Here, I am sharing just an example problem which is given in one of my textbooks: $ \ \large{ f(3x-1)=12x+5 \ , \\ x \circ f(x)= \, ? \ } \ $
And, on below of the question, the book has shown an example solution for that: $ \text{Instead of } \ \mathit{x}, \ \text{if we write the inverse function of } \ \mathbf{3x-1} \, \text{;} \\ \large{ f(x)= 12 \times \bigg( \mathbf{ \frac{x+1}{3} } \bigg) +5 \\ f(x)= 4x+4+5 \\ f(x)= 4x+9 } $ $ \large{ x \circ f(x)= f(x) =4x+9 \\ } $ So, why do we need to take the inverse function of that $ \ \large{3x-1} \ $ and make it equal to $ \ \large{ \mathbf{f(x)} } \, $ ? How about writing that equation something like this, which I have tried before to solve the problem: $ \large{ f(3x-1)= 12x+5 \\ \implies \frac{x-5}{12}= f^{-1}(3x-1) \\ \frac{3x-1-5}{12}= x \\ 12x= 3x-6 \\ 9x= -6 \\ x= \frac{-6}{9} \\ \implies \frac{-2}{3} } $ Could you please, show me my own mistakes and explain some about how should we approach to this kind of problems to solve? Thank you very much!...