Let $f$ be a function continuous on $[0, 1]$ and twice differentiable on $(0,1)$.
Suppose that:
$f(0) = f(1) = 0$ and $f(c) >0$ for some $c \in (0,1)$.
Prove that there exists $x_{0}\in(0,1)$ such that $f''(x_{0}) < 0$.
I'm not sure how to approach this question. Do I have to use the “mean value theorem” or is there actually another way to approach this question?
We can certainly use mean value theorem in it.
Question says $\exists c \in(0,1)$ such that $f(c)>0$. Let $f(c)=\alpha$ and $\alpha>0$
Now from "Lagrange's mean value theorem", $\exists x_{1}\in(0,c)$ such that $$ f'(x_{1}) = \dfrac{f(c)-0}{c-0} = \dfrac{f(c)}{c} $$ Also from "Lagrange's mean value theorem", $\exists x_{2}\in(0,c)$ such that $$ f'(x_{2}) = \dfrac{0-f(c)}{1-c}=\dfrac{-f(c)}{1-c} $$
Here we can claim $x_{1}<x_{2}.$ Also that $f'(x_{1})>0$ and $f'(x_{2})<0$.
We know that derivative of $f'(x)$ is a continuous function and value of $f'(x)$ has decreased while moving from $x_{1}$ to $x_{2}$. So there must be some interval which is the subset of the interval $(x_{1}, x_{2})$ such that f'(x) was decreasing in that interval.
Now, since we are guaranteed of an interval in $(0,1)$ such that $f'(x)$ is decreasing on that interval. At all the points in that interval it is certainly true that $f''(x)<0$.
Hence we are guaranteed of the existence of a point $x\in(0,1)$ such that $f''(x)<0$.
Let's see what Rolle’s theorem is telling us about $f$. The function has a positive absolute maximum, since it attains a positive value. Such a maximum must be at $t\in(0,1)$, as $f(0)=f(1)=0$. Therefore $f'(t)=0$.
Suppose, by way of contradiction, that $f''(x)\ge0$ for all $x\in(0,1)$. This means that $f'(x)$ is a nondecreasing function on $(0,1)$.
Since $f'(t)=0$, we have $f'(x)\ge0$ for $t<x<1$. Thus $f$ is nondecreasing on $(t,1)$ and therefore…
the definite integral of f(x) dx from 1 to 0 = f(0) = f(1) = 0.
Prove that there exists a number x0 ∈ (0,1) such that f′′(x0) = 0.
– Jason Tan Oct 25 '15 at 17:33