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

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Question

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

Sum
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Solution

We have `f (x) = x + 1/x, x in I`

Differentiating w.r.t.x, we get

`f' (x) = 1 - 1/x^2 = (x^2 - 1)/x^2`

`x^2 > 0 (1, 1), x^2 - 1 > 0 = x^2 > 1`

= `x < - 1 or x > 1`

= `x in (-oo, -1) or x in (1, oo)`

= `x in (-oo, -1) cup (1, oo) `

= `x in R -  (-1, 1)`

= f (x) is strictly increasing on I

(∵ I is an interval which is a subset of R - (-1, 1))

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Chapter 6: Application of Derivatives - Exercise 6.2 [Page 206]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 6 Application of Derivatives
Exercise 6.2 | Q 15 | Page 206

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