The minimum number of points where $f''(x)$ vanishes on $[0,2]$ is?

Question

Let $f(x)$ be a twice differential function defined on $(-\infty,\infty)$ such that $f(x)=f(2-x)$ and $f'(\frac{1}{2})=f'(\frac{1}{4})=0.$

Then, the minimum number of points where $f''(x)$ vanishes on $[0,2]$ is?

I have no idea how to start this, any hints would be helpful.

In addition question asks for,$$\int_{-1}^{1}f'(1+x)x^2e^{x^2}dx$$I know I have to use the Queen property here, but not sure how.

Answer 1

We have that

$$f'(x)=-f'(2-x)$$

Then

$$f'\left(\frac74\right)=f'\left(\frac32\right)=0$$

and also $$f'(1)=-f'(2-1)$$ which implies $f'(1)=0$.

So $f'$ has at least five zeros. Rolle's theorem implies that $f''$ has at least four zeros.

Now, if you assume that $$f'(x)=\left(x-\frac12\right)\left(x-\frac14\right)\left(x-\frac74\right)\left(x-\frac32\right)(x-1)$$

then $f''$ is a fourth degree polynomial function, that can have at most four zeros.

You need also to prove that $f(x)=f(2-x)$, but we see that $f'(x)=-f'(2-x)$, and integrating we get $f(x)=f(2-x)+C$, so it suffices to pick $C=0$.