Let $I$ be any interval disjoint from $\left( { - 1,1} \right)$. Prove that the function $f$ given by $f(x) = x + \cfrac{1}{x}$ is strictly increasing on $I$.

We have, $f(x) = x + \cfrac{1}{x},x \in I$ …(i)

Differentiating (i) w.r.t. $x$, we get $f'(x) = 1 - \cfrac{1}{{{x^2}}} = \cfrac{{{x^2} - 1}}{{{x^2}}}$

As ${x^2} > 0$ and in $I,{x^2} - 1 > 0 \Rightarrow {x^2} > 1 \Rightarrow x < - 1$ or $x > 1$

$\Rightarrow x \in ( - \infty ,\; - 1)$ or $x \in (1,\;\infty ) \Rightarrow x \in ( - \infty ,\; - 1) \cup (1,\;\infty )$

$\Rightarrow x \in R - ( - 1,1)f(x)$ is strictly increasing on $I$
[$I$ is an interval which is a subset of $R - ( - 1,1)$]