Answer
$$
f(x)=xe^{x^{2}-3x}
$$
(a)
The critical numbers : $x=\frac{1}{2} $ and $x=1 $ .
b)
The function is increasing , on intervals $(-\infty, \frac{1}{2}) , (1, \infty ). $
(c)
The function is decreasing on interval $(\frac{1}{2} ,1).$
Work Step by Step
$$
f(x)=xe^{x^{2}-3x}
$$
We find $f^{\prime}(x) $ first, using the product rule and the chain rule.
$$
\begin{aligned}
f^{\prime} &=x e^{x^{2}-3 x}(2 x-3)+e^{x^{2}-3 x} (1) \\
&=e^{x^{2}-3 x}[x(2 x-3)+1] \\
&=e^{x^{2}-3 x}\left(2 x^{2}-3 x+1\right) \\
&=e^{x^{2}-3 x}(2 x-1)(x-1) \\
\end{aligned}
$$
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
$$
\begin{aligned}
f^{\prime} &=e^{x^{2}-3 x}(2 x-1)(x-1)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
(2 x-1)(x-1)&=0\\
\Rightarrow\quad\quad\quad\quad\quad\\\
x =\frac{ 1}{2} \quad\text {or } &\quad x =1
\end{aligned}
$$
So,
(a)
The critical numbers : $x=\frac{1}{2} $ and $x=1 $ .
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
$$
(-\infty, \frac{ 1}{2} ), \quad ( \frac{ 1}{2} , 1), \quad \text {and }\quad ( 1, \infty) .
$$
(1)
Test a number in the interval $(-\infty, \frac{ 1}{2} ) $ say $0$:
$$
\begin{aligned}
f^{\prime}(0) &=e^{(0)^{2}-3 (0)}(2 (0)-1)((0)-1)\\
&=1 \\
& \gt 0
\end{aligned}
$$
to see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty, \frac{ 1}{2} )$.
(2)
Test a number in the interval $( \frac{ 1}{2}, 1 ) $ say $0.6$:
$$
\begin{aligned}
f^{\prime}(0.6) &=e^{(0.6)^{2}-3 (0.6)}(2 (0.6)-1)((0.6)-1)\\
&=e^{-1.44}(0.2)(-0.4) \\
&=-\frac{0.08}{e^{1.44}}\\
& \lt 0
\end{aligned}
$$
to see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $( \frac{ 1}{2} ,1 )$.
(3)
Test a number in the interval $( 1, \infty ) $ say $2$:
$$
\begin{aligned}
f^{\prime}(2) &=e^{(2)^{2}-3 (2)}(2 (2)-1)((2)-1)\\
&=\frac{3}{e^2}\\
& \gt 0
\end{aligned}
$$
to see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(1, \infty )$.
So,
b)
The function is increasing , on intervals $(-\infty, \frac{1}{2}) , (1, \infty ). $
(c)
The function is decreasing on interval $(\frac{1}{2} ,1).$
