Let I be any interval disjoint from (−1, 1).

We have,

$f(x)=x+\frac{1}{x}$

$\therefore f^{\prime}(x)=1-\frac{1}{x^{2}}$

Now,

$f^{\prime}(x)=0 \Rightarrow \frac{1}{x^{2}}=1 \Rightarrow x=\pm 1$

The points $x=1$ and $x=-1$ divide the real line in three disjoint intervals i.e., $(-\infty,-1),(-1,1)$, and $(1, \infty)$.

In interval (−1, 1), it is observed that:

$-1

$\Rightarrow x^{2}<1$

$\Rightarrow 1<\frac{1}{x^{2}}, x \neq 0$

$\Rightarrow 1-\frac{1}{x^{2}}<0, x \neq 0$

$\therefore f^{\prime}(x)=1-\frac{1}{x^{2}}<0$ on $(-1,1) \sim\{0\} .$

$\therefore f$ is strictly decreasing on $(-1,1) \sim\{0\}$.

In intervals $(-\infty,-1)$ and $(1, \infty)$, it is observed that:

$x<-1$ or $1

$\Rightarrow x^{2}>1$

$\Rightarrow 1>\frac{1}{x^{2}}$

$\Rightarrow 1-\frac{1}{x^{2}}>0$

$\therefore f^{\prime}(x)=1-\frac{1}{x^{2}}>0$ on $(-\infty,-1)$ and $(1, \infty)$.

$\therefore f$ is strictly increasing on $(-\infty, 1)$ and $(1, \infty)$.

Hence, function f is strictly increasing in interval I disjoint from (−1, 1).

Hence, the given result is proved.