How do you find the inverse of # f(x) = (2x-1)/(x-1)# and is it a function?

Question

How do you find the inverse of  # f(x) = (2x-1)/(x-1)# and is it a function?

Anna Allen · Accepted Answer

#f^-1(x)=(-x+1)/(2-x)#
yes, it is a function Finding the Inverse Function With the Given #f(x)=(2x-1)/(x-1)# Substitute #y# for #f(x)#. #y=(2x-1)/(x-1)# Swap the #x# and #y#. #x=(2y-1)/(y-1)# Solve for #y#. #x(y-1)=2y-1# #xy-x=2y-1# #2y-xy=-x+1# Factor out #y# from the left side. #y(2-x)=-x+1# #y=(-x+1)/(2-x)# Rewrite #y# as #f^-1(x)#. #color(green)(|bar(ul(color(white)(a/a)color(black)(f^-1(x)=(-x+1)/(2-x))color(white)(a/a)|)))# Determining Whether the Inverse Function Is a Function Graphically, #f^-1(x)=(-x+1)/(2-x)# would look like: plot{(-x+1)/(2-x) [-10, 10, -5, 5]} In the graph above, you can see that the #x# and #y# values approach the vertical and horizontal asymptotes. Since it resembles that of an exponential graph, there is only one #y# value for an #x# value. #:.#, the inverse function is a function.

Elijah Allen · Answer

undefined To find the inverse of a function, swap the roles of $ x $ and $ y $ and then solve for $ y $.

For the function $ f(x) = \frac{2x - 1}{x - 1} $:

1. Swap $ x $ and $ y $: $ x = \frac{2y - 1}{y - 1} $.

2. Solve for $ y $:

$ x(y - 1) = 2y - 1 $

$ xy - x = 2y - 1 $

$ xy - 2y = x - 1 $

$ y(x - 2) = x - 1 $

$ y = \frac{x - 1}{x - 2} $.

The inverse of $ f(x) $ is $ f^{-1}(x) = \frac{x - 1}{x - 2} $.

To determine if the inverse is a function, we need to check if every $ x $ in the domain of $ f(x) $ maps to exactly one $ y $ in the range of $ f(x) $, and vice versa. The domain of $ f(x) $ is all real numbers except $ x = 1 $, and the range of $ f(x) $ is all real numbers except $ y = 2 $. Similarly, the domain of $ f^{-1}(x) $ is all real numbers except $ x = 2 $, and the range of $ f^{-1}(x) $ is all real numbers except $ y = 1 $. Since both $ f(x) $ and $ f^{-1}(x) $ have one-to-one correspondence between their domains and ranges, they are both functions.