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Show that the local maximum value of x+1x is less than local minimum value. - Mathematics

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Question

Show that the local maximum value of `x + 1/x` is less than local minimum value.

Sum

Solution

Let y = `x + 1/x` 

⇒ `"dy"/"dx" = 1 - 1/x^2`

`"dy"/"dx"` = 0

⇒ x2 = 1

⇒ x = ± 1.

`("d"^2y)/("dx"^2) = + 2/x^3`

Therefore `("d"^2y)/("dx"^2)` (at x = 1) > 0 and `("d"^2y)/("dx"^2)` (at x = –1) < 0.

Hence local maximum value of y is at x = –1 and the local maximum value = – 2.

Local minimum value of y is at x = 1 and local minimum value = 2.

Therefore, local maximum value (–2) is less than local minimum value 2.

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Chapter 6: Application Of Derivatives - Solved Examples [Page 124]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 6 Application Of Derivatives
Solved Examples | Q 10 | Page 124

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