# Differentiation Problem

1. Sep 27, 2007

### tigerd12

The problem

The Diagram shows the graph of y=x^3-12x+17
A is the maximum point on the curve
C is the minimum point on the curve
The curve crosses the y axis at B

For the equation find dy/dx, y=x^3-12x+17 (DONE)

Heres the problem

find the gradient of the curve at B

now what am I supposed to do here?

this is what i tried but im not sure if its right

3x^2-12=17

make x the subject therefore giving

x=$$\sqrt{}29/3$$

so is that right or..?

Last edited: Sep 27, 2007
2. Sep 27, 2007

### dextercioby

For b) you need to find the value of the derivative at the point B. What are the coordinates of the point B ?

3. Sep 27, 2007

### tigerd12

im guessing the coordinates at point B are (0,17), but how does that help me in finding the gradient at that point

4. Sep 27, 2007

### HallsofIvy

Staff Emeritus
I presume you found that the derivative is 3x^2- 12. By why are you asserting that the derivative is 17? You are supposed to be finding the value of the derivative!

No, it's not right- you weren't even asked for a value of x!

You know that f'(x)= 3x^2- 12. You are asked to find its value at point B. Of course, to do that you need to know what x is there. Saying the graph crosses the x-axis tells you that y= 0. Okay y= x^3- 12x+ 17= 0. Can you determine what x is from that?

5. Sep 27, 2007

### tigerd12

Sorry, I don't seem to be understand what you are saying. Do you want me to find the solutions of the equation x^3-12x+17?

6. Oct 13, 2007

### danni7070

Remember. The curve crosses y-axis at point B. So you know the y = 0 but you have to find the x-value for the point B:

In other words. B = (x,0)

Also, you know that f'(x) = 3x^2-12

Let f'(x) = 0 and then find a value for x. Then you have got point B! :)

7. Oct 13, 2007

### HallsofIvy

Staff Emeritus
Yes, that's exactly what I said! Then find the derivative at that value of x.

?? No, the curve is y= f(x), not y= f'(x).