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Math Help - Limits # 6 - last

  1. #1
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    Limits # 6 - last

    Find the value of k so that the function defined by f(x) = (x+5)/(x+k) will be its own inverse.
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  2. #2
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    Hello, foreverbrokenpromises!

    Find the value of k so that the function defined by f(x) \:=\:\frac{x+5}{x+k} will be its own inverse.
    First, find the inverse of f(x) . . .

    We have: . y \;=\;\frac{x+5}{x+k}

    Switch x's and y's: . x \;=\;\frac{y+5}{y+k}

    Solve for y\!:\;\;xy + kx \:=\:y + 5 \quad\Rightarrow\quad xy - y \:=\:5 - kx

    Factor: . y(x-1) \:=\:5-kx \quad\Rightarrow\quad y \:=\:\frac{5-kx}{x-1}

    . . Hence: . f^{-1}(x) \;=\;\frac{5-kx}{x-1}


    Since f(x) = f^{-1}(x), we have: . \frac{x+5}{x+k} \;=\;\frac{5-kx}{x-1}

    . . This is true for: . \boxed{k \:=\:-1}

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