Answer:
C.
[tex]f^{-1}(x)=\frac{1}{x-7}+3,\text{ where x }\ne7[/tex]
Explanation:
Given the below function;
[tex]f(x)=\frac{1}{x-3}+7[/tex]
To find the inverse of the above function, we have to follow the below steps;
1. Replace f(x) with y;
[tex]y=\frac{1}{x-3}+7[/tex]
2. Switch x and y;
[tex]x=\frac{1}{y-3}+7[/tex]
3. Solve for y;
[tex]\begin{gathered} x-7=\frac{1}{y-3} \\ (y-3)(x-7)=1 \\ y-3=\frac{1}{x-7} \\ y=\frac{1}{x-7}+3 \end{gathered}[/tex]
4. Replace y with f^-1 (x);
[tex]f^{-1}(x)=\frac{1}{x-7}+3[/tex]
To find the domain of the function, we need to consider that value of x that will make the function undefined and exclude it. That value of x is 7. So x must not be equal to 7.
Therefore, the solution is as written below;
[tex]f^{-1}(x)=\frac{1}{x-7}+3,\text{ where x }\ne7[/tex]
