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

Toolbox:
• A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
• If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
• A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
Let $f(x)=x+\large\frac{1}{x}$
Differentiable w.r.t $x$,
$f'(x)=1-\large\frac{1}{x^2}$
$f'(x)=0\Rightarrow \large\frac{1}{x^2}$$=1 \Rightarrow x=\pm 1 The points x=1 and x=-1 divide the real number line into three disjoint intervals . (i.e) (-\infty,-1),(-1,1) and (1,\infty) Step 2: Consider the interval (-1,1) -1 < x < 1 \Rightarrow x^2 < 1 \Rightarrow 1 < \large\frac{1}{x^2}$$,x\neq 0$
$\Rightarrow 1-\large\frac{1}{x^2}$$<0,x\neq 0 Therefore f'(x)=1-\large\frac{1}{x^2}$$ < 0$ on $(-1,0)-\{0\}$
Therefore $f$ is strictly decreasing on $(-1,1)-\{0\}$
Step 3:
Consider the intervals $(-\infty,-1)$ and $(1,\infty)$ it can be seen that,
$x < -1$ or $1 < x$
$\Rightarrow x^2 > 1$
$\Rightarrow 1 >\large\frac{ 1}{x^2}$
$\Rightarrow 1 -\large\frac{ 1}{x^2} >$$0 Therefore f'(x)=1-\large\frac{1}{x^2}$$>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 first disjoint from $(-1,1)$.
Hence proved.
answered Jul 10, 2013