Answer
$$
f(x)=x^{2}2^{-x}
$$
(a)
The critical numbers only : $x=0$ and $x=\frac{2}{ \ln 2} \approx 2.885$ .
b)
The function is increasing , on interval $ (0,\frac{2}{ \ln 2} ). $
(c)
The function is decreasing on intervals $(-\infty ,0 )$ and $(\frac{2}{ \ln 2} , \infty ) .$
Work Step by Step
$$
f(x)=x^{2}2^{-x}
$$
We find $f^{\prime}(x) $ first, using the product rule and the chain rule.
$$
\begin{aligned}
f^{\prime}(x) &=x^{2}\left[\ln 2\left(2^{-x}\right)(-1)\right]+\left(2^{-x}\right) 2 x \\
&=2^{-x}\left(-x^{2} \ln 2+2 x\right) \\
&=\frac{x(2-x \ln 2)}{2^{x}}
\end{aligned}
$$
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
$$
\begin{aligned}
f^{\prime}(x) =\frac{x(2-x \ln 2)}{2^{x}}&=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
x(2-x \ln 2)&=0\\
\Rightarrow\quad\quad\quad\quad\quad\\\
x =0 \quad\text {or } &\quad x =\frac{2}{ \ln 2}
\end{aligned}
$$
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 : $x=0$ and $x=\frac{2}{ \ln 2} \approx 2.885$ .
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
$$
(-\infty, 0), \quad ( 0 , \frac{2}{ \ln 2} ) \quad \text {and }\quad (\frac{2}{ \ln 2}, \infty) .
$$
(1)
Test a number in the interval $(-\infty, 0) $ say $-1$:
$$
\begin{aligned}
f^{\prime}(-1) &=\frac{(-1)(2-(-1) \ln 2)}{2^{(-1)}}\\
&=-2\left(2+\ln \left(2\right)\right) \\
& \lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty, 0 )$.
(2)
Test a number in the interval $(0, \frac{2}{ \ln 2} ) $ say $1$:
$$
\begin{aligned}
f^{\prime}(1) &=\frac{(1)(2-(1) \ln 2)}{2^{(1)}}\\
&=\frac{2-\ln \left(2\right)}{2} \approx \:0.6534\\
&\gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(0, \frac{2}{ \ln 2} )$.
(3)
Test a number in the interval $( \frac{2}{ \ln 2} , \infty ) $ say $3$:
$$
\begin{aligned}
f^{\prime}(3) &=\frac{(3)(2-(3) \ln 2)}{2^{(3)}}\\
&=\frac{3\left(2-3\ln \left(2\right)\right)}{8} \approx -0.02979\\
&\lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $( \frac{2}{ \ln 2} , \infty ) $.
So,
b)
The function is increasing , on interval $ (0,\frac{2}{ \ln 2} ). $
(c)
The function is decreasing on intervals $(-\infty ,0 )$ and $
(\frac{2}{ \ln 2} , \infty ) .$
