At what value of x is the function y = x ^ 3 -36x attains its extreme…

The correct answer is Option (4) →Both  \(2\sqrt { 3 }\) and \(-2\sqrt { 3 }\)

$f(x)=x^3-36x$

and, for critical points $f'(x)=0$

$f'(x)=3x^2-36$

$⇒3x^2-36=0$

$⇒3x^2=36$

$⇒x^2=12$

$⇒x=±\sqrt{12}=2\sqrt{3}\,and \,-2\sqrt{3}$