CUET Preparation Today
The interval in which the function f given by $f(x)=x^3+\frac{1}{x^3}, x \neq 0$ is increasing: |
$(-\infty, \infty)$ $(-1,0) \cup(0,1)$ $(-\infty,-1) \cup(1, \infty)$ $(-1,1)$ |
$(-\infty,-1) \cup(1, \infty)$ |
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 $(-\infty,-1) \cup(1, \infty)$ |
