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Math Help - Finding a second derivative and solving an equation

  1. #1
    Junior Member
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    Finding a second derivative and solving an equation

    Hey,

    I need some assistance finding what's called a second derivative (I couldn't really understand the prof as he was talking about it), and then solving an equation associated with it.

    Equation to solve : f''(x) = 0

    Given:
    f(x)=\frac{x}{x-1}

    Any help greatly appreciated.
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  2. #2
    Junior Member
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    differentiate the given expression twice (you may have to use quotient rule)

    then equate the result to zero and solve for x
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Just take the derivative twice over. Like so:

    First thing I'm gonna do is rewrite so I don't have to use that icky quotient rule.


    \frac{d}{dx}[\frac{x}{x-1}]=\frac{d}{dx}x(x-1)^{-1}=-x(x-1)^{-2}+(x-1)^{-1}

    Now we do it again

    \frac{d^2}{dx}[x(x-1)^{-1}] = \frac{d}{dx}[-x(x-1)^{-2}+(x-1)^{-1}]=2x(x-1)^{-3}-(x-1)^{-2}-(x-1)^{-2}

    factoring out he least common factor

    =(x-1)^{-3}[2x-(x-1)-(x-1)]=\frac{-4x+2}{(x-1)^3}

    Set the thing to equal zero, and solve.

    \frac{-4x+2}{(x-1)^3}=0\Longleftrightarrow{-4x+2}=0

    you can take it from here.
    This is called the second derivative test. Check this out:

    Concavity and the Second Derivative Test - HMC Calculus Tutorial
    Last edited by VonNemo19; May 27th 2009 at 09:45 PM.
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