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Sketch antiderivative

  1. Aug 30, 2006 #1
    Wow, I totally forgot how to do these.

    I have a graph of a line, but it does not say the function. I have to sketch the antiderivative.

    Does anyone have any links or advice to start me out.

    TIA
     
  2. jcsd
  3. Aug 30, 2006 #2

    berkeman

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    Staff: Mentor

    Remember that the derivative dy/dx is the change in the y value per change in the x value. So the derivitave at a point is basically the slope of the line.

    Now turn that reasoning around to figure out the antiderivative. If you have a plot of the slope of the line at all points, then you can figure out what the original function's line looked like. Quiz question -- what other thing do you have to add in to the final antiderivative answer to make it correct?
     
  4. Aug 30, 2006 #3

    berkeman

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    Staff: Mentor

  5. Aug 30, 2006 #4
    Yeah, I can do the derivative if I see the graph. I just can't visualize that backwards in my head.

    These are the values I'm dealing with: f'(x)| 0 | 1 | 0 | -1 | 0
    ......................................................x |-1 | 0 | 1 | 2 | 3
     
  6. Aug 30, 2006 #5
    What kind of function is the integral of a line? A parabola right. Well, were is the vertex of that parabola? You should think about what a critical point is. Maybe the first derivative test. Does any of this help?
     
  7. Aug 30, 2006 #6
    It is going up until x=0, then goes down. It is concave down there.

    The at x = 1 it change to concave up and at x = 2 it starts increasing.

    So, I guess the critical points are x = 0, 1, 2
     
  8. Aug 30, 2006 #7
    A critical point is where f'(x) = 0. If I am reading your notation right, then the critical points are x = -1, 1, 3. So if there are 3 critical points, then the [EDIT]antiderivative should be a function of [tex]x^4[/tex]. What does this kind of function look like?
     
    Last edited: Aug 30, 2006
  9. Aug 30, 2006 #8
    Well actually, f'(x) is only 0 at x= 0 and 2, but it switches concavity at 1, so I figured that would be one too. (guess not)

    x^4, yeah I believe so, I dont have a calculator to confirm(working on it), but it has two parabolas in it.(1 upside down).
     
    Last edited: Aug 30, 2006
  10. Aug 30, 2006 #9
    Oh, Sorry, I had some trouble reading your points. Sounds like you are on the right track though.
     
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