Solve $f'(x) = 0$ and set up a sign chart for $f'$. 
I understand how my teacher got the two $x$ values, but why didn't he solve for $e^x=0$? 
I know he did
$x=0$ which is $0$
$x+2=0$ which is $-2$
so why no $e^x=0$? is there even an answer for that? I don't think there is right?

If you look closely at the definition of $e^x$ you can conclude that there is no value such that $e^x = 0$ as seen in the formula below
$$
e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!}
$$
Since when k = 0, the sum will be:
$$
e^x = 1+ \sum_{k=1}^{\infty}\frac{x^k}{k!}
$$
Since all terms when summed together will yield a result greater than zero, there is no possible way to have any real value for $x$ such that $e^x = 0$. This is even true for any value you make x, including imaginary values of x.
