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Math Help - Composition of a function Fogoh

  1. #1
    Member M670's Avatar
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    Composition of a function Fogoh


    Find .

    I understand the steps in doing this but this one I am having trouble when getting the to the last part...
    I start by writing the h
     h(x)=\sqrt[3]8x then I get \frac{\sqrt[3]8x}{\sqrt[3]8x-1}
    Buts it's adding in the last f(x) I am having trouble?
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  2. #2
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    Re: Composition of a function Fogoh

    Quote Originally Posted by M670 View Post

    Find .

    I understand the steps in doing this but this one I am having trouble when getting the to the last part...
    I start by writing the h
     h(x)=\sqrt[3]8x then I get \frac{\sqrt[3]8x}{\sqrt[3]8x-1}
    Buts it's adding in the last f(x) I am having trouble?

    If A=\frac{\sqrt[3]8x}{\sqrt[3]8x-1} then lastly \sqrt[3]{8A}
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  3. #3
    Member M670's Avatar
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    Re: Composition of a function Fogoh

    See this is what I thought I would get  \sqrt8 with \frac{\sqrt[3]8x}{\sqrt[3]8x-1} so something looking like  \sqrt8\frac{\sqrt[3]8x}{\sqrt[3]8x-1} only the whole equation under the root
    Because I would replace the X of  \sqrt8x with \frac{\sqrt[3]8x}{\sqrt[3]8x-1}
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    Re: Composition of a function Fogoh

    Quote Originally Posted by M670 View Post
    See this is what I thought I would get  \sqrt8 with \frac{\sqrt[3]8x}{\sqrt[3]8x-1} so something looking like  \sqrt8\frac{\sqrt[3]8x}{\sqrt[3]8x-1} only the whole equation under the root
    Because I would replace the X of  \sqrt8x with \frac{\sqrt[3]8x}{\sqrt[3]8x-1}


    Within f(x) we replace x with h\circ g(x).
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  5. #5
    Member M670's Avatar
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    Re: Composition of a function Fogoh

    That's what I did is it not?
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    Re: Composition of a function Fogoh

    Quote Originally Posted by M670 View Post
    That's what I did is it not?

    NO it is not.

    In post #1 it is f(x)=\sqrt{8x~}\text{ NOT}f(x)=\sqrt{8}~x

    The x is under the radical.
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