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?
Best Answer
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 $d$.