# How do you find the critical numbers for y = x * sqrt( 9 - x^2) to determine the maximum and minimum?

Apr 1, 2018

There is a local minimum at $x = - \frac{3 \sqrt{2}}{2}$, and a local maximum at $x = \frac{3 \sqrt{2}}{2}$.

#### Explanation:

This equation is in factored form, so it is easy to list the $x$ and $y$ intercepts for this graph.

The $x$-intercepts are $x = 0$, $x = - 3$, and $x = 3$. The $y$-intercept occurs at $y = 0$.

Also note that $- 3 \le x \le 3$.

Now let's take the derivative and find it's intercepts.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sqrt{9 - {x}^{2}} - {x}^{2} / \left(\sqrt{9 - {x}^{2}}\right) = \frac{9 - 2 {x}^{2}}{\sqrt{9 - {x}^{2}}}$

To find the roots of the derivative, set

$\frac{9 - 2 {x}^{2}}{\sqrt{9 - {x}^{2}}} = 0$.

This only happens when

$2 {x}^{2} = 9$

$x = \pm \frac{3 \sqrt{2}}{2}$

When $- 3 \le x \le - \frac{3 \sqrt{2}}{2}$, $y$ is decreasing because $\frac{\mathrm{dy}}{\mathrm{dx}} \le 0$.

When $- \frac{3 \sqrt{2}}{2} \le x \le 3 \frac{\sqrt{2}}{2}$, $y$ is increasing because $\frac{\mathrm{dy}}{\mathrm{dx}} \ge 0$.

This means that there is a local minimum at $x = - \frac{3 \sqrt{2}}{2}$.

When $\frac{3 \sqrt{2}}{2} \le x \le 3$, $y$ is decreasing because $\frac{\mathrm{dy}}{\mathrm{dx}} \le 0$.

This means that there is a local maximum at $x = \frac{3 \sqrt{2}}{2}$.

graph{xsqrt(9-x^2) [-4, 4, -5, 5]}