Answer
$g(t) $ has local minima at $ t = -2-\sqrt{2},\ \ t= 1$ and a local maximum at $ t= -2+\sqrt{2}$
Work Step by Step
Given $$g(t)=\left(t^{3}-2 t\right) e^{t}$$
Since
\begin{align*}
g'(x) &= \frac{d}{dt}\left(t^3-2t\right)e^t+\frac{d}{dt}\left(e^t\right)\left(t^3-2t\right)\\
&= \left(3t^2-2\right)e^t+e^t\left(t^3-2t\right)
\end{align*}
Then $g(t)$ has critical points when
\begin{align*}
g'(x)&=0\\
\left(3t^2-2\right)e^t+e^t\left(t^3-2t\right)&=0\\
e^{t}\left(t^{3}+3 t^{2}-2 t-2\right)&=0\\
e^{t}(t-1)\left(t^{2}+4 t+2\right)&=0
\end{align*}
Then $t=1,\ \ -\sqrt{2}\pm 2 $ are the critical points. To find the intervals of increasing and decreasing regions, choose
\begin{align*}
g'(-6)&<0 \\
g'(-2)& >0\\
g'( 0)&<0\\
g'(2)&>0
\end{align*}
Hence, by using the first derivative test, we see that $g(t) $ has local minima at $ t = -2-\sqrt{2},\ \ t= 1$ and a local maximum at $ t= -2+\sqrt{2}$.
