# What are the critical points of  f(x) = x*sqrt(8-x^2) ?

Feb 28, 2018

x= $\pm 2 , \pm 2 \sqrt{2}$

#### Explanation:

To find the critical points of a function, find where the first derivative equals 0.
For $f \left(x\right)$ stated above, $f ' \left(x\right) = \frac{8 - 2 {x}^{2}}{\sqrt{8 - {x}^{2}}}$

You need to use the product rule for derivatives: $u \cdot \mathrm{dv} + v \cdot \mathrm{du}$
where in this case, $u = x$, and $v = \sqrt{8 - {x}^{2}}$

$\mathrm{du}$ is the derivative of $u$ and the derivative of $x = 1$ so $\mathrm{du} = 1$.

$\mathrm{dv}$ is the derivative of $v$; the derivative of $\sqrt{8 - {x}^{2}}$ requires the use of the Chain Rule because you are taking the derivative of the function of a function (the square root of $\left(8 - {x}^{2}\right)$).
It is easier to represent $\sqrt{\left(8 - {x}^{2}\right)}$ using exponents:
$\sqrt{\left(8 - {x}^{2}\right)} = {\left(8 - {x}^{2}\right)}^{\frac{1}{2}}$.

Now use the Chain Rule to find the derivative of $v$:
$\frac{d}{\mathrm{dx}} \left({\left(8 - {x}^{2}\right)}^{\frac{1}{2}}\right) = \frac{1}{2} {\left(8 - {x}^{2}\right)}^{- \frac{1}{2}} \cdot - 2 x$
(Don't forget to find the derivative of the inside function --
that's where the $- 2 x$ comes from!)

Clean this up so $\mathrm{dv} = - \frac{x}{\sqrt{8 - {x}^{2}}}$

OKAY, put it all together now:

$u \cdot \mathrm{dv} + v \cdot \mathrm{du} =$
$x \cdot - \frac{x}{\sqrt{8 - {x}^{2}}} + \sqrt{8 - {x}^{2}} \cdot 1$
$= - {x}^{2} / \sqrt{8 - {x}^{2}} + \sqrt{8 - {x}^{2}}$

Combine these two fractions into one by getting common denominators and you get $f ' \left(x\right) = \frac{8 - 2 {x}^{2}}{\sqrt{8 - {x}^{2}}}$

Critical values occur when $f '$ = 0 or undefined (i.e. the
numerator = 0 and/or denominator = 0).
Setting the numerator = 0 and solving results in $x = \pm 2$
Setting the denominator = 0 and solving results in $x = \pm 2 \sqrt{2}$

If you use Symbolab, you can quickly get these answers, I hope my explanation helped! :)