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Find the values of x for which f(x) = xx2+1 is (a) strictly increasing (b) decreasing. - Mathematics and Statistics

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Question

Find the values of x for which f(x) = `x/(x^2 + 1)` is (a) strictly increasing (b) decreasing.

Sum
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Solution

f(x) = `x/(x^2 + 1)`

∴ f'(x) = `d/dx(x/(x^2 + 1))`

= `((x^2 + 1).d/dx(x) - xd/dx(x^2 + 1))/(x^2 + 1)^2`

= `((x^2 + 1)(1) - x(2x + 0))/(x^2 + 1)^2`

= `(x^2 + 1 - 2x^2)/(x^2 + 1)^2`

= `(1 - x^2)/(x^2 + 1)^2`

(a) f is strictly increasing if f'(x) > 0

i.e. if `(1 - x^2)/(x^2 + 1)^2 > 0`

i.e. if 1 – x2 > 0             ...[∵ (x2 + 1)2 > 0]
i.e. if 1 > x2
i.e. if x2 < 1
i.e. if – 1 < x < 1

∴ f is strictly increasing if – 1 < x < 1

(b) f is strictly decreasing if f'(x) < 0

i.e. if `(1 - x^2)/(x^2 + 1)^2 < 0`

i.e. if 1 – x2 < 0             ...[∵ (x2 + 1)2 > 0]
i.e. if 1 < x2
i.e. if x2 > 1
i.e. if  x > 1 or x < – 1

∴ f is strictly decreasing if x < – 1 or x > 1

i.e. `x ∈( - oo, - 1) ∪ (1, oo)`.

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Chapter 2: Applications of Derivatives - Exercise 2.4 [Page 90]

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