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

Toolbox:
• $f'(c)=\large\frac{f(x_2)-f(x_1)}{x_2-x_1}$
Step 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 differentiate on $(a,b)$ and $[x_1,x_2]\in (a,b)$
Therefore $f(x)$ is continuous on $[x_1,x_2]$ and differentiable on $(x_1,x_2)$
By the Lagrange's mean value theorem ,there exists $c\in (x_1,x_2)$
Such that $f'(c)=\large\frac{f(x_2)-f(x_1)}{x_2-x_1}$
Step 2:
Since $f'(x) >0$ for all $x\in (a,b)$.
So in particular $f'(c)>0$
Now $f'(c) >0$