Number of zeros: Do we use the definition of the derivative?

Question

We have that a function $f:\mathbb{R}\to \mathbb{R}$ is twice differentiable in $\mathbb{R}$ and it has infinitely many roots.

To show that the equation $f''(x)=0$ has infinitely many roots, do we use the definition of the derivative?

Answer 1

There exists a sequence $(x_n)$ of zeroes of $f$ such that $x_1 <x_2 <x_3 < ...$.

By Rolle there is a sequence $(y_n)$ of zeroes of $f'$ such that $x_1<y_1 <x_2 < y_2 <x_3 <y_3 <x_4, ...$.

Hence we have $y_1 <y_2 <y_3 < ...$.

Now repeat the above arguments with $f''$ instead of $f'$ .....

Answer 2

Just use Rolle's theorem... Between two roots of $f$ there will be at least one root of $f'$, therefore $f'$ will have an infinite number of roots. Since you have enough regularity, you can apply Rolle's theorem again, this time to $f'$ and reach the desired conclusion.

Answer 3

In order to prove that $f''$ has infinitely many zeros. We may use Rolle's theorem. If $a,b,~\text{and}~c$ are three consecutive zeros of $f$ then by Rolle's theorem their exists $x\in(a,b)$ such that $f'(x)=0$, similarly $\exists$ $y\in (b,c)$ such that $f'(y)=0$. Now since $f$ is twice differentiable so again apply Rolle's theorem to $f'$, $\exists~z\in(x,y)$ such that $f''(z)=0$. Since this holds for every three consecutive zero's of $f$, therefore $f''$ will have infinite number of zero's.