How can I find the range of the function $f(x) = 2^{x} + 2^{-x}$? How can I analytically determine the range of the function for $f(x) = 2^{x} + 2^{-x}$?
If I have an ordinary linear equation, I proceed as follows:
$$f(x) = x + 2 \Rightarrow y = x + 2 \Rightarrow  y - 2 = x$$
then the range of the function is $\mathbb{R}$, but I don't see what to do here
I speak about this range.

 A: Well, if $f: x\longmapsto 2^{x}+2^{-x}$, then ${\rm Dom}(f)=\mathbb{R}$.  The Arithmetic mean-geometric mean (AM-GM) inequality says that for non-negative real values we have $$\frac{u+v}{2}\geqslant \sqrt{uv}$$ with equality if and only if $u=v$. Hence, using AM-GM for $u(x)=2^{x}>0$ and $v(x)=2^{-x}=1/2^{x}>0$ we have
$$f(x)=2^{x}+\frac{1}{2^{x}}=u(x)+v(x)\geqslant 2\sqrt{u(x)v(x)}=2\sqrt{2^{x}\frac{1}{2^{x}}}=2\sqrt{1}=2$$
Therefore, the ${\rm Im}(f)=[2,+\infty[$.
N.B: This is simply a little more elaborate than Átila Correia's comment (which is a answer).
A: Let me I would like to say that the solution method applied based on the AM-GM inequality alone does not seem rigorous enough. Because,
$f(x):=2^x+2^{-x}≥2$ doesn't imply the range (image) of the function $f(x)$ should be $[2,+\infty)$. For instance, you can consider the following function:
$$g(x):=\frac {x^2+1}{x^2+2}+\frac {x^2+2}{x^2+1}$$
Applying the AM-GM inequality you have:
$$\frac {x^2+1}{x^2+2}+\frac {x^2+2}{x^2+1}≥2$$
Observe that, this doesn't tell us $\operatorname {Im}(g)$ is $(2,+\infty)$.
Because, the range (image) of the function $g(x)$ is exactly
$$\operatorname{Im}(g)=\left(2,\frac 52\right].$$

The possible rigorous way can be considered as follows:
$$
\begin{align}&y(x):=2^x+\frac 1{2^x},\; y>0\\
\implies &2^x=\frac 12\left(y±\sqrt {y^2-4}\right)\\
\implies &x=\log_2\left(\frac {\sqrt {y^2}±\sqrt {y^2-4}}{2}\right)\wedge y≥2\end{align}
$$
This implies that: $\forall y≥2$ there exist $x\in\mathbb R$ such that, $y=2^x+2^{-x}$ holds. This means, the image of the $f(x)$ is exactly $[2,+\infty)$.
