Question

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

Answer 1

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

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

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

f(x) is strictly decreasing in the interval for which f'(x) < 0

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

i.e. 1 – x² < 0

i.e. (1 – x)(1 + x) < 0

i.e. 1 – x < 0 and 1 + x > 0

or 

1 – x > 0 and 1 + x < 0

i.e. 1 < x and x > –1, (i.e. x > 1)

Therefore f(x) is decreasing when

x ∈ (–∞, –1) ∪ (1, ∞)