Question

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

Answer 1

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

$\therefore f^{\prime }\left(x\right)=1-\frac{1}{x^2}$
$f^{\prime }\left(x\right)=0$

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

$\Rightarrow x=\pm 1$
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:
$\Rightarrow x^2<1$
$\Rightarrow 1-\frac{1}{x^2}<0,x\ne 0$
Thus $f$ is strictly decreasing on $(-1,1)$
and $f^{\prime }\left(x\right)=1-\frac{1}{x^2}>0$ on $(-\infty ,-1)$ and $(1,\infty )$
$\therefore$ $f$ is strictly increasing on $(-\infty ,-1)\text{and}(1,\infty )$
Hence, function $f$ is strictly increasing in interval $I$ disjoint from $(-1,1)$.
Hence, the given result is proved.