1. Jan 28, 2009

1. The problem statement, all variables and given/known data

1. Calculate the gradient of the curve y = 2x3 - 5x2 + 46x + 87 at the point where it crosses the x-axix.

2. Show by differentiation and solving a quadratic equation, that there are no points on the above curve where the gradient is zero.

2. Relevant equations

y = 2x3 - 5x2 + 46x + 87

3. The attempt at a solution

dy/dx = 6x2 - 10x + 46

It crossess at x-axis where y = 0 ( it it right....?) I then calculate the roots of the equation, and the two roots of x are the points where it crossess the x-axis.

How do I do the part 2 problem...?

Last edited: Jan 29, 2009
2. Jan 28, 2009

### danago

You have found the derivative of the function, so how about setting it equal to zero. If you then solve it, you will find all the x values where the derivative is zero (if they exist). What should you expect if no zero derivatives exist?

3. Jan 29, 2009

what do you mean by "What should you expect if no zero derivatives exist?" What do I do?
My question for the problem is are both part1 and part 2 the same thing. I mean part 1 says calculate and part 2 says show by differentiation and solving a quadratic equation. Do I have to differentiate(for part 2) one more time..?

4. Jan 29, 2009

### danago

The question asks you to show that the gradient of y = 2x3 - 5x2 + 46x + 87 is never zero. You have already found the derivative of the function, dy/dx = 6x2 - 10x + 46, which is the first step.

You have been asked to show that there exists no x such that the gradient is zero i.e. there are no such x values that make dy/dx = 0, or 6x2 - 10x + 46 = 0. So your task then becomes to show that no real x values satisfy the equation 6x2 - 10x + 46 = 0 i.e. the equation has no solutions. How can you show it has no real solutions?

5. Jan 29, 2009