Let $I\subset \mathbb{R}$ be an open interval and let > $f(x):I\to\mathbb{R}$ be a twice differentiable function in $a\in I$ s.a $f''(a)>0$.

Question

Let $I\subset \mathbb{R}$ be an open interval and let $f(x):I\to\mathbb{R}$ be a twice differentiable function in $a\in I$ s.a $f''(a)>0$.

Prove there exist $x_1,x_2\in I$, $x_1<a<x_2$ s.a:

$f'(a)=\frac{f(x_2)-f(x_1)}{x_2-x_1}$

Attempt: I tried to use MVT but I need to prove it for a specific $a$ and not just for some $x_1<c<x_2$

Answer 1

Note that $f''(a) > 0$. This means, that in a neighbourhood of $a$, $f'$ is increasing i.e. there is some $h > 0$ such that for all $0 < h' \leq h$, we have $f'(a-h') < f'(a) < f(a+h')$. Let $h' = \frac h2$, and define $g : [a-h',a]$ by $g(x) = \frac{f(x+ h')-f(x)}{h'}$.

Note that $g(a-h') = f'(c)$ for some $c \in (a-h',a)$ by the mean value theorem. Therefore, $g(a-h') = f'(c) < f'(a)$ since $a-h < a-h' <c < a$. Similarly, $g(a) = f'(d)$ for some $d \in (a,a+h')$, and again, $a+h > a+h'>c>a$, so that $f'(d) > f'(a)$. Therefore, $g(a) > f'(a) > g(a-h')$.

It follows by continuity of $g$ and the intermediate value theorem that there exists some $y \in (a-h',a)$ such that $g(y) = f'(a)$. But then, $g(y) = \frac{f(y+h') - f(y)}{(y+h')-y}$, where $y < a$ and $y+h' > a-h'+h' > a$. Therefore, put $x_1 = y$ and $x_2 = y+h'$ to get your result. Something stronger has been shown to hold, though.

If $f$ is twice differentiable and $f''(a) > 0$ for some $a$, then for all sufficiently small $h > 0$, there exist two points $x_1< a < x_2$ which are separated by distance $h$, such that $\frac{f(x_2)-f(x_1)}{x_2-x_1} = f'(a)$.

Answer 2

If we translate the problem in geometric language then we are supposed to find two points on graph of $f$ for which the chord is parallel to tangent at $a$. This is clearly true if the graph is convex or concave at $a$ ie $f''(a) \neq 0$. Here we are supposed to deal with $f''(a)>0$.

Consider the equation of tangent at $a$ given by $$Y-f(a) =(X-a) f'(a) $$ and any line parallel to it is given by $$Y-f(a) =(X-a) f'(a) +c, c\neq 0$$ We just need to show that there is a value of $c$ for which this line intersects the graph of $f$ at two points (one to the left of $a$ another to the right of $a$). Since $f''(a) >0$ the ratio $(f(x) - f(a)) /(x-a)=f' (d) $ is greater than $f'(a) $ for all values of $x$ sufficiently near $a$ and greater than $a$. Let one such value of $x$ be $q$ so that $$\frac{f(q) - f(a)} {q-a} >f'(a), q>a\implies A=f(q) - f(a)-(q-a)f'(a) >0$$ and similarly there is a $p$ such that $$\frac{f(p) - f(a)} {p-a} <f'(a), p<a\implies B=f(p) - f(a) - (p - a)f'(a) >0$$ Let $c$ be chosen such that $$0<c<\min(A,B)$$ and let $$g(x)=f(x) - f(a) - (x-a)f'(a) - c$$ Then $g$ is continuous and $g$ changes sign in $[p, a] $ hence vanishes at some $x_1\in (p, a) $ and again it changes sign in $[a, q] $ so it vanishes at $x_2\in(a,q)$. Thus the line $$Y-f(a) =(X-a) f'(a) +c$$ intersects the graph of $f$ at these points $x_1,x_2$.