Question

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

Answer 1

We have,

f(x)=x+ \(\frac{1}{x}\)

f'(x)=1=\(\frac{1}{x}^2\)

Now,

f'(x)=0=\(\frac{1}{x}^2\)=1x⇒≠1

The points x = 1 and x = −1 divide the real line in three disjoint intervals i.e.,

(-∞,-1).(-1,1), and (1.∞).

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

-1<x<1

⇒ x2<1

⇒ 1<\(1<\frac{1}{x}^2,\)x≠0

⇒ \(1-1\frac{1}{x}2<0,\) x≠0

∴ f'(x)=1-1/x2<0 on (-1,1)∼{0}.

∴ f is strictly decreasing on (-1,1){0}.

In intervals(-∞,-1) and (1,∞),it is observed that:

x<-1 or 1<x

⇒x2>1

⇒\(1>\frac{1}{x}^2\)

⇒ \(1-\frac{1}{x}^2\)>0

∴ f'(x)=1=\(\frac{1}{x}^2\) on (-∞,1) and (1,∞).

∴ f is strictly increasing on (-∞,1) and (1,-∞).

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