The question is as follows
Consider a non-constant thrice differentiable function $f:\mathbb R \to \mathbb R$, such that $\,f(x) = f(6-x) \, \forall\, x \in \mathbb R $.
And f'(0)=f'(2)=f'(5) = 0. If n denotes the minimum no. of roots of the equation $$ [f''(x)]^2+f'(x)f'''(x) =0$$ in [0,6]. Then the minimum value of n is?
The answer is 12 and the approach used requires us to find the $x \in [0,6]$ where $f'(x) = 0$, the solution only used the equation
$f(x) = f(6-x)$
Differentiating both sides
$f'(x) = -f'(6-x)$
Using this equation and the fact that $f'(0)=f'(2)=f'(5) = 0$, it deduces that $f'(x) = 0$ at $x =0,1,2,3,4,5,6$.
My question is can we use the fact $f(x) = f(6-x)$ and when x = 0, $f(0) = f(6)$, and then
by rolle's theorem $\exists \, c \in (0,6)$ such that f'(c) = 0. and this c isn't necessarily in {1,2,3,4,5}.
And we could probably extend this and find distinct $c \in \mathbb R$ for every $x \in (0,3)$ by using f(x) = f(6-x).
Is my assertion correct and does it work in this question as in since we require minimum n should such c's be ignored?
Edit: 1.)I came across this question similar to the one, I have asked question, and even the accepted answer under this question doesn't use Rolle's theorem to deduce existence of a possible root for f'(x) different from {0,1,2,3,4,5,6}
2.) Corrected the equation $[f''(x)]^2-f'(x)f'''(x) =0$ to
$[f''(x)]^2+f'(x)f'''(x) =0$
Consider $f(x)=\cos(n\pi x)$ with $n$ a positive integer. Then $f(x) = f(6-x)$, $f'(0)=f'(2)=f'(5) = 0$, and $$[f''(x)]^2+f'(x)f'''(x)=\pi^4n^4(\cos(n\pi x)^2-\sin(n\pi x)^2)=\pi^4n^2\cos(2n\pi x)=0$$ has exactly $12n$ roots in $[0,6]$. Note that $f'(x)=-n\pi\sin(n\pi x)$ has $6n+1$ roots in $[0,6]$. So if $n>1$ then $f'$ has more than $7$ roots in $[0,6]$ beside those at $0,1,2,3,4,5,6$.