Find the minimum roots of $f'(x)\cdot f'''(x)+(f''(x))^2 =0$ given certain conditions on $f(x)$. Problem:
Let $f(x)$ be a thrice differentiable function satisfying:
$$|f(x) - f(4-x)| + |f(4-x)-f(4+x)| = 0, \forall x \in R$$
If $f'(1)=0$, then find the minimum number of roots of $f'(x)\cdot f'''(x)+(f''(x))^2 =0$, on $x \in [0,6]$
My attempt:
We know: $f(x)=f(4-x)$ and $f(4-x)=f(4+x)$
So, $f(x)=f(x+4)$. That is, the period of the given function is $4$.
It can also be noted that the function is symmetric about $2$ and $4$.
I also know that the second equation is nothing but $\frac{d}{dx}(f'(x) \cdot f''(x))$
I don't know how to proceed from here.
 A: Since $f(x) = f(4-x) = f(4+x)$ we have $f'(x) = -f'(4-x) = f'(4+x)$ for all $x \in \mathbb{R}$.
Since $f'(1) = 0$, using the above identity for $x = 1$ gives us $f'(1) = f'(3) = f'(5) = 0$.
Also, using the above identity for $x = 2$ yields $f'(2) = -f'(2) = f'(6)$, so $f'(2) = f'(6) = 0$.
Similarly, for $x = 0$, we get $f'(0) = -f'(4) = f'(4)$, so $f'(0) = f'(4) = 0$. 
By Rolle's theorem, for $n = 1,2,3,4,5,6$, there exists a $x_n \in (n-1,n)$ such that $f''(x_n) = 0$.
Thus, the function $g(x) = f'(x)f''(x)$ has zeros at $x = 0,x_1,1,x_2,2,x_3,3,x_4,4,x_5,5,x_6,6$. 
By using Rolle's theorem again, $g'(x) = f'(x)f'''(x)+f''(x)^2$ has at least $12$ zeros in $[0,6]$.

The function $f(x) = \cos\pi x$ satisfies $f(x) = f(4-x) = f(4+x)$ for all $x \in \mathbb{R}$ and $f'(1) = 0$. 
For this function, $f'(x)f'''(x)+f''(x)^2 = (-\pi \sin \pi x)(\pi^3\sin\pi x)+(-\pi^2\cos\pi x)^2$ $= \pi^4(\cos^2\pi x - \sin^2\pi x) = \dfrac{\pi^4}{2}\cos 2\pi x$, which has $12$ zeros on $[0,6]$.

Therefore, the minimum number of zeros of $f'(x)f'''(x)+f''(x)^2$ on $[0,6]$ is $12$.
