Question

Let f be a function defined on [a, b] such that f'(x) > 0, for all $xϵ$ [a, b]. Then prove that f is an increasing function on [a, b].

Answer 1

Given, f'(x) > 0 on [a, b]
$\therefore$ f is differentiable function on [a, b]
Also, every differentiable function is continuous, therefore f is continuous on [a, b].
Let $x_1,x_2ϵ[a,b]$ and $x_2>x_1$, then by LMV theorem, there exist $cϵ[a,b]$ such that
$f^{\prime }\left(c\right)=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}\Rightarrow f\left(x_2\right)-f\left(x_1\right)=(x_2-x_1)f\left(c\right)\Rightarrow f\left(x_2\right)-f\left(x_1\right)>0as{x}_2>x_{1}and{f}^{\prime }\left(x\right)>0\Rightarrow f\left(x_2\right)>f\left(x_1\right)\therefore For{x}_1<x_1\Rightarrow f\left(x_1\right)<f\left(x_2\right)$
Hence, f is an increasing function on (a, b).