$f(2-x)=f(2+x)$ and $f'(1/2)=0=f'(1)$. Find minimum number of roots of $f''(x)=0$. 
Let $f$ be a non constant twice differentiable function satisfying $f(2-x)=f(2+x)$ and $f'(1/2)=0=f'(1)$. Find the minimum number of roots of $f''(x)=0$ in the interval $(0,4)$.
Answer: $4$

I managed to rewrite the given equation as $f(x)=f(4-x)$. Using this, I obtained
$$f'(x)+f'(4-x)=0$$
And
$$f''(x)=f''(4-x)$$
So it suffices to show that $f''(x)=0$ has at least $2$ roots in the interval $(0,2)$ but I'm unsure what's gonna happen at $x=2$ here. Also, I have $f'(7/2)=0=f'(3)$ from the given info and the first equation I obtained, but I don't know how to use this.
Any help would be great.
 A: Your function is symmetric about the line $x=2$. One property of a symmetric continuous function is that at the line of symmetry, the slope must be zero (can you see why?)
Hence, we have
$f'(1/2) = f'(1) = f'(2) = 0$
Now, using Rolle's theorem, there exists at least one point each in the intervals $(1/2,1)$ and $ (1,2)$ such that $f''(x) = 0$. For the absolute minimum, we can have 2 zeros. Now since that would be repeated for the mirror image, the answer should be $4$
A: From the equation
$$f’(x)+f’(4-x)=0$$
you may plug in $x=2$ and deduce $f’(2)=0$. Now you know that
$$f’(1/2)=f’(1)=f’(2)=0$$
By the mean value theorem, you can deduce that $f’’(c_1)=0$ for some $c_1\in (1/2,1)$, and $f’’(c_2)=0$ for some $c_2\in (1,2)$. Thus, $c_1,c_2$ are the inflection points you were looking for in the interval $(0,2)$, and there may be more.
However, in order to show that $4$ is the minimum number of inflection points that such a function could have, you must now find a function $f$ that has exactly $4$ inflection points and satisfies all other constraints. Can you take care of this part?
A: To prove that $f''(x)=0$ has at least $ 4$ roots, it is sufficient to show that $f'(x)=0$ has a least $5$ roots.
we already know that
$$f'(\frac 12)=f'(1)=f'(3)=f'(\frac 72)=0$$
because
$$f'(x+2)=-f'(2-x)$$
on the other hand
$$f(2-\frac 12)=f(2+\frac 12)$$
thus there exists (By Roole's Theorem) $c\in (\frac 32,\frac 52) $ such that
$$f'(c)=0$$
A: To add to Dhanvi Sreenivasan's answer, whenever you are given a functional equation it is usually good to simply just put different values of $x$ and see what sorts of relationships you can find. Here, once you start substituting in values of $x$ like $0,2,4$ etc., you will notice the fact that the function is symmetric about $x=2$. In this case it was relatively easy to notice this relation by inspection alone but if you come across a more complicated functional equation, going by this method will probably be more helpful.
