## Filters

Q&A - Ask Doubts and Get Answers
Q

# Let I be any interval disjoint from [–1, 1]. Prove that the function f given by f of x is equal to x plus 1 by x is increasing on I.

15) Let I be any interval disjoint from [–1, 1]. Prove that the function f given by $f ( x) = x + 1/x$  is increasing on I.

Views

Given function is,
$f ( x) = x + 1/x$
$f^{'}(x) = 1 - \frac{1}{x^2}$
Now,
$f^{'}(x) = 0\\ 1 - \frac{1}{x^2} = 0\\ x^{2} = 1\\ x = \pm1$

So, intervals are from  $(-\infty,-1), (-1,1) \ and \ (1,\infty)$
In interval $(-\infty,-1), (1,\infty)$  ,  $\frac{1}{x^2} < 1 \Rightarrow 1 - \frac{1}{x^2} > 0$
$f^{'}(x) > 0$
Hence, $f ( x) = x + 1/x$ is increasing   in  interval  $(-\infty,-1)\cup (1,\infty)$
In interval (-1,1) ,  $\frac{1}{x^2} > 1 \Rightarrow 1 - \frac{1}{x^2} < 0$
$f^{'}(x) < 0$
Hence, $f ( x) = x + 1/x$ is decreasing   in  interval  (-1,1)
Hence, the function f given by $f ( x) = x + 1/x$  is increasing on I  disjoint from [–1, 1]

Exams
Articles
Questions