Prove that if $f$ is continuous in $[0,1]$, and $f(0)=f(1)= \frac{1}{3}$, then there is $c \in (0,1)$ so that $f(c)=c^2$ Prove that if $f$ is continuous in $[0,1]$, and $f(0)=f(1)= \frac{1}{3}$, then there is $c \in (0,1)$ so that $f(c)=c^2$
Stuck here. Any assistance will be appreciated! 
Also, please help me with hints only, don't want to be spoon-fed.
Thank you!   

Let $h$ the function defined by $h(x)=f(x)-x^2$ then $h$ is continuous and
$$h(0)h(1)<0$$
hence by the intermediate value theorem the result follows.
