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Math Help - Simplifying a complex rational

  1. #1
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    Simplifying a complex rational

    I am having a rather frustrating time simplifying this expression...


    {[1/(x+h)^2]-[1/x^2]}/h

    my first step as instructed by my pre-calculus textbook is to multiply [x^2(x+h)^2] to both the numerator and denominator as a result I see:

    {[x^2(x+h)^2]/(x+h)^2}-[x^2(x+h)^2]/x^2

    I think my treatment of fractions is the problem because my next action is to eliminate (x+h)^2 from the num. and denom. of the left expression and x ^2 from the num. and denom. of the right expression leaving me with:

    [x^2-(x+h)^2]/[hx^2(x+h)^2]

    this is my dead end and I know that the only reason I am here is because I cannot see past the error in my mind, please help me.
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  2. #2
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    Do you not multiply the h in the denominator by x^2(x+h)^2 as well?
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  3. #3
    MHF Contributor Unknown008's Avatar
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    Wait, I did a mistake, sorry...

    \dfrac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \dfrac{\frac{x^2 - (x+h)^2}{x^2(x+h)^2}}{h}

    Now you multiply by x^2(x+h)^2 on both numerator and denominator.

    \dfrac{\frac{x^2 - (x+h)^2}{x^2(x+h)^2}}{h} = \dfrac{x^2 - (x+h)^2}{hx^2(x+h)^2}
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  4. #4
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    yes and that is where I see no further...I am confused as to how to get to the answer in the book which is stated to be:

    -{[2x+h]/[x^2(x+h)^2]} with restrictions such that x and h {-=} 0 and x{-=} -h

    the majority of my confusion lies in the origin of the number 2...
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Right. Expand the numerator. What do you get?
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  6. #6
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    \displaystyle \frac{1}{h}\left[\frac{1}{(x+h)^2} - \frac{1}{x^2}\right]<br />

    \displaystyle \frac{1}{h}\left[\frac{x^2 - (x+h)^2}{x^2(x+h)^2}\right]<br />

    ... note that (x+h)^2 = x^2 + 2xh + h^2

    \displaystyle \frac{1}{h}\left[\frac{x^2 - (x^2+2xh+h^2)}{x^2(x+h)^2}\right]

    \displaystyle \frac{1}{h}\left[\frac{-2xh-h^2}{x^2(x+h)^2}\right]

    \displaystyle \frac{1}{h}\left[\frac{-h(2x+h)}{x^2(x+h)^2}\right]

    \displaystyle -\frac{2x+h}{x^2(x+h)^2}
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  7. #7
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    Thank you, I apologize for my ignorance of mathematical expression.
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  8. #8
    MHF Contributor Unknown008's Avatar
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    It's okay, we're here to learn
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