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Calculus assignment

  1. Mar 20, 2007 #1
    1.Show that x^3 - 3kx + 1 = 0 does not have two distinct roots when k<0

    2.Locate the points of discontinuity for the functions :
    a.) f(x)= 2u(x-3)-u(x-4)
    b.) g(x)= [u(x)-u(x-pie/2)

    u(x) is a unit step function.

    Could u guys help me out, I have no idea how to solve this..
     
  2. jcsd
  3. Mar 20, 2007 #2
    The statement in number 1 is false if take into account complex roots. Otherwise:

    1. For a polynomial of degree 3, there are 3 possibilities. One root, two roots or three roots.

    For one root, the limits as the function goes to infinity and minus infinity are opposite in sign (these limits are either infinity or minus infinity). Also, the derivative has either 0 or 1 root.

    For two roots, the limits as the function goes to infinity and minus infinity have the same sign (these limits are either infinity or minus infinity). Also, the derivative has one root.

    For one root, the limits as the function goes to infinity and minus infinity are opposite in sign (these limits are either infinity or minus infinity). Also, the derivative has two roots.

    Now, with those criteria, can you find your answer?

    As for the second question, I think there's an infinite number of points at which the functions are discontinuous... number a is pretty obvious since the numbers in the displacement transformations are integers. As for b), look at u(1) and u (pi/2 - pi/2). It should help.
     
    Last edited: Mar 20, 2007
  4. Mar 20, 2007 #3
    I'm aware of those criterias, but I'm still not sure how to show that it doesn't have 2 distinct roots..
     
  5. Mar 20, 2007 #4
    ignoring imaginary roots?
     
  6. Mar 20, 2007 #5

    HallsofIvy

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    Assuming that by "does not have two distinct roots" you mean "has only one real root", then you need to show that the function is monotonic- always increasing or always decreasing. Do you know how to use the derivative to do that?

    For
    what is the definition of "point of discontinuity"? How would you recognize one? Are you able to draw the graphs of these functions? You say "u(x) is a unit step function". What does that mean? Can you graph u(x)?

    There are several reasons for asking people to show what work they have done on problems. One, of course, is that the more work you put into a problem yourself, the more you will learn from it- far more than seeing how another person solved the problem. Another is that we are able to see what techniques you have for solving problems- there are typically many ways to solve any given problem.

    If you are given a homework problem where you honestly have no idea at all how to even start, you have a serious problem that we can't help you with- go to your teacher and throw yourself on his/her mercy!
     
  7. Mar 20, 2007 #6
    Thanks Ivy, I think I kinda got the idea of how to solve the first question..

    About the 2nd one, I'm not sure what's unit step function, my lecturer has never mentioned it, even my friends r not sure about it..That's y I'm not sure how to approach the question when it comes to dealing with step function..Point of discontinuity is the point x where the line on the graph is not continuous..

    I really appreciate the replies..
     
  8. Mar 20, 2007 #7

    HallsofIvy

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    The unit step function (also called the Heaviside function) is 0 for x< 0,
    1 for x>= 0. It is continuous everywhere except at 0. Of course, u(x-3) is 0 for all x-3< 0 and 1 for all x-3>=0. In other words for x< 3 and x>=3. u(x-3) is discontinuous at x- 3= 0 or x= 3.
     
  9. Mar 20, 2007 #8
    ChocHeartz, when derivate the function, you find a second degree polynomial. Then you check for roots.
     
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