Question

The points on the curve $y=12x-x^3$ at which the gradient is zero are

• A. (0, 2), (2,16)
• B. (0, – 2), (2, – 16)
• C. (2, –16), (– 2, 16)
• D. (2, 16), (– 2, – 16)

Answer 1

The correct option is D (2, 16), (– 2, – 16)

Given function $y=12x-x^3$
Since, slope of the tangent of given function is $\frac{dy}{dx}=12-3x^2$
Since, gradient is zero, so slope must be zero.
$\Rightarrow 12-3x^2=0$
$\Rightarrow 12=3x^2$
$\Rightarrow x^2=4$
$\therefore x=\pm 2$
When $x=2$ then $y=12\left(2\right)-2^3=24-8=16$
When$x=-2$ then $y=12(-2)-(-2{)}^3=-24+8=-16$
Hence, points are $(2,16)$ and $(–2,–16)$.