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

Question:

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).

Solution:

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)>0f^{\prime}(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 $(a, b)$.