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I am trying to solve the following question:

Question: If $f$ is a thrice differentiable non negative function and has two distinct roots in the interval $(0, 1)$ then there exist $c \in (0, 1)$ such that $f^{\prime\prime\prime}(c) = 0$, where $f^{\prime\prime\prime}$ denotes third derivative of the function $f$.

I am trying to solve this problem by using Rolle's theorem. Given that function $f$ is thrice differentiable on $(0, 1)$ and have two distinct roots, say $\alpha$, and $\beta$ such that $f(\alpha) = f(\beta) = 0$. Thus, all the condition for Rolle's theorem are satisfied hence there must exist $c \in (0, 1)$ such that $f^{\prime}(c) = 0$.

I am not sure how to proceed from here to show that $f^{\prime\prime\prime}(c) = 0$. I am not able to apply the given condition that the function $f$ is non negative on the given interval and how this is going to help me to solve the given problem. Also, is there any other way to solve this problem.

Thank you for your time.

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    I think the idea is to prove there are two values such that $f''(t_1)=f''(t_2)$ – 2017-02-14
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    @EugenCovaci Thanks for the comment. – 2017-02-15

1 Answers 1

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Hint. Since $f$ is non-negative in $(0,1)$ and $\alpha,\beta\in(0,1)$ are such that $f(\alpha) = f(\beta) = 0$ then, by Fermat's theorem, $f'(\alpha) = f'(\beta) = 0$. Moreover, between these two minimum points, $f$ should have a maximum point where the derivative is zero.

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    Thank you very much for the answer. I am just not sure why $f^{\prime} (\alpha) = f^{\prime} (\beta) = 0$ if $f$ is non negative. I suppose that after that repeated application of Rolle's theorem will prove the result. Thanks again – 2017-02-15
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    If $f(x)\geq 0$ in $(0,1)$ and $f(\alpha)=0$ with $\alpha\in (0,1)$ then $\alpha$ is an internal minimum point. Then use Fermat's theorem (https://en.wikipedia.org/wiki/Fermat's_theorem_(stationary_points)). The same for $\beta$. – 2017-02-15
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    Thank you very much. I fully understood now. Really happy with this answer. :) – 2017-02-15