The function \(f(x)=\frac{x-1}x\) is defined on the domain \((-\infty,0)\cup(0,\infty)\). Is \(f(x)\) invertible? If an inverse function exists, find a formula for the inverse and state its domain and range. If an inverse does not exist, explain why not.

Guest Jun 21, 2017

#1**+2 **

This function is invertible since it's one-to-one

To find the inverse......write as

y = (x -1) / x multiply both sides by x

xy = x - 1 subtract x from both sides

xy - x = -1 factor the left side

x ( y - 1) = -1 divide both sides by (y - 1)

x = -1 / (y - 1) and we can write

x = 1 / ( 1 - y) "swap" x and y

y = 1 / (1 - x) and for y, write f^{-1} (x)

f^{-1}(x) = 1 / (1 - x)

The domain for this function will be (-inf, 1) U ( 1, inf)

The range will be (-inf, 0) U (0, inf )

CPhill Jun 21, 2017