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Homework Help: Interesting piece-wise problem

  1. May 6, 2010 #1
    I hate a piece-wise function

    f(x)= [ 1 if 0 </eq x </1
    [ 0 if 1 < x < 2
    [ -2x+4 if x >/eq 2

    If I let F(x)= integral (from 0->x) f(t) dt, how can i find a formula for F(x) without using integrals? I think it may have something to do with the 2nd definition of the Fundamental Theorem of Calculus, but I'm not sure.
     
  2. jcsd
  3. May 6, 2010 #2

    Mark44

    Staff: Mentor

    Use <= instead of </eq, and >= instead of >/eq.
    Have you graphed this function? After you do this, you should be able to come up with a formula for F(x), using nothing more sophisticated than geometry.
     
  4. May 6, 2010 #3
    Should I be using limits? I actually did graph it. I know what the F(x) should be, I'm just not sure how to rigorously show it using a theorem, formula, etc
     
  5. May 6, 2010 #4

    Mark44

    Staff: Mentor

    No, you don't need limits. This is mostly a geometry problem. Your formula for F(x) will be defined in a piecewise fashion.
    To get you started,
    F(x) = x, 0 <= x < 1

    You will need two more definitions, one for each of the two additiona intervals. The formula for F(x) for 1 <= x < 2 is very easy.
     
  6. May 6, 2010 #5
    Is this correct? I'm worried the last one should be more complicated?

    F(x) = x, 0 <= x <= 1
    = 0, 1 < x < 2
    = -x^2+4x, x >= 2.
     
  7. May 6, 2010 #6
    Well it should actually be F(x)=-x^2+4x-4 for x>= 2 right?
     
  8. May 6, 2010 #7
    Just one??? I hate ALL piece-wise functions!
     
  9. May 6, 2010 #8
    Also, is there a way for me a find a formula for F'? How do I first find where F is differentiable?
     
  10. May 6, 2010 #9

    Mark44

    Staff: Mentor

    The second and third parts are wrong. Remember that F(x) is defined as
    [tex]\int_0^x f(t) dt[/tex]

    So, in particular, F(1.5) [itex]\neq[/itex] 0, as you have, and F(2) [itex]\neq[/itex] 4, as you have. F(2) should actually be equal to 1.
     
  11. May 6, 2010 #10

    Mark44

    Staff: Mentor

    I divide functions into three categories:
    • the ones I hate
    • the ones I love
    • the ones that I feel neutral about

    :smile:
     
  12. May 6, 2010 #11
    so should F(x)=x also if 1<x<2?
    F(x)= int (1,2)= int(0,1) 1 dx + int (1,2) 0 dx = x + 0 = 0

    is that correct?
     
  13. May 6, 2010 #12
    Actually shouldn't F(x)= 1, 0<=x<=1, F(x)=1, 1<x<2, and F(x)=-x^2+4x-3, x>=2
     
  14. May 6, 2010 #13

    Mark44

    Staff: Mentor

    Edit: This refers to post 11.
    No, that's not it either. For example, what is F(1.5)? It's not 1.5.
    I'm going to translate part of the line above into actual mathematics.
    [tex]F(2) = \int_0^1 1~dx + \int_1^2 0~dx [/tex]
    The value of the first integral is not x, and x + 0 is not 0.
     
  15. May 6, 2010 #14

    Mark44

    Staff: Mentor

    Edit:Changed my answer.
    Well, you're close. The first part should be F(x) = x, 0 <= x <= 1.
     
    Last edited: May 6, 2010
  16. May 6, 2010 #15
    But we're evaluating it from (0,1)... So once we integrate and get x|, then we have 1-0=1. This isn't correct?
     
  17. May 6, 2010 #16
    Oh and in Post #11, I meant to say:
    F(x)= int (1,2)= int(0,1) 1 dx + int (1,2) 0 dx = x + 0 = 1
     
  18. May 6, 2010 #17
    Oh and in Post #11, I meant to say:
    F(x)= int (1,2)= int(0,1) 1 dx + int (1,2) 0 dx = x + 0 = 1
     
  19. May 6, 2010 #18

    Mark44

    Staff: Mentor

    A lot of this is gibberish. F(x) is not the integral from 1 to 2 and it's not the integral from 0 to 2, and it's not x, which is not 1.

    F(x) needs to be defined in piecewise fashion.
     
  20. May 6, 2010 #19
    okay well thanks, anyways
     
  21. May 6, 2010 #20

    Mark44

    Staff: Mentor

    I'm not trying to discourage you (although I might be succeeding), but I am trying to get you to write things that make sense.
     
  22. May 6, 2010 #21
    F(x)= x if 0 <= x <= 1.

    Then, for 1< x <2, we're taking the integral of 0, so I should just get a constant?
    I just don't know how I am supposed to write that part of the third part whre x>=2
     
  23. May 6, 2010 #22

    Mark44

    Staff: Mentor

    [tex]F(x) = \int_0^x f(t) dt[/tex]
    If 0 <= x <= 1, F(x) = x
    If 1 < x <= 2, F(x) = 1 + 0 = 1. The 1 comes from the rectangle above the interval [0, 1].
    For the 2nd and 3rd intervals you have to add the contribution(s) from the previous interval(s).

    It's really important to have a good graph of the function F, showing the three parts. The first part is a line with slope 1 between (0, 0) and (1, 1). the second part is a horizontal line between (1, 1) and (2, 1). The third part is the right half of a parabola that opens downward and whose vertex is at (2, 1). At the point where the parabola crosses the x-axis, F of that x is zero. From that point onward, F(x) < 0.
     
  24. May 6, 2010 #23
    I think I actually understood that better than I thought and was just getting confused by my own writing. Thank you so much. I know you've already helped alot, but do you have any ideas about how I would then find a formula for F'(x) wherever F is differentiable? My test is tomorrow, that is the only reason I'm trying to learn this stuff now.
     
  25. May 6, 2010 #24

    Mark44

    Staff: Mentor

    The graph of F is continuous for x >= 0. The only points where it is not differentiable are at x = 1 and x = 2. The formula for F'(x) will have three parts.

    For 0 < x < 1, F(x) = x, so F'(x) = 1. Do the same for the other two parts, but keep in mind that F isn't differentiable at x = 1 or x = 2 (or at x = 0). Here's where you might start thinking about the Fund. Theorem of Calculus.
     
  26. May 6, 2010 #25
    But doesn't the Fundamental Theorem of Calculus tell us that since our F is continuous everywhere, that it should be differentiable there? How do I show that x=1,2 are not differentiable? I know if the limit from the left and right are equal, it is continuous at that point.
     
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