1. **State the problem:** Determine if the functions $f(x) = 3x + 1$ and $g(x) = x - \frac{1}{3}$ are inverses of each other. 2. **Recall the rule for inverse functions:** Two functions $f$ and $g$ are inverses if and only if $f(g(x)) = x$ and $g(f(x)) = x$ for all $x$ in the domains. 3. **Calculate $f(g(x))$:** $$f(g(x)) = f\left(x - \frac{1}{3}\right) = 3\left(x - \frac{1}{3}\right) + 1 = 3x - 1 + 1 = 3x - \cancel{1} + \cancel{1} = 3x$$ 4. Since $f(g(x)) = 3x \neq x$, the first condition for inverses is not satisfied. 5. **Calculate $g(f(x))$:** $$g(f(x)) = g(3x + 1) = (3x + 1) - \frac{1}{3} = 3x + 1 - \frac{1}{3} = 3x + \frac{3}{3} - \frac{1}{3} = 3x + \frac{2}{3}$$ 6. Since $g(f(x)) = 3x + \frac{2}{3} \neq x$, the second condition for inverses is also not satisfied. 7. **Conclusion:** Since neither $f(g(x)) = x$ nor $g(f(x)) = x$, the functions $f$ and $g$ are not inverses.