Here's the function: $f(x) = x^2 e^{-x/2}, [-2,8]$
Sorry for asking this question again, but I can't seem to move forward. Can I get some help again? So I graphed the function as shown below! I then get stuck finding the local extrema by finding $f'(x)=0$, which I think is $f'(x)=2x(0)$ since the derivative of $e=0$. I'm not sure how to finish through the end. http://i.stack.imgur.com/oqC0H.png
Best Answer
$$f(x)=x^{2}e^{-\frac{x}{2}}, x\in [-2,8]$$
By the Extreme Value Theorem, the function is continuous on the interval and hence could have maximum and minimum at the end points. Moreover, the function is differentible on the open interval, so the extreme values can occur either at the end points of $[-2,8]$ or when $f^{'}(x)=0$.
$$f^{'}(x)=2xe^{-\frac{x}{2}}-\frac{1}{2}x^{2}e^{-\frac{x}{2}}=0\Rightarrow 4x-x^{2}=0\Rightarrow (x=0) or (x=4)$$
Hence, now you have four points to be checked: $$x=0\ and \ x=4$$ $$x=-2\ and \ x=8$$
The rest is easy.