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Math Help - inverse functions

  1. #1
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    inverse functions

    Ok need to find the inverse and i need to show that
    f ^(-1) o f= x= f o f^(-1)

    p(x) =

    x
    -----
    x - 1

    i did

    y=
    x
    ---
    y -1 = (y-1) x = y then xy -1x = y -xy
    then

    -1x = y(-x)
    ----- -------
    -x -x

    which means x = y ?

    then i guess i need to find the inverse of that? i dont get it?
    the inverse of x =y is y=x ?? i'm lost???
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  2. #2
    MHF Contributor red_dog's Avatar
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    y=\frac{x}{x-1}\Rightarrow yx-y=x\Rightarrow x=\frac{y}{y-1}

    Then f^{-1}(y)=\frac{y}{y-1}
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  3. #3
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    Hello, b_austintx!

    Your algebra is off . . .


    Find the inverse and verify that: . f\circ f^{-1} \,=\,x

    . . p(x) \:=\:\frac{x}{x-1}

    We have: . y \:=\:\frac{x}{x-1}

    Switch variables: . x \:=\:\frac{y}{y-1}

    Solve for y\!:\;\;x(y-1) \:=\:y \quad\Rightarrow\quad xy - x \:=\:y \quad\Rightarrow\quad xy-y \:=\:x

    . . Factor: . y(x-1) \:=\:x \quad\Rightarrow\quad y \:=\: \frac{x}{x-1}

    Therefore: . p^{-1}(x) \:=\:\frac{x}{x-1}

    . .
    Interesting . . . p(x) is its own inverse!


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    We will verify that: .  p\circ p^{-1} \:=\:x


    p\circ p^{-1} \;=\;p\left(p^{-1}\right) \;= \;p\left(\frac{x}{x-1}\right) \;=\; \frac{\dfrac{x}{x-1}}{\dfrac{x}{x-1} - 1}

    Multiply by \frac{x-1}{x-1}\!:\quad \frac{(x-1)\left(\dfrac{x}{x-1}\right)} {(x-1)\left(\dfrac{x}{x-1} - 1\right)} \;=\;\frac{x}{x - (x-1)} \;=\; \frac{x}{1} \;=\;x
    . . Done!

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