm \frac{5}{\sqrt{2}}$.

These are the local extrema of your function.