Let $ f $ be a continuous function of $[0, 1]$ into itself, prove that there exists c, an element of $[0, 1]$ such that $f(c)$ = $c^2$

Question

"Let $ f $ be a continuous function of $[0, 1]$ into itself, prove that there exists c, an element of $[0, 1]$ such that $f(c)$ = $c^2$". It seems I need to use the intermediate value theorem for the case $f(0)\not=f(1)$ and create a function $g(x)=f(x)-x^2$ but I'm struggling to work that out into a proof, even when considering $g(0)$ and $g(1)$. Any help is appreciated.

Answer 1

Hint: let $g(x)=f(x)-x^2$, $g(0)\geq 0, g(1)\leq 1$.

$g(0)=f(0)-0^2=f(0)\geq 0$ and $g(1)=f(1)-1^2\leq 0$ since $f(1)\in [0,1]$.

Answer 2

$g(0)=f(0)-0\geq 0$ because range of $f$ is contained in $[0,1]$. Also $g(1)=f(1)-1\leq 0$ for the same reason. If one of these two are equalities there is nothing to prove. Otherwise intermediate value property of continuous functions tells you that $g(c)=0$ for some $c \in (0,1)$ which gives $f(c)=c^{2}$.