# Stuck in calculus and derivative question

Let $$f:[a,b] \to \mathbb{R}$$ continues function such that $$f(b)>f(a)$$ and $$f$$ is not linear (meaning $$f \not= c x +d$$)

And $$f$$ is differential in $$(a,b)$$ , prove that there is $$c \in (a,b)$$ such that :

$$f'(c) > \frac{f(b)-f(a)}{b-a}$$

By Lagrange theorem i know that there is $$t \in (a,b)$$ such that $$f'(t) = \frac{f(b)-f(a)}{b-a}$$ but how to get strictly bigger and not just equal?

• Could it happen that $f'(t)\le\frac{f(b)-f(a)}{b-a}$ for all $t\in (a,b)$? – Ted Shifrin Dec 23 '18 at 17:02

## 1 Answer

Let$$\begin{array}{rccc}g\colon&[a,b]&\longrightarrow&\mathbb R\\&x&\mapsto&f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a).\end{array}$$Then $$g(a)=g(b)=0$$. On the other hand, $$g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}$$ and so asserting that there is no such $$c$$ is equivalent to asserting that $$g'(x)\leqslant0$$ for each $$x\in[a,b]$$. But then $$g$$ is decreasing. The only way that a decreasing function from $$[a,b]$$ to $$\mathbb R$$ has zeros at $$a$$ and $$b$$ is that $$g$$ is the null function. But then$$(\forall x\in[a,b]):f(x)=f(a)+\frac{f(b)-f(a)}{b-a}(x-a).$$So, $$f$$ would be linear.