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I received the following question during a Calculus $2.0$ course in my university. I am not a native speaker, so please excuse my English.

The question is as follows:

Let $f$ be a function with derivative of order $2$ in $\mathbb{R}$ with $f(0) \not = 0$ and there exists $M>0$ so that for all $x\in\mathbb{R}$, $|f''(x)|\leq M$. Define $h=\sqrt{\frac{2|f(0)|}{3M}}$ and prove that if $f$ is not monotonic in $(-h,h)$, then there is no $x\in (-h,h)$ so that $f(x)=0$.

Our professor told us that all we need from the assumption of $f$ being non-monotonic in $(-h,h)$ is that there is a point $x\in (-h,h)$ so that $f' (x)=0$.

${\bf \text{What I did}}$: After proving the existence of such a point, $x\in (-h,h)$ so that $f'(x)=0$, I used Taylor's Theorem on $f$, around that point to express $f$ as Taylor series of order $1$, plus remainder in the Lagrange form (Taylor Expansion). Then I managed to prove the required result but only for a radius of $\sqrt{2} h$ around that point. This means that I only need to take care of what happens when that point is close to the edges, $h$ or $-h$. This, in turn, means, that I need to prove that closer to the other edge, there is no $x\in (-h,h)$ so that $f(x)=0$. I was not able to do that.

That being said, I know that there cannot be two points, such that $f(x)=0$, because then, based on Rolle's theorem, there must be a point between these points where, again, $f'(x)=0$, and that contradicts what I found regarding what happens in radius of $\sqrt{2} h$, around such a point, since there was already a small continuous section of $(-h,h)$, where there could be $x\in (-h,h)$ so that $f(x)=0$.

I also thought about using a different point to build the Taylor Series around (say, around $x=0$. Then it's a Maclaurin series), and then carefully using something like the formula for the roots of a quadratic equation (the 'careful' part was due to the fact that the coefficient of the Remainder in the Lagrange form is a function), but it didn't amount to much.

Any help will be much appreciated. Sorry for the English.

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  • $\begingroup$ I fixed the formatting of the question. Please check that I did not mess up anything. See this page for how to write nice looking math formulas on this site. $\endgroup$ – Winther May 24 '15 at 17:12

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