$f$ is having maxima at $\frac12$ show that $f\circ f$ is having minima at $\frac12$.

Question

$f:[0,1]\to [0,1]$

$f(0)=0=f(1)$

$f$ is having local maxima at $x=\frac12$.

Show that $f\circ f(x)$ is having local minima at $x=\frac12$.

Using chain rule I was only able to find that $(f\circ f)'(\frac12)=0$

Now it is sure that at $x=\frac12$ we have either maxima or minima. How to show that $f\circ f(x)$ at $x=\frac12$ is a point of minima?