Before we find [tex]f^{-1}[/tex], let's think for a minute about what [tex]f[/tex] does, and what it means to find the inverse of a function. A function [tex]f[/tex] essentially takes a value [tex]x[/tex] from the domain and maps it to another value [tex]y[/tex] in that function's range according to a set of rules. Here, those rules are defined by the formula
[tex]y= \frac{4x+2}{3x-1} , x \neq \frac{1}{3} [/tex]
All the inverse does is swap the domain and range of the function. Now, instead of trying to map [tex]x[/tex] to [tex]y[/tex], we're trying to find a set of rules that'll map [tex]y[/tex] back to [tex]x[/tex]. To find those rules, all we have to do is solve the above equation for [tex]x[/tex].
First, we'll multiply both sides of the equation by [tex]3x-1[/tex] to get it out of the denominator:
[tex](3x-1)y=( \frac{4x+2}{3x-1})(3x-1) [/tex]
Cancelling on the right side and distributing on the left, we get:
[tex]3xy-y=4x+2[/tex]
Next, we collect all of our x terms on one side, and all of our non-x terms on the other:
[tex](3xy-y)+y=(4x+2)+y\\(3xy)-4x=(4x+2+y)-4x\\3xy-4x=2+y[/tex]
Let's rearrange the right side and factor out an x on the left:
[tex]x(3y-4)=y+2[/tex]
And finally, we divide both sides by [tex]3y-4[/tex] to obtain our answer:
[tex][x(3y-4)]/(3y-4)=(y+2)/(3y-4)\\\\x= \frac{y+2}{3y-4} [/tex]
This equation gives us the "rules" for mapping any given y in the range back to an x in the domain. If we swap the domain and the range, we can define our function
[tex]f^{-1}(x)= \frac{x+2}{3x-4}[/tex]
