Question

use a graphing calculator to find the intervals on which the function is increasing or decreasing.
f(x)=x\sqrt{9 - x^{2}}, for - 3\leq x\leq3

determine the interval(s) on which the function is increasing. select the correct choice below and fill in any answer boxes in your choice.

a. the function is increasing on the interval(s)
(type your answer in interval notation. round to three decimal places as needed. use a comma to separate answers as needed.)
b. the function is never increasing.

Explanation:

Step1: Find the derivative

First, use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = x$ and $v=\sqrt{9 - x^{2}}=(9 - x^{2})^{\frac{1}{2}}$.
$u^\prime=1$, and $v^\prime=\frac{1}{2}(9 - x^{2})^{-\frac{1}{2}}\times(-2x)=\frac{-x}{\sqrt{9 - x^{2}}}$.
Then $f^\prime(x)=\sqrt{9 - x^{2}}+x\times\frac{-x}{\sqrt{9 - x^{2}}}=\frac{9 - x^{2}-x^{2}}{\sqrt{9 - x^{2}}}=\frac{9 - 2x^{2}}{\sqrt{9 - x^{2}}}$.

Step2: Find critical points

Set $f^\prime(x) = 0$, so $\frac{9 - 2x^{2}}{\sqrt{9 - x^{2}}}=0$. The numerator $9 - 2x^{2}=0$ gives $x^{2}=\frac{9}{2}$, then $x=\pm\frac{3}{\sqrt{2}}\approx\pm2.121$. Also, the derivative is undefined when $9 - x^{2}=0$, i.e., $x = \pm3$.

Step3: Test intervals

We have the intervals $[-3,-\frac{3}{\sqrt{2}}]$, $[-\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}}]$, $[\frac{3}{\sqrt{2}},3]$.
Choose test points: for $[-3,-\frac{3}{\sqrt{2}}]$, let $x=-2.5$, $f^\prime(-2.5)=\frac{9-2\times(-2.5)^{2}}{\sqrt{9 - (-2.5)^{2}}}<0$.
For $[-\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}}]$, let $x = 0$, $f^\prime(0)=\frac{9-2\times0^{2}}{\sqrt{9 - 0^{2}}}=3>0$.
For $[\frac{3}{\sqrt{2}},3]$, let $x = 2.5$, $f^\prime(2.5)=\frac{9-2\times(2.5)^{2}}{\sqrt{9 - (2.5)^{2}}}<0$.

Answer:

A. The function is increasing on the interval(s) $[-2.121,2.121]$