Q: Show that there exist $x_{0}\in[0,1]$ such that $f(x_{0})=g(x_{0})$ Let $f,g:[0,1]\rightarrow[0,\infty)$ be continuous  such that
$\smash{\displaystyle\max_{x \in [0,1]}} f(x) = \smash{\displaystyle\max_{x \in [0,1]}} g(x)$.
Show that  there exist $x_{0}\in[0,1]$ such that $f(x_{0})=g(x_{0})$.
Hint: Consider the function $f-g$

A: Hint: Let $u$ such that $f(u)= max_{x\in [0,1]}f(x)$ and $v$ such that $f(v)=max_{x\in [0,1]}g(v)$. $u$ and $v$ exist since $[0,1]$ is compact. If $u=v$ done. If not we can suppose $u<v$, $f(u)-g(u)\geq 0$ and $f(v)-g(v)\leq 0$ apply the intermediate value theorem.

A: I would suggest to begin with the trivial case where $f(0) = g(0)$.
That said, try to draw a picture of the other case where $f(0) > g(0)$ and respect the condition about maximums. I'm sure it will help you to get the proof.
