# Prove that there exists a point $d \in (a,b)$ such that $f''(d) > 0$

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 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?

• The function $f$ is continuous and differentiable or is it continuously differentiable? Feb 8, 2020 at 12:57
• @Fakemistake continuous and differentiable Feb 8, 2020 at 12:58
• A differentiable function is always continuous, so it must be continuously differentiable. Feb 8, 2020 at 13:00
• @Fakemistake maybe it's a mistake in the question then since I can't see any other way Feb 8, 2020 at 13:00
• You have to assume, that $f$ is a $C^2$-function, otherwise you cannot prove this statement. Feb 9, 2020 at 7:50

Since $$c_2 and the function $$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$$.
• @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. Feb 8, 2020 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 math.stackexchange.com/a/829936/365780 the plot of $f$ and $f'$.