# What are the critical points for f(x) = xsqrt(16 - x^2)?

Jun 26, 2015

The critical numbers are $\pm 2 \sqrt{2}$, and $\pm 4$.

#### Explanation:

Critical numbers for a function, $f$, are numbers in the domain $f$ at which $f ' \left(x\right) = 0$ or $f ' \left(x\right)$ does not exist.

For, $f \left(x\right) = x \sqrt{16 - {x}^{2}}$, the derivative is:

$f ' \left(x\right) = \sqrt{16 - {x}^{2}} + x \frac{1}{2 \sqrt{16 - {x}^{2}}} \cdot \left(- 2 x\right)$

$= \frac{\sqrt{16 - {x}^{2}}}{1} - {x}^{2} / \sqrt{16 - {x}^{2}}$

$= \frac{16 - 2 {x}^{2}}{\sqrt{16 - {x}^{2}}}$

Derivative is 0
$f ' \left(x\right) = 0$ when $16 - 2 {x}^{2} = 0$.
Which happens at $x = \pm \sqrt{8} = \pm 2 \sqrt{2}$.

Both $\sqrt{8}$ and $- \sqrt{8}$ are in the domain of $f$, so they are both critical numbers.

Derivative does not exist
$f ' \left(x\right)$ does not exist when $16 - {x}^{2} = 0$, which happens at $\pm 4$.
Both $4$ and $- 4$ are in the domain of $f$, so they are both critical numbers.