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Defining a function and a delta limit problem

  1. Jun 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Im having a lot of dificulties evaluating this function, I really need some easy to understand explanation about how to evaluate it by the given values, I would appreciate any help.


    -Problem one

    given the function:

    U(x) =

    0 if x < 0
    1 if 0 ≤

    define the function by

    a.) U(x-1)
    b.) U(x) - 1
    c.) U(x) - U(x-1)



    The solutions are

    a.) U(x-1) =

    0 if x < 1
    1 if 1 ≤ x


    b.) U(x) - 1 =

    -1 if x < 0
    0 if 0 ≤ x


    c.) U(x) - U(x-1) =

    0 if x < 0
    1 if 0 ≤ x < 1
    0 if 1 ≤ x







    -Problem 2
    given f(x), a, L and epsilon, determine that delta > 0


    the given function is x^2 -2x +1 , a = 2 , L = 1 and epsilon = 0.4



    so I tried getting f(x1) and f(x2)

    f(x1) = 0.6 and f(x2) = 1.4



    but then I got stuck with the algebra when I had to get x1 and x2.

    (x1)^2 - 2(x1) + 1 = o.6


    and


    (x2)^2 - 2(x2) + 1 = 1.4



    how can I solve these problems?


    thanks a lot in advance.
     
  2. jcsd
  3. Jun 2, 2009 #2

    Mark44

    Staff: Mentor

    For #1, have you sketched a graph of the function? The graph consists two horizontal lines. If you know what the graph of y = f(x) looks like, the graph of y = f(x -c) (where c is a positive number) is the translation to the right of y = f(x) by c units.

    The graph of y = f(x) - c, (again c > 0), is the translation down of the graph of y = f(x) by c units.

    For #2, the problem as you state it doesn't make much sense:
    From the rest of what you show, it appears that you are trying to establish that
    [tex]\lim_{x \rightarrow 2} x^2 - 2x + 1 = 1[/tex]

    Draw a graph of y = x2 - 2x + 1 if you haven't already done so. You are given an epsilon of .4, so you are trying to find a value for delta, so that if |x - 2| < delta, then |f(x) - 1| < .4.


    The equations x12 - 2x1 + 1 = .6 and x22 - 2x2 + 1 = 1.4 are quadratic equations. Surely you have had practice in solving quadratic equations if you are starting in on delta-epsilon proofs of limits. After you have found x1 and x2, take delta to be the smaller of the differences |x1 - 2| and |x2 - 2|.
     
  4. Jun 2, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    In the first one, you have just replaced x by x-1 so in the conditions x< 0, 0< x, replace that x by x-1 also: x-1< 0 and 0< x-1. What does that simplify to?
     
  5. Jun 3, 2009 #4
    Thanks a lot guys I already have the graphs for every problem. Im still having dificulties with problem 2.

    Im not sure about how to solve this:


    (x1)^2 - 2(x1) + 1 = 0.6


    and


    (x2)^2 - 2(x2) + 1 = 1.4



    I tried doing this but I dont think that it is right.



    (x1)^2 - 2(x1) + 1 = 0.6

    (x1)^2 - 2(x1) = -0.4

    (x1)^2 = 0.2

    x1 = square root of 0.2

    x1 = 0.447




    and I couldnt solve (x2)^2 - 2(x2) + 1 = 1.4
     
  6. Jun 3, 2009 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It would help a lot to recognize that [itex]x^2- 2x+ 1= (x- 1)^2[/itex].
     
  7. Jun 3, 2009 #6

    Mark44

    Staff: Mentor

    Or it would help if you reviewed how to solve quadratic equations.
    How did you get this? What happened to the -2x1 term and the -.4 term?
    If you're trying to solve the equation x12 -2x1 + 1 = .6, HallsOfIvy's suggestion of recognizing the perfect square trinomial on the left side is the quickest way. If you don't recognize that, you can always use the quadratic formula, after a bit of rearrangement.

    x12 - 2x1 +1 = 0.6
    ==> x12 - 2x1 + 0.4 = 0

    Now you can use the quadratic formula, with a = 1, b = -2, and c = 0.4.
     
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