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Given that f is a continuous, differentiable function such that $f(a) = f(b) = 0,$ and that $\exists\ c \in (a,b)$ such that $f(c)<0$, prove $\exists\ d \in (a,b)$ such that $f''(d) > 0$

Here's what I have tried:

Since there is a point $f(c)<0$, we can use Weirstrass Theorem to conclude that the function does attain a minimum, which we will call $c_1$

Then using Lagrange's twice we can say:

$\exists\ c_2 \in (a,c_1)$ such that $f'(c_2)=\frac{f(a) - f(c_1)}{a - c_1}<0$

$\exists\ c_3 \in (c_1,b)$ such that $f'(c_3)=\frac{f(c_1) - f(b)}{c_1 - b}>0$

Thus we have $f'(c_2)<0<f'(c_3)$ which is where I am currently stuck.

I'm wondering if it's possible to use Lagrange Theorem again on $f'(c_2), f'(c_3)$ or maybe something else such as Darboux Theorem?

Fakemistake
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GaussBiggestFanboi
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1 Answers1

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Now argue as follows:

Since $c_2<c_3$ and the function $f'$ is continuous in $(c_2,c_3)$, it must have a root inside of it (intermediate value theorem) with a sign change from $-$ to $+$.

Name it $d$.

Fakemistake
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  • Hey thanks for the reply. I see the intuition here but I'm unsure exactly how to prove it's increasing and has a root – GaussBiggestFanboi Feb 08 '20 at 12:50
  • In this case, the intermediate value theorem for derivatives is the Darboux Theorem I believe – GaussBiggestFanboi Feb 08 '20 at 12:54
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    @S.Miller I made a small edit, because we cannot say it is increasing in $(c_2,c_3)$. But there's one point missing. The function is assumed to be $C^1$-function, so we cannot say, the second derivative exists. – Fakemistake Feb 08 '20 at 12:55
  • I think $f$ must be at least $C^2$ in an open subset of $(a, b)$. Otherwise take $f(x) = \int g(x)$ where $g(x)= \sum_{n \in \mathbb{N}}\frac{cos(3^n\pi x)}{2^n}$ is a Weistrass function. $f$ is $C^1$ but $f' = g$ respect the hypothesis but it's nowhere differentiable. You can see here https://math.stackexchange.com/a/829936/365780 the plot of $f$ and $f'$. – Ker Feb 08 '20 at 15:55