1. **State the problem:** Find the domain and range of the function $$f(x) = 3 \cos^{-1}\left(\frac{1}{2x - 1}\right) - 2.$$\n\n2. **Find the domain:** The function involves the inverse cosine, $$\cos^{-1}(y)$$, which is defined only for $$y$$ values in the interval $$[-1,1].$$\n\nSo, we need to find all $$x$$ such that $$\frac{1}{2x - 1} \in [-1,1].$$\n\n3. **Solve the inequality:**\n\n$$-1 \leq \frac{1}{2x - 1} \leq 1.$$\n\nWe consider two inequalities separately:\n\n(i) $$\frac{1}{2x - 1} \leq 1,$$\n\n(ii) $$\frac{1}{2x - 1} \geq -1.$$\n\n4. **Solve (i):**\n\n$$\frac{1}{2x - 1} \leq 1.$$\n\nMultiply both sides by $$2x - 1$$, but be careful about the sign of $$2x - 1$$:\n\n- If $$2x - 1 > 0$$ (i.e., $$x > \frac{1}{2}$$), then multiplying preserves inequality:\n$$1 \leq 1(2x - 1) \Rightarrow 1 \leq 2x - 1 \Rightarrow 2x \geq 2 \Rightarrow x \geq 1.$$\n\n- If $$2x - 1 < 0$$ (i.e., $$x < \frac{1}{2}$$), then multiplying reverses inequality:\n$$1 \geq 1(2x - 1) \Rightarrow 1 \geq 2x - 1 \Rightarrow 2 \geq 2x \Rightarrow x \leq 1.$$\n\nSince $$x < \frac{1}{2}$$ in this case, the solution here is $$x < \frac{1}{2}$$ (which is automatically $$\leq 1$$).\n\n5. **Solve (ii):**\n\n$$\frac{1}{2x - 1} \geq -1.$$\n\nAgain, multiply both sides by $$2x - 1$$ considering sign:\n\n- If $$2x - 1 > 0$$ (i.e., $$x > \frac{1}{2}$$), inequality preserved:\n$$1 \geq -1(2x - 1) \Rightarrow 1 \geq -2x + 1 \Rightarrow 0 \geq -2x \Rightarrow x \geq 0.$$\n\n- If $$2x - 1 < 0$$ (i.e., $$x < \frac{1}{2}$$), inequality reverses:\n$$1 \leq -1(2x - 1) \Rightarrow 1 \leq -2x + 1 \Rightarrow 0 \leq -2x \Rightarrow x \leq 0.$$\n\n6. **Combine solutions:**\n\nFrom (i): $$x \geq 1$$ or $$x < \frac{1}{2}.$$\nFrom (ii): $$x \geq 0$$ or $$x \leq 0.$$\n\nDomain is intersection of these conditions and $$x \neq \frac{1}{2}$$ (denominator zero).\n\n- For $$x > \frac{1}{2}$$, domain requires $$x \geq 1$$ and $$x \geq 0$$, so $$x \geq 1.$$\n- For $$x < \frac{1}{2}$$, domain requires $$x < \frac{1}{2}$$ and $$x \leq 0$$, so $$x \leq 0.$$\n\nTherefore, domain is $$(-\infty, 0] \cup [1, \infty).$$\n\n7. **Find the range:**\n\nThe range of $$\cos^{-1}(z)$$ is $$[0, \pi].$$\n\nSince $$z = \frac{1}{2x - 1}$$ takes values in $$[-1,1],$$ the inner function $$\cos^{-1}\left(\frac{1}{2x - 1}\right)$$ ranges from $$0$$ to $$\pi$$.\n\nMultiply by 3 and subtract 2:\n\n$$f(x) = 3 \cos^{-1}\left(\frac{1}{2x - 1}\right) - 2,$$\n\nso range is $$3 \times [0, \pi] - 2 = [ -2, 3\pi - 2].$$\n\n**Final answers:**\n\n- Domain: $$(-\infty, 0] \cup [1, \infty)$$\n- Range: $$[-2, 3\pi - 2].$$