Determine the points in $[0, π]$ at which $f(x) = e^{2\sin(x)−x}$ attains its maximum and minimum values.
For this question I differentiated $f(x)$ and got $f'(x)= e^{ 2\sin (x)−x} (2\cos(x)-1).$
But I am not sure where to go from here. Usually for a simple equation you would factorise and get values for $x$ and then sub in these values into the original equation to find the max and min values. But I am not sure what I would do in this case.
Any help is appreciated!
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So we want to find the maximum and minimum values of the function $$f(x)=e^{2\sin(x)-x}.$$ A differentiable function $f$ has a maximum (or minimum) if $f'=0$.
Since $e^x$ and $2\sin(x)-x$ are differentiable, we know that their composition is differentiable and the derivative of such a composition is given by the chain rule and thus we find $$f'(x)=e^{2\sin(x)-x}(2\cos(x)-1).$$ Above leads to the equation $$e^{2\sin(x)-x}(2\cos(x)-1)=0.$$ This maybe looks scary or complicated, but remember that $e^x>0$ for all $x\in \mathbb{R}$ and thus you just have to solve $$2\cos(x)-1=0.$$ I don't want to spoil everything and I think that you can try to catch up from here. Please ask if you need more help. Good luck!
Once you differentiate $f(x)$ and find $f'(x)=(2\cos(x)-1)e^{2\sin(x)-x}$, you should determine where $f'(x)$ is positive.
The inequality $(2\cos(x)-1)e^{2\sin(x)-x}> 0$ can be re-written as $2\cos(x)-1> 0$, since any exponential is positive. In $[0, \pi]$ one has $\cos(x)> \frac{1}{2}$ for $x\in[0, \frac{\pi}{3})$.
This means that $f$ is (strictly) increasing in $[0, \frac{\pi}{3}]$ and (strictly) decreasing in $[ \frac{\pi}{3}, \pi]$. We deduce that $f$ attains its maximum value at $x=\frac{\pi}{3}$, $f\left(\frac{\pi}{3}\right)=e^{\sqrt{3}-\frac{\pi}{3}}$.
There are no stationary points other than $\frac{\pi}{3}$, which means the minimum of $f$ bust me attained either at $x=0$ or at $x=\pi$. Computing the values of $f$ at these points yields $f(\pi)=e^{-\pi}<1=f(0),$ so the minimum of $f$ on $[0,\pi]$ is $f(\pi)=e^{-\pi}$.