. Lel $f$ 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 $(a,b).$

Let ${x_1},{x_2} \in (a,\;b)$ such that ${x_1} < {x_2}.$

Let the sub-interval be $[{x_1},{x_2}].$

Since $f(x)$ is differentiable on $(a,\;b)$ and $[{x_1},{x_2}] \subset (a,\;b)$

Therefore, $f(x)$ is continuous in $[{x_1},\;{x_2}]$ and differentiable in $({x_1},\;{x_2})$ .

Therefore by using L.M.V. theorem, there exists $c \in ({x_1},\;{x_2})$ such that $f'(c) = \cfrac{{f({x_2}) - f({x_1})}}{{{x_2} - {x_1}}}$

Now, $f'(x) > 0\;for\;all\;x \in (a,\;b) \Rightarrow f'(c) > 0$
$\Rightarrow \cfrac{{f({x_2}) - f({x_1})}}{{{x_2} - {x_1}}} > 0 \Rightarrow f({x_2}) - f({x_1}) > 0$

$\Rightarrow f({x_1}) < f({x_2})$ if ${x_1} < {x_2}$

Hence, $f$ is increasing in $(a,\;b)$. [ are arbitrary]