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Math Help - Composite function

  1. #1
    Senior Member Stroodle's Avatar
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    Composite function

    Can someone please show me the working for:

    f(x)=\frac{1}{2}\left (\frac{1}{x}+1\right )

    g(x)=\frac{1}{2x-1}

    Find f\circ g

    Cheers
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Stroodle View Post
    Can someone please show me the working for:

    f(x)=\frac{1}{2}\left (\frac{1}{x}+1\right )

    g(x)=\frac{1}{2x-1}

    Find f\circ g

    Cheers
    Note f\circ g=f\!\left(g\!\left(x\right)\right)=\frac{1}{2}\le  ft(\frac{1}{g\!\left(x\right)}+1\right).

    Now substitute what g\!\left(x\right) is into this expression, and simplify. It will work out very nicely. (Check spoiler to see if you're right)

    Spoiler:
    f\!\left(g\!\left(x\right)\right)=x
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  3. #3
    Senior Member Stroodle's Avatar
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    Thanks. I'm just having trouble with simplifying it though.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Stroodle View Post
    Can someone please show me the working for:

    f(x)=\frac{1}{2}\left (\frac{1}{x}+1\right )

    g(x)=\frac{1}{2x-1}

    Find f\circ g

    Cheers
    All this wants is for you to substitute g(x) everywhere you see x in f(x).



    f(x)=\frac{1}{2}\left (\frac{1}{\text{insert g(x) here}}+1\right )

    f(x)=\frac{1}{2}\left (\frac{1}{\frac{1}{2x-1}}+1\right )=\frac{1}{2}(2x-1+1)=\frac{2x}{2}=x

    remember that \frac{1}{\frac{1}{2x-1}}=\frac{1}{1}*\frac{2x-1}{1}=2x-1. (flip it and multiply)

    It appears that these functions are inverses.
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  5. #5
    Senior Member Stroodle's Avatar
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    Awesome. Got it. THanks
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Stroodle View Post
    Thanks. I'm just having trouble with simplifying it though.
    No problem.

    Note that when you do the substitution, you have f\!\left(g\!\left(x\right)\right)=\frac{1}{2}\left  (\frac{1}{\frac{1}{2x-1}}+1\right).

    Note that \frac{1}{\frac{1}{2x-1}}=2x-1. Thus, we now have f\!\left(g\!\left(x\right)\right)=\frac{1}{2}\left  (2x-1+1\right).

    Can you finish the simplification process?
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  7. #7
    Senior Member Stroodle's Avatar
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    Yeah. I just didn't work out that \frac{1}{\frac{1}{2x-1}}=2x-1, but I understand it now. Thanks
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