Answer
$$
f(x)=xe^{-3x}
$$
(a)
The critical numbers : $x=\frac{1}{3}$.
b)
The function is increasing , on interval $(-\infty, \frac{1}{3}).$
(c)
The function is decreasing on interval $(\frac{1}{3} , \infty ).$
Work Step by Step
$$
f(x)=xe^{-3x}
$$
We find $f^{\prime}(x) $ first, using the product rule and the chain rule.
$$
\begin{aligned}
f^{\prime}(x) &=e^{-3 x}+x\left(-3 e^{-3 x}\right) \\
&=(1-3 x) e^{-3 x} \\
&=\frac{1-3 x}{e^{3 x}}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime}(x)=0 $ , here $f^{\prime}(x)=0 $ when $x=\frac{1}{3}. $ 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 $e^{3 x}$ does not equal to 0. So, $x=\frac{1}{3} $ is the only critical number.
(a)
The critical numbers : $x=\frac{1}{3}$.
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. We observe, the number $0$ divides the number line into two intervals: $(-\infty, \frac{1}{3})$ and $(\frac{1}{3}, \infty)$. use a test point in each of these intervals to find that sign of $f^{\prime}$as follows:
(1) Test a number in the interval $(-\infty, \frac{1}{3})$ say $0$:
$$
\begin{aligned}
f^{\prime}(0) &=\frac{1-3 (0)}{e^{3 (0)}} \\
&=1 \\
&\gt 0
\end{aligned}
$$
we see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty, \frac{1}{3}).$
(2) Test a number in the interval $(\frac{1}{3}, \infty )$ say $1$:
$$
\begin{aligned}
f^{\prime}(1) &=\frac{1-3 (1)}{e^{3 (1)}} \\
&=-\frac{2}{e^3} \\
&\lt 0
\end{aligned}
$$
we see that $ f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(\frac{1}{3} , \infty ).$
So,
b)
The function is increasing , on interval $(-\infty, \frac{1}{3})$
(c)
The function is decreasing on interval $(\frac{1}{3} , \infty ).$
