Question

The point on the curve $y=12 x-x^{3}$ 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

Hint:
In the given question, gradient is zero and it is calculated by dividing the change in y coordinate by change in x coordinate. First we differentiate the given equation by applying the derivative power rule. After differentiating, we need to put the given point in the differentiated equation. Then we can find the point of the given curve.

Formula used:
If $y=f(x)+g(x)$, then $\dfrac{dy}{dx}=f'(x)+g'(x)$
$\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$

 Complete Step by step solution:
Given that
$y=12 x-x^{3}$
Differentiate both the sides
We get
$\dfrac{dy}{dx}=\dfrac{d}{dx}(12x-{{x}^{3}})$
$\dfrac{d y}{ d x}=12-3 x^{2}$
Differentiation is the ratio of a slight change in one quantity to a little change in another that depends on the first quantity. If $y = f(x)$ and $f(x)$ are differentiable, then $f'(x)$ or $dy/dx$ is used to indicate the differentiation.
Equating gradient to zero.
$\dfrac{d y}{ d x}=12-3 x^{2}=0$
$x=\pm 2$
At these value of x, value of y:
$y=12x-{{x}^{3}}$
Put $x=+2$
$y=12(2)-{{(2)}^{3}}$
$y=24-8=16$
Now put $x=-2$
$y=12(-2)-{{(-2)}^{3}}$
$y=-24-(-8)$
$y=-16$
 The points are $(2, 16)$ and $(-2, -16)$.

So the correct answer is option(D)

Note:
Remember that the gradient is positive when m is greater than zero and it is negative when m is less than zero. If m is equal to zero then it is a constant function. The gradient of two parallel lines is equal and also the product of the gradient of two perpendicular lines is -1.