![f^(-1)(x)=(x)/(3)+(5)/(3)\text{ }](https://img.qammunity.org/2023/formulas/mathematics/college/ctqmisiivi2g2fqwwft24b9f37akjzykxx.png)
1) To find the inverse function of a linear one, we must follow some steps:
2) Take the given function, and swap the variables so
![y=3x-5](https://img.qammunity.org/2023/formulas/mathematics/college/3zcpto78boxc2pleelsl9ft31ukl57wh2j.png)
It's going to become in the process:
![x=3y-5](https://img.qammunity.org/2023/formulas/mathematics/college/vtq9a2bpfeav0h9hefg1tnn3ueppjwphe9.png)
3) And rewrite it putting y on the left side:
![\begin{gathered} -3y=-x-5 \\ 3y=x+5 \\ (3y)/(3)=(x)/(3)+(5)/(3) \\ y=(x)/(3)+(5)/(3)\text{ or} \\ f^(-1)(x)=(x)/(3)+(5)/(3)\text{ } \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/p9i2kfg8g26o6mdesj6vuvqcacc0rgi334.png)
So that's our inverse function whose Range is the Domain of the original one and the Domain is the Range of the original one.