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Antiderivative and Extrema

  1. Nov 11, 2013 #1

    Qube

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    1. The problem statement, all variables and given/known data

    https://scontent-b-mia.xx.fbcdn.net/hphotos-prn2/v/1388504_10201044108366607_730785214_n.jpg?oh=9e67700cd15429886ee87ce2eed63328&oe=528397C9

    2. Relevant equations

    F(x) = ∫f(x).

    We can apply the second derivative test.

    F''(x) = f'(x)

    3. The attempt at a solution

    F''(x) is negative at x = -2 since the slope of f(x) is negative at x = -2. f'(x) doesn't exist at the corner at x = 0. f'(x) is negative at x = 3. Therefore there are two local maxes (x = -2 and x = 3) determined through the second derivative test.

    We can apply the first derivative test at the point x = 0. F'(x) = f(x). f(x) however fails to change signs across x = 0; it remains negative across x = 0, so x = 0 cannot be an extrema of any kind.
     
  2. jcsd
  3. Nov 11, 2013 #2

    Simon Bridge

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    So far so good - did you have a question?

    Note: it works better to do the 1st derivative test first, to identify the critical points.
    What are the critical points of F(x)?
     
  4. Nov 11, 2013 #3

    Qube

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    I guess my question was if I missed anything. And nice catch! Thanks; I should consider critical points. These are where the first derivative of the anti derivative or just f(x) = 0 or fails to exist. These points would therefore be -2, 1, and 3.
     
  5. Nov 12, 2013 #4

    Simon Bridge

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    That's it - the critical points of F are where f=0:
    ... which means you can eliminate options (a) and (d) right away.
    (a) because x=2 is not a critical point, and (d) because x=0 is not a critical point.

    (e) is just "none of the above" so it is the answer if the others don't fit.
    The remaining two options only mention x=-2 and x=3.

    The 2nd derivative test can be figured out just by the slope of f(x) at these points - which you can get by inspection. You seem to have figured that out already.

    f'(-2)<0 and f'(3)<0 - is that correct?

    What does that mean for these critical points? Concave-up or concave-down?
     
  6. Nov 12, 2013 #5

    Qube

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    Thanks for the insight. I shouldn't have overlooked the central importance of critical points. The slope is indeed negative at x = -2 and x = 3, which means these critical points are local maximums according to the second derivative test. Also it means that at these points the graph of F(x) is concave down since the second derivative is negative.
     
  7. Nov 12, 2013 #6

    Simon Bridge

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    Well done - now you have enough information to answer the question. You can stop ;)

    See how that was less work and the results clearer?
    This sort of approach helps you gain confidence in your results.

    Consider:
    x=0 is no sort of extrema simply because f(x=0)≠0. Same with x=2. .... nuff said.
    No need to do any more analysis. Stop there.

    For the suggested method (drawing the number lines) you didn't need the entire number line - just the bits that tell you the answers. So the annoying bit about x=1, where the slope of f(x) changes abruptly, is not needed: it's a red herring - the kind of thing put there to waste your time in an exam.

    However - for completeness: to sit between two local maxima, x=0 has to be some sort of minima - or an asymptote. If F had an assymptote at x=1, what would f look like there?

    By looking at the concavity, you can figure what happens to F at x=0 and x=2 as well.
    These are just the transition points in the curvature of F.
     
    Last edited: Nov 12, 2013
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