Inverse Function E11481

1. **State the problem:** We have the function $$f(x) = \frac{2x+1}{2x-1}$$ with domain $$x < \frac{1}{2}$$. We need to find the inverse function $$f^{-1}(x)$$ and its domain. 2. **Recall the inverse function definition:** The inverse function $$f^{-1}(x)$$ satisfies $$f(f^{-1}(x)) = x$$. To find $$f^{-1}(x)$$, we start by setting $$y = f(x)$$ and then solve for $$x$$ in terms of $$y$$. 3. **Set up the equation:** $$y = \frac{2x+1}{2x-1}$$ 4. **Solve for $$x$$:** Multiply both sides by $$2x - 1$$: $$y(2x - 1) = 2x + 1$$ Distribute $$y$$: $$2xy - y = 2x + 1$$ Group terms with $$x$$ on one side: $$2xy - 2x = y + 1$$ Factor out $$x$$: $$x(2y - 2) = y + 1$$ Simplify the factor: $$x \cdot 2(y - 1) = y + 1$$ Divide both sides by $$2(y - 1)$$: $$x = \frac{y + 1}{2(y - 1)}$$ Show cancellation explicitly: $$x = \frac{\cancel{y + 1}}{2\cancel{(y - 1)}}$$ (no cancellation here, so no slashes needed) 5. **Write the inverse function:** Replace $$y$$ by $$x$$ to express the inverse function: $$f^{-1}(x) = \frac{x + 1}{2(x - 1)}$$ 6. **Determine the domain of $$f^{-1}$$:** The denominator $$2(x - 1)$$ cannot be zero, so: $$x - 1 \neq 0 \implies x \neq 1$$ Also, since the original function $$f(x)$$ has domain $$x < \frac{1}{2}$$, the range of $$f$$ (which is the domain of $$f^{-1}$$) excludes the value that $$f$$ approaches as $$x \to \frac{1}{2}^-$$. 7. **Find the range of $$f$$ for $$x < \frac{1}{2}$$:** As $$x \to \frac{1}{2}^-$$, denominator $$2x - 1 \to 0^-$$, so $$f(x) \to -\infty$$ or $$+\infty$$ depending on the sign. Check the horizontal asymptote: $$\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{2x+1}{2x-1} = 1$$ Since the graph approaches $$y=1$$ from below on the left, the range is: $$(-\infty, 1)$$ Therefore, the domain of $$f^{-1}$$ is: $$(-\infty, 1)$$ --- **Final answers:** $$f^{-1}(x) = \frac{x + 1}{2(x - 1)}$$ Domain of $$f^{-1}$$: $$x < 1$$ --- **Graphical note:** The graph of $$f^{-1}(x)$$ is the reflection of $$f(x)$$ about the line $$y = x$$, which is the mirror line.