1. **State the problem:** Find the range of values of $x$ satisfying the inequality $$|x - 1| - |2x + 1| \geq -4.$$\n\n2. **Recall the definition of absolute value:** For any real number $a$, $$|a| = \begin{cases} a & \text{if } a \geq 0 \\ -a & \text{if } a < 0 \end{cases}.$$\n\n3. **Identify critical points:** The expressions inside the absolute values change sign at points where they are zero.\n- For $|x - 1|$, zero at $x = 1$.\n- For $|2x + 1|$, zero at $x = -\frac{1}{2}$.\n\n4. **Split the real line into intervals based on these points:**\n- Interval 1: $x < -\frac{1}{2}$\n- Interval 2: $-\frac{1}{2} \leq x < 1$\n- Interval 3: $x \geq 1$\n\n5. **Analyze each interval:**\n\n**Interval 1: $x < -\frac{1}{2}$**\n- $x - 1 < 0 \Rightarrow |x - 1| = -(x - 1) = 1 - x$\n- $2x + 1 < 0 \Rightarrow |2x + 1| = -(2x + 1) = -2x - 1$\nInequality becomes:\n$$|x - 1| - |2x + 1| \geq -4 \Rightarrow (1 - x) - (-2x - 1) \geq -4.$$\nSimplify:\n$$1 - x + 2x + 1 \geq -4 \Rightarrow x + 2 \geq -4 \Rightarrow x \geq -6.$$\nSince we are in $x < -\frac{1}{2}$, the solution in this interval is $$-6 \leq x < -\frac{1}{2}.$$\n\n**Interval 2: $-\frac{1}{2} \leq x < 1$**\n- $x - 1 < 0 \Rightarrow |x - 1| = 1 - x$\n- $2x + 1 \geq 0 \Rightarrow |2x + 1| = 2x + 1$\nInequality becomes:\n$$ (1 - x) - (2x + 1) \geq -4.$$\nSimplify:\n$$1 - x - 2x - 1 \geq -4 \Rightarrow -3x \geq -4 \Rightarrow x \leq \frac{4}{3}.$$\nSince $x < 1$ in this interval, the solution is $$-\frac{1}{2} \leq x < 1.$$\n\n**Interval 3: $x \geq 1$**\n- $x - 1 \geq 0 \Rightarrow |x - 1| = x - 1$\n- $2x + 1 \geq 0 \Rightarrow |2x + 1| = 2x + 1$\nInequality becomes:\n$$ (x - 1) - (2x + 1) \geq -4.$$\nSimplify:\n$$x - 1 - 2x - 1 \geq -4 \Rightarrow -x - 2 \geq -4 \Rightarrow -x \geq -2 \Rightarrow x \leq 2.$$\nSince $x \geq 1$, solution is $$1 \leq x \leq 2.$$\n\n6. **Combine all solutions:**\n$$[-6, -\frac{1}{2}) \cup [-\frac{1}{2}, 1) \cup [1, 2] = [-6, 2].$$\n\n7. **Final answer:** The values of $x$ satisfying the inequality are $$x \in [-6, 2].$$\n\n**This corresponds to option (A).