# [SOLVED] Differentiation Question

• Apr 25th 2008, 11:05 AM
looi76
[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 $x$ ?
• Apr 25th 2008, 11:16 AM
Moo
Hello,

Perfect thing for the first question ! (Clapping)

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 :)
• Apr 25th 2008, 11:18 AM