Maximum and Minimum Value Let $f : [0, 2\pi] \to \mathbb{R} $ be defined by $f(x) :=e^x\sin(x)$. Find the maximum and minimum value of $f$ over the interval $[0, 2\pi]$. You must fully justify your answer.
Sorry for the bad use of notation, I have no idea where to start such a question, any help step by step would be much appreciated!
 A: The derivative of $f(x)=e^x \sin x$ by the product rule is $f'(x)=e^x \sin x+e^x \cos x$. 
So, $f'(x)=0$, only when $e^x=0$, which is not possible, or $\sin x+\cos x=0$. This means that $\sin x=-\cos x$, or $\tan x=-1$, when $\cos x\ne 0$, but if $\cos x=0$, then $\sin x \ne 0$, hence we are forced to conclude that $\tan x =-1$. This is only possible in $[0,2\pi]$ when $x=\frac{3\pi}4$ or $x=\frac{7\pi}4$.
Now use the closed interval test, which says we only need to check the values of $f(0), f(\frac{3\pi}4), f(\frac{7\pi}4), f(2\pi)$. 
Now $f(0)$ and $f(2\pi)$ would obviously be $0$. Note that $f(\frac{3\pi}4)$ would be positive and $f(\frac{7\pi}4)$ would be negative (why?), hence $f(\frac{3\pi}4)$ would be the maximum and $f(\frac{7\pi}4)$ would be the minimum. Its surprising that we have found them to be the extremas without even knowing their values!
A: Well looking by derivative $$f'(x)=e^x\sin x+e^x\cos x\\f'(x)=0=e^x(\sin x+\cos x)\implies \sin x+\cos x=0\\\sin x=-\cos x\\x=\frac{3\pi}{4}+k\pi\\x_{1,2}=\frac{3\pi}{4},\frac{7\pi}{4}$$
One of those is a local minimum and one is a maximum,since $e^x>0$ and $\sin\frac{7\pi}{4}<0,\sin\frac{3\pi}{4}>0$
This implies that $\frac{7\pi}{4}$ is a minimum and $\frac{3\pi}{4}$
