Question
Find the intervals in which the function f given by $\text{f}\text{(x)}=\text{x}^3+\frac{1}{\text{x}^3},\text{x}\neq0\text{ is}$ (i) increasing (ii) decreasing.

Answer

$\text{f}\text{(x)}=\text{x}^3+\frac{1}{\text{x}^3}$
$\therefore\ \text{f}'\text{(x)}=3\text{x}^2-\frac{3}{\text{x}^4}=\frac{3\text{x}^6-3}{\text{x}^4}$
Then,$\text{f}'\text{x}=0\Rightarrow3\text{x}^6-3=0\Rightarrow\text{x}^6=1\Rightarrow\text{x}\pm1$
Now, the points x = 1 and x = -1 divide the real line into three disjoint intervals i.e.,
$(-\infty,-1),(-1,1),\text{ and }(1,\infty).$
In intervals $(-\infty,-1) \text{ and } (1,\infty)$ i.e., when x < -1 and x > 1, f'(x) > 0.
Thus, when x < -1 and x > 1, f is increasing.
In interval (-1, 1) i.e., when -1 < x < 1, f'(x) <0.
Thus, when -1 < x < 1, f is decreasing.

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