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Math Help - Predicting second derivative

  1. #1
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    Predicting second derivative

    What do I use to figure this out? Not sure what to do with this one either.

    Suppose the graph of a function f(x) has something spilled on it, as shown. If we know the function and all of it's derivatives are continuous, is the integral of f ''(x) from 0 to 2 positive, negative or zero and explain why.
    Attached Thumbnails Attached Thumbnails Predicting second derivative-graph.gif  
    Last edited by ebonyscythe; November 29th 2007 at 06:05 AM.
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  2. #2
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    f''(x) shows the concavity of the function. You can tell there is an inflection point between zero and one due to the curvature (not a quadratic curve, slows down as it comes to next critical point.

    Nevertheless, the 1st derivative is zero at zero and increases and keeps on increasing till x=0.5. You can tell there is an inflection point there because the slope reverts back to zero at x=1. So, from (0,0.5), f''(x) is positive, from (0.5, 1), f''(x) is negative because the graph of f'(x) has a negative slope at that portion. From one to two, the slope goes from zero to a large negative # with an inflection point at what looks to be two, so the second derivative is negative.

    Therefore, f''(x) is positive from (0, 0.5) and negative from (0.5, 2).

    I hope this is right, usually I am better at this when I have a chalkboard in front of me!
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  3. #3
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    But at x=1 the graph is blacked out, so how do we know it's zero?

    Each tick mark on the graph is .2, I should have mentioned that. And it's my rendition of a graph I couldn't scan...
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  4. #4
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    Quote Originally Posted by ebonyscythe View Post
    What do I use to figure this out? Not sure what to do with this one either.

    Suppose the graph of a function f(x) has something spilled on it, as shown. If we know the function and all of it's derivatives are continuous, is the integral of f ''(x) from 0 to 2 positive, negative or zero and explain why.
    Well, by the fundamental theorem
    \int_a^b f^{\prime}(x)~dx = f(b) - f(a)

    so to apply this to your case we have
    \int_0^2 f^{\prime \prime}(x)~dx = f^{\prime}(2) - f^{\prime}(0)

    I'll leave it to you to decide the values of the derivatives.

    -Dan
    Last edited by topsquark; November 29th 2007 at 06:32 AM.
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    Oh, that makes sense, though I didn't know you could switch up that theorem and use it for the second derivative as well.

    But wouldn't it f '(a) be f '(0) in the second integral?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ebonyscythe View Post
    Oh, that makes sense, though I didn't know you could switch up that theorem and use it for the second derivative as well.

    But wouldn't it f '(a) be f '(0) in the second integral?
    I fixed the typo. Thanks for pointing that out.

    Yes, you can use it for derivatives (they are, after all, just functions), so long as the functions meet the requirements for the fundamental theorem (which I'm too lazy to look up right now.)

    -Dan
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