1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gradient and differentiation

  1. Jan 28, 2009 #1
    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. jcsd
  3. Jan 28, 2009 #2

    danago

    User Avatar
    Gold Member

    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?
     
  4. Jan 29, 2009 #3
    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..?
     
  5. Jan 29, 2009 #4

    danago

    User Avatar
    Gold Member

    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?
     
  6. Jan 29, 2009 #5
    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.....?
     
  7. Jan 29, 2009 #6

    danago

    User Avatar
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?