How to find the maximum and minimum value of $2^{\sin x}+2^{\cos x}$ My try: Let $y$ =  $2^{\sin x}+2^{\cos x}$


*

*Applying AM GM inequality I get 
$y$ $> 2.2^{(\sin x+\cos x)/2}$. 
Now, the highest value of R.H.S is $2^{\frac{\left(2+\sqrt{2}\right)}{2}}$.


Should this mean that $y$ is always greater than $2^{\frac{ \left ( 2+\sqrt{2}\right ) }{2}}$?
But this is not true (we can see in the graph).


*Calculus method:
$dy/dx$ = $\ln\left(2\right){\cdot}{2}^{\sin\left(x\right)}\cos\left(x\right){-\ln\left(2\right){\cdot}{2}^{\cos\left(x\right)}\sin\left(x\right)}$


When $dy/dx$ =0,
$\tan x = 2^{\sin x- \cos x}$ and I am stuck here.
https://www.desmos.com/calculator/p3zfvkq2mn

Let $f(x)=2^{\sin x}+2^{\cos x},\;$ so $f^{\prime}(x)=(2^{\sin x}\cos x-2^{\cos x}\sin x)(\ln 2)$.
$\textbf{A)}$ To find the maximum of $f(x)$, it is enough to consider $f$ on $[0,\frac{\pi}{2}]$.
$\;\;\;$Then $f(0)=3$ and $f(\frac{\pi}{2})=3$, and 
$\;\;\;f^{\prime}(x)=0\iff 2^{\sin x}\cos x=2^{\cos x}\sin x\iff\frac{2^{\sin x}}{\sin x}=\frac{2^{\cos x}}{\cos x}\iff x=\frac{\pi}{4}$,
$\;\;\;$since if $\displaystyle g(t)=\frac{2^t}{t},\;$ $\displaystyle g^{\prime}(t)=\frac{2^t(t\ln 2-1)}{t^2}<0$ for $0< t<1$ so $g$ is 1-1 on $(0,1]$.
$\;\;\;$Since $f(\frac{\pi}{4})=2\cdot2^{\sqrt{2}/2}=2^{1+\frac{\sqrt{2}}{2}}>3,\;$ $f(\frac{\pi}{4})=2^{1+\frac{\sqrt{2}}{2}}$ is the maximum value.
$\textbf{B)}$ Similarly, to find the minimum of $f(x)$ it is enough to consider $f$ on $[\pi, \frac{3\pi}{2}]$.
$\;\;\;$Then $f(\pi)=\frac{3}{2}$ and $f(\frac{3\pi}{2})=\frac{3}{2}$, and  $f^{\prime}(x)=0\iff\frac{2^{\sin x}}{\sin x}=\frac{2^{\cos x}}{\cos x}\iff x=\frac{5\pi}{4}$ 
$\;\;\;$ since $\displaystyle g(t)=\frac{2^t}{t}$ is 1-1 on $[-1,0)$ as above.
$\;\;\;$Since $f(\frac{5\pi}{4})=2\cdot2^{-\frac{\sqrt{2}}{2}}=2^{1-\frac{\sqrt{2}}{2}}<\frac{3}{2},\;\;$ $f(\frac{5\pi}{4})=2^{1-\frac{\sqrt{2}}{2}}$ is the minimum value.
