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Thread: Limits # 6 - last

  1. #1
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    Limits # 6 - last

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

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

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

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

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

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

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


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

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

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