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Thread: inverse function proof

  1. #1
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    inverse function proof

    f:x -> 2x+1 g:x -> 1/x+1
    show that {f[g(x)]}^-1 = g^-1[f^-1(x)]

    i get {f[g(x)]}^-1 =x-3

    and i get g^-1[f^-1(x)] = 4-2x//x-1
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  2. #2
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    Re: inverse function proof

    If $g(x)=\dfrac{1}{x+1}$ ...

    $\{f[g(x)]\}^{-1}=g^{-1}[f^{-1}(x)]=\dfrac{3-x}{x-1}$

    When are you going to start using LaTeX?
    Last edited by skeeter; Jun 8th 2017 at 04:47 AM.
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  3. #3
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    Re: inverse function proof

    I do not understand your answer.
    here is a picture of the question
    inverse function proof-win_20170608_15_18_24_pro.jpgClick image for larger version. 

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    Im stuck on question 1(iv) i understand how to get i,ii,iii but part iv will not work out for me
    i get part iii to be x-3

    and for my questions I do not see the benefit in latex?
    If you really want me to use it though I will?
    Last edited by markosheehan; Jun 8th 2017 at 07:01 AM.
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  4. #4
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    Re: inverse function proof

    $f(g(x)) = 2\left(\dfrac{1}{x+1}\right)+1$

    Solve for $x$:

    $x = \dfrac{3-f(g(x))}{f(g(x))-1}$

    So, $[f(g(x))]^{-1} = \dfrac{3-x}{x-1}$

    So, you got part (iii) wrong. Skeeter understood and was telling you that you had it wrong.
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  5. #5
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    Re: inverse function proof

    its working out for me now. cheers
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  6. #6
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    Re: inverse function proof

    and for my questions I do not see the benefit in latex?
    you may see no benefit, but someone reading your posts would ... the whole idea is to effectively communicate mathematics.

    f:x -> 2x+1 g:x -> 1/x+1
    if you're not going to use latex, at least use proper grouping symbols to make clear your expressions. you've been told this before.

    1/(x+1)
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