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Math Help - Zeroes of a function

  1. #1
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    Zeroes of a function

    First of all, apologies if this is in the wrong section It's a calculus problem but I'm stuck on the Algebra of it.

    I'm supposed to find the critical numbers of f(x)=x^(3/4)-2x^(1/4).
    I found the derivative to be f'(x)= (3/4)x^(-1/4)-(1/2)x^(-3/4), but for the life of me I can't remember how to go about finding it's zeroes.
    I know one of them is x=0, and the book says another is x=4/9, but how do I get that? Thanks for any and all help in advance.
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  2. #2
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    Re: Zeroes of a function

    Your differentiation is fine.

    f'(x)= \frac{3}{4}x^{\frac{-1}{4}}-\frac{1}{2}x^{\frac{-3}{4}}

    0=\frac{3}{4}x^{\frac{-1}{4}}-\frac{1}{2}x^{\frac{-3}{4}}

    0=3x^{\frac{-1}{4}}-2x^{\frac{-3}{4}}

    For a clearer understanding of what we have:

    0=\frac{3}{\sqrt[4]{x}}-\frac{2}{\sqrt[4]{x^3}}

    \frac{2}{\sqrt[4]{x^3}}=\frac{3}{\sqrt[4]{x}}

    2\sqrt[4]{x}=3\sqrt[4]{x^3}

    Raising both sides to the 4th power:

    16x=81x^3

    81x^3-16x=0

    x(81x^2-16)=0
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  3. #3
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    Re: Zeroes of a function

    How did you go from 2/x^(3/4)=3/x^(1/4) to 2x^(1/4)=3x^(3/4)?
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  4. #4
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    Re: Zeroes of a function

    Quote Originally Posted by TheDarkNight93 View Post
    How did you go from 2/x^(3/4)=3/x^(1/4) to 2x^(1/4)=3x^(3/4)?
    I would say he cross-multiplied the fractions ...

    if \frac{a}{b} = \frac{c}{d} , then ad = bc
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  5. #5
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    Re: Zeroes of a function

    Thanks, thats a lot of help. It's embaressing that I don't remember this (3-4 years ago). Could anyone point me towards a basic algebra review to help refresh my memory on this kind of stuff?
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