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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.
Solution
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).
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