# Cubic differenation problem

1. Jan 2, 2008

### thomas49th

Hi

The curve C has equation y = f(x) and the point P(3,5) lies on C

Given that

f'(x) = 3x² - 8x + 6

The point Q also lies on C, and the tangent to C at Q is parallel to the tangent to C at P

Find the x-coordinate of Q

So

if there parallel the gradients are equal

but i cant really seem to get anywhere after that :(

2. Jan 2, 2008

### CompuChip

Indeed, the gradients of y = f(x) at P and Q are equal. What is the value of that slope at P? Then, what other x gives the same slope?

3. Jan 3, 2008

### thomas49th

You factorise the derative of f(x) which is f'(x)
giving you
(3x+1)(x-3)

x = 3 or x = -1/3

we already know x = 3 so the other is -1/3

i believe that is correct?

4. Jan 3, 2008

### HallsofIvy

Staff Emeritus
Yes, that is correct but the way you phrased it confused me a bit! You did not factorize f '(x). You can easily calculate that the derivative at x= 3 is 9 so at the other point, you must have $3x^2- 8x+ 6= 9$ which then gives $3x^3- 8x- 3= 0$. That is what you factored, not the derivative of f.

Last edited: Jan 4, 2008