To determine whether $f(x)$ and $g(x)$ are inverse functions, we need to verify if:

1. $f(g(x)) = x$
2. $g(f(x)) = x$

Let’s start with the first condition:

Step 1: Check if $f(g(x)) = x$

Substitute $g(x) = \frac{x - 1}{3}$ into $f(x) = 3x + 1$:

$f(g(x)) = 3\left(\frac{x - 1}{3}\right) + 1$

Simplifying this:

$f(g(x)) = (x - 1) + 1 = x$

So, $f(g(x)) = x$.

Step 2: Check if $g(f(x)) = x$

Substitute $f(x) = 3x + 1$ into $g(x) = \frac{x - 1}{3}$:

$g(f(x)) = \frac{(3x + 1) - 1}{3} = \frac{3x}{3} = x$

So, $g(f(x)) = x$.

Conclusion:

Since both $f(g(x)) = x$ and $g(f(x)) = x$, the functions $f(x) = 3x + 1$ and $g(x) = \frac{x - 1}{3}$ are indeed inverses of each other.

Would you like further details or have any questions?

Here are 5 related questions:

1. How do you find the inverse of more complex functions?
2. What happens if only one of $f(g(x)) = x$ or $g(f(x)) = x$ holds?
3. How can you graphically verify that two functions are inverses?
4. Can inverse functions exist for non-linear functions?
5. How do you handle finding inverses when dealing with trigonometric functions?

Tip: The graph of a function and its inverse are reflections over the line $y = x$.