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Math Help - Using the difference quotient

  1. #1
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    Using the difference quotient

    Can somebody please help me out with this?

    Let f(x)=(2/x)-4, then the expression

    [f(x+h)-f(x)]/h can be written in the form

    (A)/(Bx^2+Cxh), where A, B, and C are constants.

    Find A, B, and C.

    Any help would be greatly appreciated. Thanks!
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  2. #2
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    Quote Originally Posted by jtowery View Post
    Can somebody please help me out with this?

    Let f(x)=(2/x)-4, then the expression

    [f(x+h)-f(x)]/h can be written in the form

    (A)/(Bx^2+Cxh), where A, B, and C are constants.

    Find A, B, and C.

    Any help would be greatly appreciated. Thanks!
    \frac{\frac{2}{x+h}-4-(\frac{2}{x}-4)}{h}
    the 4s cancel
    \frac{\frac{2}{x+h}-\frac{2}{x}}{h}
    divide the h into each part
    \frac{2}{h(x+h)}-\frac{2}{hx}
    make each denominator the same
    \frac{2hx}{h^2x(x+h)}-\frac{2h(x+h)}{h^2x(x+h)}
    combine
    \frac{2hx-2h(x+h)}{h^2x(x+h)}
    1 h on top cancels with 1 h on bottom + factor out 2
    \frac{2(x-x-h)}{hx(x+h)}
    x's on top cancel leaving another h to cancel with the h below
    \frac{-2}{x(x+h)}
    expand denominator
    \frac{-2}{x^2+xh}
    \therefore a = -2
    b=1
    c = 1
    Last edited by 1005; January 28th 2010 at 01:15 PM.
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  3. #3
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    nvm?
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  4. #4
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    Quote Originally Posted by 1005 View Post
    nvm
    ??

     <br />
\frac{f(x+h)-f(x)}{h}=\frac{\left(\frac{2}{x+h}-4\right)-\left(\frac{2}{x}-4\right)}{h}=\dots<br />

    Now simplify.
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  5. #5
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    Awesome. Thank you both so much
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