![g^(-1)(x)=\pm(\sqrt[]{x-36+6x})](https://img.qammunity.org/2023/formulas/mathematics/college/4w0wqihaf3uy229puym65dfvfv6xr2oq4w.png)
Step-by-step explanation

The inverse function returns the original value for which a function gave the output
Step 1
swap x and y and solve for y
so

now, isolate y
![\begin{gathered} x=y^2-6x+36 \\ \text{subtract 36 in both sides} \\ x-36=y^2-6x+36-36 \\ x-36=y^2-6x \\ \text{add 6x in both sides} \\ x-36+6x=y^2-6x+6x \\ x-36+6x=y^2 \\ get\text{ the square root in both sides} \\ √(x-36+6x)=√(y^2) \\ \pm(\sqrt[]{x-36+6x})=y \\ y=\pm(\sqrt[]{x-36+6x}) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6sjveevpff5s021r28kcwfe12kasn2bnkl.png)
note, there is a negative answer , it is because when gettin the square we get y , so
the answer is
![g^(-1)(x)=\pm(\sqrt[]{x-36+6x})](https://img.qammunity.org/2023/formulas/mathematics/college/4w0wqihaf3uy229puym65dfvfv6xr2oq4w.png)
I hope this helps you