Answer
$g(x)$ has the absolute maximum value of $1/e$ at $x=1$ and the absolute minimum value of $-e$ at $x=-1$ in the defined domain.
Work Step by Step
$$g(x)=xe^{-x}\hspace{1cm}-1\le x\le1$$
1) Find the derivative $g'(x)$:
$$g'(x)=(x)'e^{-x}+x(e^{-x})'$$ $$g'(x)=e^{-x}+xe^{-x}(-x)'$$ $$g'(x)=e^{-x}-xe^{-x}$$
For $g'(x)=0$, we have $$e^{-x}-xe^{-x}=0$$ $$xe^{-x}=e^{-x}$$ $$x=1$$
As $x=1$ lies in the domain $[-1,1]$, $x=1$ is the critical point of $g(x)$.
2) Evaluate $g(x)$ at the critical points and endpoints:
- For $x=-1$: $$g(-1)=-1\times e^{-(-1)}=-e^1=-e$$
- For $x=1$: $$g(1)=1\times e^{-1}=\frac{1}{e}$$
This means $g(x)$ has the absolute maximum value of $1/e$ at $x=1$ and the absolute minimum value of $-e$ at $x=-1$ in the defined domain.