Finding the Gradient and Solving for Zero Points on a Cubic Curve

[maobadi]

Homework Statement

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.

Homework Equations

y = 2x³ - 5x² + 46x + 87

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...?

 

[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?

 

[maobadi]

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..?

 

[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?

 

[maobadi]

solving the quadratic equation i get x = (5+18.5i) and (5-18.5i) which are the complex roots and does this mean that it will never touch the x-axis...?

 

[danago]

solving the quadratic equation i get x = (5+18.5i) and (5-18.5i) which are the complex roots and does this mean that it will never touch the x-axis...?

Right. By showing that the equation has only non-real solutions, you have shown that there exists no real value of x that makes dy/dx equal zero i.e. the curve never has a zero gradient.