Let f be a function defined on [a, b] such that


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


Let $x_{1}, x_{2} \in(a, b)$ such that $x_{1}

Consider the sub-interval $\left[x_{1}, x_{2}\right] .$ Since $f(x)$ is differentiable on $(a, b)$ and $\left[x_{1}, x_{2}\right] \subset(a, b)$.

Therefore, $\mathrm{f}(\mathrm{x})$ is continous on $\left[x_{1}, x_{2}\right]$ and differentiable on $\left(x_{1}, x_{2}\right)$.

By the Lagrange's mean value theorm, there exists $c \in\left(x_{1}, x_{2}\right)$ such that

$f^{\prime}(c)=\frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}} \quad \ldots(1)$

Since $f(x)>0$ for all $x \in(a, b)$, so in particular, $f(c)>0$


$\Rightarrow \frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}>0 \quad[\operatorname{Using}(1)]$

$\Rightarrow f\left(x_{2}\right)-f\left(x_{1}\right)>0 \quad\left[\because x_{2}-x_{1}>0\right.$ when $\left.x_{1}

$\Rightarrow f\left(x_{2}\right)>f\left(x_{1}\right) \Rightarrow f\left(x_{1}\right)

Since $x_{1}, x_{2}$ are arbitrary points in $(a, b)$.

Therefore, $x_{1}

Hence, $f(x)$ is increasing on $(a, b)$.

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