1. **Stating the problem:** We have a function $f$ defined by $f(x) = \frac{x+3}{2x-1}$ with domain $x \in \mathbb{R}, x \neq \frac{1}{2}$. We need to show that $f(f(x)) = x$ and then find the inverse function $f^{-1}(x)$. 2. **Show that $f(f(x)) = x$:** We start by substituting $f(x)$ into $f$ itself: $$f(f(x)) = f\left(\frac{x+3}{2x-1}\right) = \frac{\frac{x+3}{2x-1} + 3}{2\left(\frac{x+3}{2x-1}\right) - 1}$$ 3. **Simplify numerator:** $$\frac{x+3}{2x-1} + 3 = \frac{x+3}{2x-1} + \frac{3(2x-1)}{2x-1} = \frac{x+3 + 6x - 3}{2x-1} = \frac{7x}{2x-1}$$ 4. **Simplify denominator:** $$2\left(\frac{x+3}{2x-1}\right) - 1 = \frac{2(x+3)}{2x-1} - \frac{2x-1}{2x-1} = \frac{2x + 6 - (2x - 1)}{2x-1} = \frac{7}{2x-1}$$ 5. **Put numerator and denominator together:** $$f(f(x)) = \frac{\frac{7x}{2x-1}}{\frac{7}{2x-1}} = \frac{7x}{2x-1} \times \frac{2x-1}{7} = \frac{\cancel{7}x}{\cancel{2x-1}} \times \frac{\cancel{2x-1}}{\cancel{7}} = x$$ This shows that $f(f(x)) = x$. 6. **Find the inverse function $f^{-1}(x)$:** Since $f(f(x)) = x$, the function $f$ is its own inverse, so: $$f^{-1}(x) = f(x) = \frac{x+3}{2x-1}$$ **Answer:** $$f^{-1}(x) = \frac{x+3}{2x-1}$$