We have [tex]f(x)=\dfrac{x+5}{3x-1}[/tex].
To find inverse function [tex]f^{-1}(x)[/tex] we substitute x with [tex]f^{-1}(x)[/tex] and vice-versa to get
[tex]x=\dfrac{f^{-1}(x)+5}{3f^{-1}(x)-1}[/tex]
Now solve for [tex]f^{-1}(x)[/tex]. Note that I will use [tex]j[/tex] instead.
[tex]
x=\dfrac{j+5}{3j-1} \\
x(3j-1)=j+5 \\
3jx-x=j+5 \\
3jx-x-j-5=0 \\
3jx-j=x+5 \\
j(3x-1)=x+5 \\
j=\dfrac{x+5}{3x-1}
[/tex]
So we find that [tex]f(x)=f^{-1}[/tex].
Hope this helps.
Answer:
f-1(x) = (x + 5)/(3x - 1).
Step-by-step explanation:
Let y = (x + 5)/ (3x - 1)
3xy - y = x + 5
3xy - x = y + 5
x(3y - 1) = y + 5
x = (y + 5)/(3y - 1)
So f-1(x) = (x + 5)/(3x - 1).
