Answer
$$
f(x)=x 2^{-x^{2}}
$$
(a)
The critical numbers only : $x=\frac{1} {\sqrt {2\ln 2}}\approx 0.8493 $ and $x=-\frac{1} {\sqrt {2\ln 2}} $ .
b)
The function is increasing , on interval $ (- \frac{1} {\sqrt {2\ln 2}} , \frac{1} {\sqrt {2\ln 2}} ). $
(c)
The function is decreasing on intervals $(-\infty , - \frac{1} {\sqrt {2\ln 2}} ) $ and $ ( \frac{1} {\sqrt {2\ln 2}} , \infty ).$
Work Step by Step
$$
f(x)=x 2^{-x^{2}}
$$
We find $f^{\prime}(x) $ first, using the product rule and the chain rule.
$$
\begin{aligned}
f^{\prime}(x) &=x\left[\ln 2\left(2^{-x^{2}}(-2 x)\right)\right]+\left(2^{-x^{2}}\right) \\
&=2^{-x^{2}}\left(-2 x^{2} \ln 2+1\right) \\
&=\frac{1-2 x^{2} \ln 2}{2^{x}}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime} =0 $. Now, solve the equation $f^{\prime} =0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =\frac{1-2 x^{2} \ln 2}{2^{x}}&=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
1-2 x^{2} \ln 2 &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
2 x^{2} \ln 2 &=1
\\
\Rightarrow\quad\quad\quad\quad\quad\\
x^{2} &=\frac{1}{2\ln 2}
\\
\Rightarrow\quad\quad\quad\quad\quad\\
x =\frac{1} {\sqrt {2\ln 2}} \quad\text {or } &\quad x =-\frac{1} {\sqrt {2\ln 2}}
\end{aligned}
$$
Next, we find any values of $x$ where $f^{\prime}(x) $ fails to exist. This occurs whenever the denominator of $f^{\prime}(x) $ is $0$, here we observe that the denominator $2^{ x}$ does not equal to 0. So,
(a)
The critical numbers only : $x=\frac{1} {\sqrt {2\ln 2}}\approx 0.8493 $ and $x=-\frac{1} {\sqrt {2\ln 2}}$ .
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. Now, check the sign of $f^{\prime}(x)$ in the intervals
$$
(-\infty, - \frac{1} {\sqrt {2\ln 2}} ), \quad (- \frac{1} {\sqrt {2\ln 2}} , \frac{1} {\sqrt {2\ln 2}} ) \quad \text {and }\quad ( \frac{1} {\sqrt {2\ln 2}} , \infty) .
$$
(1)
Test a number in the interval $(-\infty, - \frac{1} {\sqrt {2\ln 2}} ) $ say $-1$:
$$
\begin{aligned}
f^{\prime}(-1) &=\frac{1-2 (-1)^{2} \ln 2}{2^{(-1)}}\\
&=2\left(1-2\ln \left(2\right)\right) \\
& \approx -0.77258\\
& \lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty, - \frac{1} {\sqrt {2\ln 2}} )$.
(2)
Test a number in the interval $(- \frac{1} {\sqrt {2\ln 2}}, \frac{1} {\sqrt {2\ln 2}} ) $ say $0$:
$$
\begin{aligned}
f^{\prime}(0) &=\frac{1-2 (0)^{2} \ln 2}{2^{(0)}}\\
&=1 \\
&\gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(- \frac{1} {\sqrt {2\ln 2}}, \frac{1} {\sqrt {2\ln 2}} ) $.
(3)
Test a number in the interval $( \frac{1} {\sqrt {2\ln 2}} , \infty ) $ say $1$:
$$
\begin{aligned}
f^{\prime}(1) &=\frac{1-2 (1)^{2} \ln 2}{2^{(1)}}\\
&=\frac{1-2\ln \left(2\right)}{2} \\
& \approx -0.1931\\
&\lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $ ( \frac{1} {\sqrt {2\ln 2}} , \infty ) $.
So,
b)
The function is increasing , on interval $ (- \frac{1} {\sqrt {2\ln 2}} , \frac{1} {\sqrt {2\ln 2}} ). $
(c)
The function is decreasing on intervals $(-\infty , - \frac{1} {\sqrt {2\ln 2}} ) $ and $ ( \frac{1} {\sqrt {2\ln 2}} , \infty ).$
