Continuous function and maximum value

Question

Given function $f : (0; 10] \rightarrow \mathbb{R}$ is continuous and $f(3)$ is its maximum value. I have to prove that there is $x_0$ such $f(x_0)=f(3x_0)$. How to do that?

Answer 1

Consider the function $g(x)=f(x)-f(3x)$. Solutions to $g(x_0)=0$ correspond to values $f(x_0)=f(3x_0)$. Now $g(3)=f(3)-f(9)\ge0$ since $f(3)\ge f(9)$, and $g(1)=f(1)-f(3)\le 0$ since $f(1)\le f(3)$. Then by the intermediate value theorem applied to $g$ on $[1,3]$, there is an $x\in(1,3)$ with $g(x)=0$.