TITLE: Maximum value of $f(x)$ if $f(x)+f\left(1/x\right)=x$
QUESTION [9 upvotes]: What is the maximum value of $f(x)$ if $f(x)+f\left(\frac{1}{x}\right)=x$ for all $x$ in the domain of this real valued function?  
Now $f(x)+f\left(\frac{1}{x}\right)=x$ and by symmetry $f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x}$. Therefore $x=\frac{1}{x}$ $\implies$ $x=1$ or $-1$.
Does it mean that the domain of the function is only $1$ or $-1$ ?
And therefore substituting $x=1$ in the equation we get maximum value of $f(x)$ as $1/2$?

REPLY [3 votes]: Most of this has already been said in the comments.
Your argument is correct that if $x$ is in the domain of $f$, then $x=1$ or $x=-1$. The question does not specify what the domain of $f$ is, so there are four possibilities for the domain: $\varnothing$, $\{1\}$, $\{-1\}$, or $\{1,-1\}$.
You computed that if $f$ is defined at $x=1$ then $f(1)=\frac{1}{2}$. Similarly, if $f$ is defined at $x=-1$ then $f(-1)=-\frac{1}{2}$. So we have three possibilities for the maximum value:

If the domain of $f$ is $\{1\}$ or $\{1,-1\}$, then $\max f=\frac{1}{2}$.
If the domain of $f$ is $\{-1\}$, then $\max f=-\frac{1}{2}$.
If the domain of $f$ is $\varnothing$, then $\max f$ is not defined (some people write $\max f=-\infty$ in this case).