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Math Help - Another issue with rationalizing

  1. #1
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    Another issue with rationalizing

    I ran into another rationalizing problem and it is messing me up. it is f(x) = 1/sqrtx


    so when I take my f(a+h) - f(a) I get {[1/sqrt(x+h)] - [1/sqrtx]}/h but inlike the question before that dealt with indeterminate form I am finding rationalizing is getting a bit messy. I end up with [(sqrt(x+h) + sqrtx)/h]/h



    I will get this algebra someday. Haha
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  2. #2
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    Re: Another issue with rationalizing

    I'm not sure what your question is or exactly HOW you "rationalized". The point of rationalizing is that (a- b)(a+ b)= a^2- b^2 so if a or b involved a square root, we will have gotten rid of it by squaring.

    I could see two ways of approaching \frac{1}{\sqrt{x+h}}- \frac{1}{\sqrt{x}}.

    a) Multiply by \frac{\frac{1}{\sqrt{x+h}}+ \frac{1}}{\sqrt{x}}{\frac{1}{\sqrt{x+h}}- \frac{1}{\sqrt{x}}} to get
    \frac{\frac{1}{x+ h}- \frac{1}{x}}{\frac{1}{\sqrt{x+h}}+ \frac{1}{\sqrt{x}}}
    so the numerator can be written as \frac{x- x- h}{x(x+ h)}= \frac{-h}{x(x+h)} and then, divided by h,
    \frac{-1}{x(x+ h)} so that, with the denominator, we have
    \frac{-1}{(x(x+h)(\frac{1}{\sqrt{x+h}}+ \frac{1}{\sqrt{x}}}

    The limit, as h goes to 0 is \frac{-1}{x^2(\frac{2}{\sqrt{x}})}= \frac{-\sqrt{x}}{2x^2}= -\frac{1}{2}x^{-3/2}

    Personally, I would combine the fractions first: \frac{1}{\sqrt{x+h}}- \frac{1}{\sqrt{x}}= \frac{\sqrt{x}- \sqrt{x+h}}{\sqrt{x+h}\sqrt{x}}
    Now, multiply numerator and denominator by \frac{\sqrt{x}+ \sqrt{x+h}} to get
    \frac{x- x- h}{(\sqrt{x}+ \sqrt{x+ h})(\sqrt{x+h}\sqrt{x})}= \frac{-h}{(\sqrt{x}+ \sqrt{x+ h})(\sqrt{x+h}\sqrt{x})}

    Dividing by h and taking the limit as h goes to 0 gives the same as before.
    Last edited by HallsofIvy; September 22nd 2013 at 01:24 PM.
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  3. #3
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    Re: Another issue with rationalizing

    Quote Originally Posted by Jonroberts74 View Post
    I ran into another rationalizing problem and it is messing me up. it is f(x) = 1/sqrtx
    so when I take my f(a+h) - f(a) I get {[1/sqrt(x+h)] - [1/sqrtx]}/h but inlike the question before that dealt with indeterminate form I am finding rationalizing is getting a bit messy. I end up with [(sqrt(x+h) + sqrtx)/h]/h
    To make the "algebra" less messy write it as:
    f(x)=\frac{1}{\sqrt{x}}=\frac{\sqrt{x}}{x}.
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  4. #4
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    Re: Another issue with rationalizing

    Thank you, I see what I was doing wrong.
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