Question

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

Answer 1

Let $x_1,x_2\in (a,b)$ such that $x_1<x_2$.
Consider the sub-interval [$x_1,x_2$]. Since f (x) is differentiable on (a, b) and $[x_1,x_2]\subset (a,b)$.
Therefore, f(x) is continous on [$x_1,x_2$] and differentiable on $(x_1,x_2)$.
By the Lagrange's mean value theorm, there exists $c\in (x_1,x_2)$ such that
$f\text{'}\left(c\right)=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}...\left(1\right)$
Since f'(x) > 0 for all $x\in (a,b)$, so in particular, f'(c) > 0
$$\begin{gathered} f\text{'}\left(c\right)>0 \\ \Rightarrow \frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}>0[\mathrm{Using}(1\left)\right] \end{gathered}$$
$\Rightarrow f\left(x_2\right)-f\left(x_1\right)>0$ [∵ $x_2-x_1>0\mathrm{when}{x}_1<x_2$]
$\Rightarrow f\left(x_2\right)>f\left(x_1\right)\Rightarrow f\left(x_1\right)<f\left(x_2\right)$
Since $x_1,x_2$ are arbitrary points in $(a,b)$.
Therefore, $x_1<x_2\Rightarrow f\left(x_1\right)<f\left(x_2\right)\mathrm{for}\mathrm{all}{x}_1,x_2\in (a,b)$
Hence, f (x) is increasing on (a,b).