1. ## [SOLVED] Differentiation Question

Question:
The equation of a curve is $y = 6x^2 - x^3$. 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 $x$ which $6x^2 - x^3$ is a decreasing function of $x$.

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

Attempt:

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

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

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

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

$12x - 3x^2$
$3x(4 - x)$

$3x = 0$
$x = 0$

$4 - x = 0$
$-x = -4$
$x = 4$

$x = 0$ and $x = 4$

$y = 6(0)^2 - (0)^3 = 0$
$y = 6(4)^2 - (4)^3 = 32$

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

How to find the decreasing function of $x$ ?

2. Hello,

Perfect thing for the first question !

For the decreasing : a function decreases when its derivative is negative.

Thus, you have to solve for $3x(4-x)<0$.

A quick way to do is to say that it's negative if and only if the signs for 3x and 4-x are different, that is to say that if 3x is positive, then 4-x has to be negative, and if 3x is negative, 4-x has to be positive.

Hence the solutions are :

$\left(3x>0 \text{ AND } 4-x<0\right) {\color{red}\text{ or }} \left(3x<0 \text{ AND } 4-x>0\right)$

This should give you the set of values for x that you want

3. well another method is to actually find the gradient right before and after the stationary points. Negative value will mean the curve is moving downwards while positive means moving upwards.

just substitute an x value right after and before the stationary point to obtain y and compare it to the coordinate of the stationary point too see whether the value of y is increasing or decreasing