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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?

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

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