Let $f(x)$ be a non-constant twice differentiable function defined on $(-\infty,\infty)$ such that $f(x)=f(4-x)$ and $f(x)=0$ has at least two distinct repeated roots in $(2,4)$, then find the minimum number of roots of $f''(x)=0$ in $[0,4]$.
My Attempt
Clearly $f(x)$ is symmetrical about $x=2$. So $f'(2)=0$.
On $(2,4)$ we have two points ,say $x=\alpha$ and $x=\beta$ where $f(\alpha)=f'(\alpha)=f(\beta)=f'(\beta)=0$.
So between $x=\alpha$ and $x=\beta$ we will have some $x=\gamma$ for which $f'(\gamma)=0$.
So $f'(x)=0$ has $4$ roots on $[2,4]$. So $f''(x)=0$ has at least three roots in $(2,4)$
Since $f(x)$ is symmetrical about $x=2$, there will be at least three roots in $(0,2)$
So in all there should be at least $6$ roots of $f''(x)=0$ on $[0,4]$
But the given answer is $4$. Am I overcounting some case