## Calculus (3rd Edition)

$g(t)$ has local minima at $t = -2-\sqrt{2},\ \ t= 1$ and a local maximum at $t= -2+\sqrt{2}$
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 $t= -2+\sqrt{2}$.