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I'm having some trouble proving the following:

Let $f$ be a polynomial with exactly 3 distinct real zeros. Prove that there exists $x_0 \in \mathbb R$ such that $f''(x_0)=0$.

Visually I understand why this would be true, but I don't really know how to prove this rigorously.

I'm thinking that we can use the fact that if $f$ is a polynomial, then $f'$ is also a polynomial and so will be $f''$. So, because $f''$ is a polynomial it's continuous and then maybe use the Intermediate value theorem to prove that there is a point $x_0$ such that $f''(x_0) = 0$, But I'm not so sure how to do it.

How can I prove this?

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3 Answers 3

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It does not even need to be a polynomial. Just assume $f$ to be twice differentiable (which, of course, applies to every polynomial). Let $y_0 < y_1 < y_2 \in \mathbb{R}$ be the zeros of $f$. Then $f(y_0) = f(y_1) = f(y_2) = 0$. So by Rolle's theorem, there exist $z_0 \in (y_0, y_1)$ and $z_1 \in (y_1, y_2)$ such that $f'(z_0) = f'(z_1) = 0$. Then apply Rolle's theorem again to get $x_0 \in (z_0, z_1)$ such that $f''(x_0) = 0$.

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    $\begingroup$ Wow, that proof is so nice, I never thought of using Rolle's theorem twice. +1 $\endgroup$ Apr 1, 2021 at 11:01
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    $\begingroup$ Thx. A standard proof, that is worth being memorized $\endgroup$ Apr 1, 2021 at 11:02
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Let $f$ have three roots $\alpha\lt \beta \lt \gamma$, then clearly $f(\alpha)=f(\beta)=f(\gamma)=0$
By Rolle's theorem, there exist $y_1,y_2$ such that $y_1\in (\alpha, \beta)$ and $y_2\in (\beta, \gamma)$ such that $f'(y_1)=f'(y_2)=0$. Again apply Rolle's theorem on $[y_1,y_2]$.

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You may write $f(x)=(x-a)(x-b)(x-c)g(x)$ , where $a<b<c$ and $g$ has no real roots so it retains it's sign. Now $f''(a)$, $f''(c)$ have different signs and as $f''$, is continous you get the result.

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