Question

Find the intervals in which the function f given by `f(x) = x^3 + 1/x^3 x != 0`, is (i) increasing (ii) decreasing.

Answer 1

We have `f (x) = x^3 + 1/x^3`

Differentiating w.r.t x, we get

`f' (x) = 3x^2 - 3/x^4`

(i) For f(x) to be increasing function of x,

`f' (x) > 0`

⇒ x⁶ - 1 > 0

⇒ (x³ - 1) (x³ + 1) > 0

Either x³ - 1 > 0 or x³ + 1 > 0

⇒ x³ > 1 or x³ > -1 

⇒ x > 1 and x > -1

⇒ x > 1 

⇒ x ∈ (1, ∞)

or x³ - 1 < 0 and x³ + 1 < 0

x³ < 1 and x³ < -1 

⇒ x < 1 and x < -1

⇒ x < -1

⇒ x ∈ (-∞, -1)

Hence, f(x) is increasing in (-∞, -1) ∪ (1,∞)

(ii) For f (x) to be decreasing function of x, 

f' (x) < 0

⇒ `3 (x^2 - 1/x^4) < 0`

⇒ `x^2 - 1/x^4 < 0`

⇒ x⁶ - 1 < 0

⇒ (x³ - 1) (x³ + 1) < 0

Either x³ - 1 > 0 and x³ + 1 < 0

⇒ x³ > 1 and x³ < -1

⇒ x > 1 and x < -1

Which is not possible

or x³ - 1 < 0 and x³ + 1 > 0

⇒ x³ < 1 and x³ > -1 

⇒ x < 1 and x > -1

⇒ -1 < x < 1

Hence, f (x) is decreasing in (-1,1).