# Thread: Limits # 6 - last

1. ## 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.

2. 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: .$\displaystyle \boxed{k \:=\:-1}$