Answer
$$
f(x) =3 x e^{x}+2
$$
(*)
A relative minimum occurs at $x=-1,$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(-1) &=3 (-1) e^{(-1)}+2\\
&=-\frac{3}{e}+2 \\
& \approx 0.896
\end{aligned}
$$
Work Step by Step
$$
f(x) =3 x e^{x}+2
$$
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=3 x e^{x}+3 e^{x} \\
&=3 e^{x}(x+1)
\end{aligned}
$$
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
$$
\begin{aligned}
f^{\prime}(x) =3 e^{x}(x+1)&=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
e^{x}(x+1) &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
(x+1)=0 & \quad \quad or \quad \quad e^{x} =0 \\
x=-1 & \quad \quad \text {since } e^{x} \text {never equal zero} .\\
\end{aligned}
$$
So
The only critical number is: $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
$$
(-\infty, -1 ), \quad\quad (-1 ,\infty ) .
$$
(1)
Test a number in the interval $(-\infty, -1 )$ say $-2$:
$$
\begin{aligned}
f^{\prime}(-2) &=3 e^{(-2)}((-2)+1)\\
&=-\frac{3}{e^2} \\
&\approx-0.406\\
&\lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty ,-1 )$.
(2)
Test a number in the interval $(-1, \infty )$ say $0 $:
$$
\begin{aligned}
f^{\prime}(0) &=3 e^{(0)}((0)+1)\\
&=3\\
&\gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-1, \infty )$.
From (1), (2) we find that:
(*)
The function $f$ is decreasing on interval $ ( -\infty , -1) , $ and it is increasing on interval $ (-1 ,\infty ) . $ So, a relative minimum occurs at $x=-1,$ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(-1) &=3 (-1) e^{(-1)}+2\\
&=-\frac{3}{e}+2 \\
& \approx 0.896
\end{aligned}
$$
