CUET Preparation Today
At what value of x is the function y = \( { x }^{ 3 } -36x\) attains its extreme values? |
$2\sqrt { 3 }$ $-2\sqrt { 3 }$ Both \(3\sqrt { 2 }\) and \(-3\sqrt { 2 }\) Neither \(2\sqrt { 3 }\) nor \(-2\sqrt { 3 }\) |
Both \(3\sqrt { 2 }\) and \(-3\sqrt { 2 }\) |
The correct answer is Option (3) → Both \(3\sqrt { 2 }\) and \(-3\sqrt { 2 }\) $f(x)=x^3-36x$ and, for critical points $f'(c)=0$ $⇒3x^2-12=0$ $⇒x^2=12$ $⇒x=±\sqrt{12}=3\sqrt{2}\,or\,-3\sqrt{2}$ |
