The point on the curve y=12x-x3 at which the gradient is zero are
(0,2),(2,16)
(0,-2),(2,-16)
(2,-16),(-2,16)
(2,16),(-2,-16)
Compute the required points:
Given: y=12x-x3………………1
Since the gradient is zero
So, dydx=0
⇒12-3x2=0
⇒ 3x2=12
⇒ x2=4
⇒ x=±2
Substitute the value of x in equation 1,
ix=2⇒y=24-8⇒y=16
iix=-2⇒y=-24+8⇒y=-16
Therefore, the points are 2,16 and -2,-16.
Hence option D is the correct answer.
The points on the curve y=12x−x3 at which the gradient is zero are