1. **State the problem:** We have a function $$f(x) = \frac{2x + 1}{g(x)}$$ where $$g(x) = \frac{1}{3}x - 1$$. We want to find the value of $$k$$ such that the range of $$f$$ is $$R = \{y : y \neq k\}$$ with $$k \in \mathbb{Z}^+$$. 2. **Understand the problem:** The range excludes a particular value $$k$$. This happens when $$f(x)$$ cannot take the value $$k$$ for any $$x$$. We need to find this $$k$$. 3. **Set up the equation:** Suppose $$f(x) = y$$. Then: $$y = \frac{2x + 1}{\frac{1}{3}x - 1}$$ 4. **Rewrite the equation to solve for $$x$$:** Multiply both sides by $$\frac{1}{3}x - 1$$: $$y\left(\frac{1}{3}x - 1\right) = 2x + 1$$ 5. **Distribute $$y$$:** $$\frac{y}{3}x - y = 2x + 1$$ 6. **Group terms with $$x$$ on one side:** $$\frac{y}{3}x - 2x = y + 1$$ 7. **Factor out $$x$$:** $$x\left(\frac{y}{3} - 2\right) = y + 1$$ 8. **Solve for $$x$$:** $$x = \frac{y + 1}{\frac{y}{3} - 2} = \frac{y + 1}{\frac{y - 6}{3}}$$ 9. **Simplify the denominator:** $$x = \frac{y + 1}{\frac{y - 6}{3}} = (y + 1) \times \frac{3}{y - 6} = \frac{3(y + 1)}{y - 6}$$ 10. **Analyze the domain of $$x$$:** For $$x$$ to be defined, the denominator $$y - 6 \neq 0$$, so $$y \neq 6$$. 11. **Interpretation:** The function $$f(x)$$ can take any value $$y$$ except $$y = 6$$ because if $$y = 6$$, the denominator in the expression for $$x$$ becomes zero, making $$x$$ undefined. 12. **Conclusion:** The range of $$f$$ is all real numbers except $$6$$. Since $$k \in \mathbb{Z}^+$$, we have $$k = 6$$. **Final answer:** $$k = 6$$.