Answer
$$
y=x\sqrt {9-x^{2}}
$$
(a)
The critical numbers are $x=3,-3, -\frac{ 3}{\sqrt {2}} $ and $\frac{ 3}{\sqrt {2}} $ .
(b)
The function is increasing , on interval $(-\frac{ 3}{\sqrt {2}} , \frac{ 3}{\sqrt {2}} ) $.
(c)
The function is decreasing on intervals $(-3 , -\frac{ 3}{\sqrt {2}})$ , and $(\frac{ 3}{\sqrt {2}} , 3)$ .
Work Step by Step
$$
y=x\sqrt {9-x^{2}}
$$
Notice that the function $f$ does not exist for $9-x^{2} \lt 0 $ ,So the function $f$ is defined on interval $[-3,3]$.
We find $y^{\prime} $ first, using the product rule.
$$
\begin{aligned}
y^{\prime}=&(1)\left(9-x^{2}\right)^{1 / 2} \\
&+\frac{1}{2}\left(9-x^{2}\right)^{-1 / 2}(-2 x)(x) \\
=&\left(9-x^{2}\right)^{1 / 2}-x^{2}\left(9-x^{2}\right)^{-1 / 2} \\
=&\left(9-x^{2}\right)^{-1 / 2}\left(9-x^{2}-x^{2}\right) \\
=&\left(9-x^{2}\right)^{-1 / 2}\left(9-2 x^{2}\right) \\
=& \frac{9-2 x^{2}}{\sqrt{9-x^{2}}}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $y^{\prime} =0 $.
Now, solve the equation $y^{\prime} =0 $ to get
$$
\begin{aligned}
y^{\prime} &=\frac{9-2 x^{2}}{\sqrt{9-x^{2}}}=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
9-2 x^{2}&=0\\
\Rightarrow\quad\quad\quad\quad\quad\\\
x =\frac{ 3}{\sqrt {2}} \quad\text {or } &\quad x =\frac{ -3}{\sqrt {2}}
\end{aligned}
$$
Next, we find any values of $x$ where fails to exist. This
occurs whenever the denominator of $y^{\prime}$ is $0$, so set the denominator equal to $0$. and solve.
$$
\begin{aligned}
\sqrt{9-x^{2}}&=0\\
9-x^{2}&=0\\
x^{2}&=9\\
\Rightarrow\quad\quad
x =3 \quad\text {or } &\quad x =-3
\end{aligned}
$$
So,
(a)
The critical numbers are $x=3,-3, -\frac{ 3}{\sqrt {2}} $ and $\frac{ 3}{\sqrt {2}} $ .
Now, we can use the first derivative test. Check the sign of $y^{\prime}(x)$.in the intervals
$$
(-3, -\frac{ 3}{\sqrt {2}} ), \quad ( -\frac{ 3}{\sqrt {2}} , \frac{ 3}{\sqrt {2}}), \quad \text {and }\quad ( \frac{ 3}{\sqrt {2}}, 3) .
$$
(1)
Test a number in the interval $(-3, -\frac{ 3}{\sqrt {2}} ) $ say $-\frac{5}{2}$:
$$
\begin{aligned}
f^{\prime}(-\frac{5}{2}) &=\frac{9-2(-\frac{5}{2})^{2}}{\sqrt{9-(-\frac{5}{2})^{2}}}\\
&=-\frac{7\sqrt{11}}{11} \approx -2.1106 \\
& \lt 0
\end{aligned}
$$
to see that $y^{\prime}(x)$ is negative in that interval, so $y(x)$ is decreasing on $(-3, -\frac{ 3}{\sqrt {2}} )$.
(2)
Test a number in the interval $(-\frac{ 3}{\sqrt {2}} , \frac{ 3}{\sqrt {2}} ) $ say $0$:
$$
\begin{aligned}
f^{\prime}(0) &=\frac{9-2(0)^{2}}{\sqrt{9-(0)^{2}}}\\
&=3\\
&\gt 0
\end{aligned}
$$
to see that $y^{\prime}(x)$ is positive in that interval, so $y(x)$ is increasing on $(-\frac{ 3}{\sqrt {2}} , \frac{ 3}{\sqrt {2}} ) $.
(3)
Test a number in the interval $(\frac{ 3}{\sqrt {2}} , 3 ) $ say $\frac{5}{2}$:
$$
\begin{aligned}
f^{\prime}(\frac{5}{2}) &=\frac{9-2(\frac{5}{2})^{2}}{\sqrt{9-(\frac{5}{2})^{2}}}\\
&=-\frac{7\sqrt{11}}{11} \approx -2.1106 \\
& \lt 0
\end{aligned}
$$
to see that $y^{\prime}(x)$ is negative in that interval, so $y(x)$ is decreasing on $(\frac{ 3}{\sqrt {2}} , 3)$.
So,
(b)
The function is increasing , on interval $(-\frac{ 3}{\sqrt {2}} , \frac{ 3}{\sqrt {2}} ) $.
(c)
The function is decreasing on intervals $(-3 , -\frac{ 3}{\sqrt {2}})$ , and $(\frac{ 3}{\sqrt {2}} , 3)$ .
