The interval in which the function f given by $f(x)=x^3+\frac{1}{x^3}, x \neq 0$ is increasing:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) - $(-\infty,-1) \cup(1, \infty)$
$f(x)=x^3+\frac{1}{x^3}$
$f'(x)=3x^2-\frac{3}{x^4}=0$
$⇒\frac{3(x^6-1)}{x^4}=0$
$x=±1$
$f'(x)>0$ in $(-\infty,-1) \cup(1, \infty)$