1. **State the problem:** Find the slope of the tangent line to the function $$f(x) = \frac{4x - 1}{x}$$ at $$x = 1$$. 2. **Recall the formula:** The slope of the tangent line at a point is the derivative of the function evaluated at that point. So we need to find $$f'(x)$$ and then compute $$f'(1)$$. 3. **Rewrite the function:** Simplify $$f(x)$$ before differentiating: $$f(x) = \frac{4x - 1}{x} = \frac{4x}{x} - \frac{1}{x} = 4 - x^{-1}$$ 4. **Differentiate:** Using the power rule and constant rule: $$f'(x) = 0 - (-1)x^{-2} = x^{-2} = \frac{1}{x^2}$$ 5. **Evaluate at $$x=1$$:** $$f'(1) = \frac{1}{1^2} = 1$$ 6. **Interpretation:** The slope of the tangent line to the curve at $$x=1$$ is $$1$$. **Final answer:** $$\boxed{1}$$