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


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*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
 A: Using AM GM inequality $$2^{\sin x}+2^{\cos x}\ge2\cdot2^{\frac{\sin x+\cos x}2}$$
Now the the minimum value i.e., equality occurs if $2^{\sin x}=2^{\cos x}\iff \sin x=\cos x$ 
$\implies\tan x=1\implies\sin x=\cos x=\pm\dfrac1{\sqrt2}$
So, the minimum value $$=2^{1-\frac{1+1}{\sqrt2}}$$
Set $y=\pi+x$ to find the maximum value
A: 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.
A: Let $f(u,v)=2^u+2^v;\;$ we want to find the extrema of $f$ on the circle $u^2+v^2=1$.
Using Lagrange multipliers, $\;\;2^u\ln2=\lambda(2u)$ and $2^v\ln 2=\lambda(2v)\;$ so $\;\displaystyle\frac{2^u\ln2}{2u}=\frac{2^v\ln2}{2v}$.
Then $\displaystyle\frac{2^u}{u}=\frac{2^v}{v}$ and therefore $u=v$, 
$\;\;\;$since the function $\displaystyle g(t)=\frac{2^t}{t}$ is 1-1 on $[-1,0)$ and $(0,1]$
$\;\;\;$because $\displaystyle g^{\prime}(t)=\frac{2^t(t\ln 2-1)}{t^2}<0$ for $t\in(-1,0)\cup(0,1)$.
Then $u^2+v^2=1\implies 2u^2=1\implies u=v=\pm\frac{\sqrt{2}}{2}$.
Thus $\displaystyle f\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)=2^{1+\frac{\sqrt{2}}{2}}$ is the maximum, and $\displaystyle f\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)=2^{1-\frac{\sqrt{2}}{2}}$ is the minimum.
A: Write $x:={\pi\over4}+t$, and put ${\displaystyle{1\over\sqrt{2}}\log 2}=:\lambda\doteq0.490$. Then
$$2^{\cos x}+2^{\sin x}=2 e^{\lambda\cos t}\>\cosh(\lambda\sin t)=: g(t)\ .$$
Since $g$ is a smooth $2\pi$-periodic function its global extrema are at the zeros of its derivative
$$g'(t)=2\lambda e^{\lambda\cos t}\bigl(-\sin t\>\cosh(\lambda\sin t)+\cos t\>\sinh(\lambda\sin t) \bigr)\ .$$
When $\cos t=0$ one necessarily has $g'(t)\ne0$. We may therefore consider the equation
$$\tan t=\tanh(\lambda \sin t)\ .\tag{1}$$
For $0< t<{\pi\over2}$ one has $$\tanh(\lambda \sin t)<\lambda\sin t<t<\tan t\ .$$
Symmetry considerations then imply that $t=0$ and $t=\pi$ are the only solutions of $(1)$ mod $2\pi$. Therefore
$$\bigl\{g(0), g(\pi)\bigr\}=\{2e^\lambda, 2e^{-\lambda}\}=\bigl\{2\cdot 2^{1/\sqrt{2}}, \> 2\cdot 2^{-1/\sqrt{2}}\bigr\}$$
are the maximal and the minimal value of $g$, and whence of $2^{\cos x}+2^{\sin x}$, on ${\mathbb R}$.
