Question:

The equation of a curve is $\displaystyle y = 6x^2 - x^3$. Find the coordinates of the two stationary points on the curve, and determine the nature of each of these stationary points.

State the set of values of $\displaystyle x$ which $\displaystyle 6x^2 - x^3$ is a decreasing function of $\displaystyle x$.

The gradient at the point $\displaystyle M$ on the curve is$\displaystyle 12$. Find the equation of the tangent to the curve at $\displaystyle M$.

Attempt:

$\displaystyle y = 6x^2 - x^3$

$\displaystyle \frac{dy}{dx} = 6\times2^{2-1} - 3x^{3-1}$

$\displaystyle \frac{dy}{dx} = 12x - 3x^2$

$\displaystyle \frac{dy}{dx} = 0$

$\displaystyle 12x - 3x^2$

$\displaystyle 3x(4 - x)$

$\displaystyle 3x = 0$

$\displaystyle x = 0$

$\displaystyle 4 - x = 0$

$\displaystyle -x = -4$

$\displaystyle x = 4$

$\displaystyle x = 0$ and $\displaystyle x = 4$

$\displaystyle y = 6(0)^2 - (0)^3 = 0$

$\displaystyle y = 6(4)^2 - (4)^3 = 32$

So, the two coordinates are $\displaystyle (0,0)$ and $\displaystyle (4,32)$.

How to find the decreasing function of $\displaystyle x$ ?