Find the intervals in which the function f given by $f (x) = x^{3} + \frac{1}{x^{3}}, x \neq 0$ is
(i) increasing
(ii) decreasing.

Solution

$f (x) = x^{3} + \frac{1}{x^{3}}, x \neq 0 \Rightarrow f' (x) = 3 x^{2} - 3 x^{- 4} = 3 (x^{2} - \frac{1}{x^{4}}) \Rightarrow f' (x) = 3 x^{2} - 3 x^{- 4} = \frac{3}{x^{4}} (x^{6} - 1) \Rightarrow f' (x) = \frac{3}{x^{4}} (x^{2} - 1) (x^{4} + x^{2} + 1) \Rightarrow f' (x) = 3 (\frac{x^{4} + x^{2} + 1}{x^{4}}) (x^{2} - 1)$

(i) For an increasing function, we should have,

$f' (x) > 0 \Rightarrow 3 (\frac{x^{4} + x^{2} + 1}{x^{4}}) (x^{2} - 1) > 0 \Rightarrow (x^{2} - 1) > 0 [∵ 3 (\frac{x^{4} + x^{2} + 1}{x^{4}}) > 0] \Rightarrow (x - 1) (x + 1) > 0 \Rightarrow x \in (- \infty, - 1) \cup x \in (1, \infty)$

So, f( x ) is increasing on $(- \infty, - 1) \cup (1, \infty)$

(ii) For a decreasing function, we should have f ( x ) $<$ 0

$f' (x) < 0 \Rightarrow 3 (\frac{x^{4} + x^{2} + 1}{x^{4}}) (x^{2} - 1) < 0 \Rightarrow (x^{2} - 1) < 0 [∵ 3 (\frac{x^{4} + x^{2} + 1}{x^{4}}) > 0] \Rightarrow (x - 1) (x + 1) < 0 \Rightarrow x \in (- 1, 0) \cup x \in (0, 1)$

So, f ( x ) is decreasing on ( -1, 0 ) $\cup$ ( 0, 1 )

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Application Of Derivatives

Find the intervals in which the function f given by $f (x) = x^{3} + \frac{1}{x^{3}}, x \neq 0$ is
(i) increasing
(ii) decreasing.

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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Application Of Derivatives

Find the intervals in which the function f given by f ( x ) = x3 + 1x3, x ≠ 0 is(i) increasing(ii) decreasing.

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find the intervals in which the function f given by $f (x) = x^{3} + \frac{1}{x^{3}}, x \neq 0$ is
(i) increasing
(ii) decreasing.