Let’s check if $f^{-1}(x) = 3 + \sqrt{x}$ is indeed the inverse of $g(x) = 6 - 2x$.

Steps:

1. Rewrite $g(x) = 6 - 2x$ in terms of $x$: Let $y = g(x) = 6 - 2x$. Solve for $x$ in terms of $y$: $y = 6 - 2x \implies 2x = 6 - y \implies x = \frac{6 - y}{2}.$ So, the inverse of $g(x)$ is: $g^{-1}(x) = \frac{6 - x}{2}.$

2. Compare with $f^{-1}(x) = 3 + \sqrt{x}$: Clearly, $g^{-1}(x) = \frac{6 - x}{2}$ is not the same as $3 + \sqrt{x}$.

Conclusion:

The inverse function of $g(x) = 6 - 2x$ is not $f^{-1}(x) = 3 + \sqrt{x}$. Instead, it is $g^{-1}(x) = \frac{6 - x}{2}$.

Would you like a deeper dive into inverse functions or verification steps? Here are 5 related questions:

1. How do you find the inverse of more complex functions?
2. Can all functions have an inverse? Why or why not?
3. How do you verify if two functions are inverses of each other?
4. What is the graphical relationship between a function and its inverse?
5. How do restrictions on the domain affect the existence of an inverse?

Tip: To verify an inverse function, always check if $f(g(x)) = x$ and $g(f(x)) = x$.