To solve the inequality $|2x - 1| \geq |x - 1|$, we need to consider different cases based on the properties of absolute values. Absolute values split into different cases depending on whether the expressions inside them are positive or negative.

Step-by-Step Solution

1. Identify critical points: The critical points are the values of $x$ that make each expression inside the absolute value equal to zero.

   • For $|2x - 1|$, the critical point is when $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$.
   • For $|x - 1|$, the critical point is when $x - 1 = 0 \Rightarrow x = 1$.

   These points divide the real number line into three intervals: $(-\infty, \frac{1}{2})$, $(\frac{1}{2}, 1)$, and $(1, \infty)$.

2. Analyze each interval:

   Let's solve the inequality in each interval.

   • Interval 1: $x < \frac{1}{2}$

     In this interval:

     • $2x - 1 < 0 \Rightarrow |2x - 1| = -(2x - 1) = -2x + 1$
     • $x - 1 < 0 \Rightarrow |x - 1| = -(x - 1) = -x + 1$

     The inequality becomes: $-2x + 1 \geq -x + 1$

     Simplify this: $-2x + 1 \geq -x + 1$ $-2x + 1 - 1 \geq -x + 1 - 1$ $-2x \geq -x$ Divide both sides by -1 (and reverse the inequality): $2x \leq x$ Subtract $x$ from both sides: $x \leq 0$

     So, in the interval $(-\infty, \frac{1}{2})$, the inequality holds for $x \leq 0$.

   • Interval 2: $\frac{1}{2} \leq x < 1$

     In this interval:

     • $2x - 1 \geq 0 \Rightarrow |2x - 1| = 2x - 1$
     • $x - 1 < 0 \Rightarrow |x - 1| = -(x - 1) = -x + 1$

     The inequality becomes: $2x - 1 \geq -x + 1$

     Simplify this: $2x - 1 \geq -x + 1$ $2x + x \geq 1 + 1$ $3x \geq 2$ Divide both sides by 3: $x \geq \frac{2}{3}$

     So, in the interval $(\frac{1}{2}, 1)$, the inequality holds for $x \geq \frac{2}{3}$.

   • Interval 3: $x \geq 1$

     In this interval:

     • $2x - 1 \geq 0 \Rightarrow |2x - 1| = 2x - 1$
     • $x - 1 \geq 0 \Rightarrow |x - 1| = x - 1$

     The inequality becomes: $2x - 1 \geq x - 1$

     Simplify this: $2x - x \geq -1 + 1$ $x \geq 0$

     Since we are already in the interval where $x \geq 1$, and $x \geq 0$ is always true in this interval, the inequality holds for all $x \geq 1$.

3. Combine the solutions:

   The solutions from each interval are:

   • $x \leq 0$ in $(-\infty, \frac{1}{2})$
   • $x \geq \frac{2}{3}$ in $(\frac{1}{2}, 1)$
   • $x \geq 1$ in $(1, \infty)$

   Combining these, the solution set for the inequality $|2x - 1| \geq |x - 1|$ is: $x \in (-\infty, 0] \cup \left[\frac{2}{3}, \infty\right)$

Conclusion

The solution to the inequality $|2x - 1| \geq |x - 1|$ is:

$x \in (-\infty, 0] \cup \left[\frac{2}{3}, \infty\right)$

Do you want details or have any questions?

Here are 5 related questions to explore further:

1. How do you solve an inequality involving absolute values in general?
2. What are the graphical interpretations of absolute value inequalities?
3. How would the solution change if the inequality was $|2x - 1| \leq |x - 1|$?
4. Can you solve an absolute value inequality by squaring both sides? Why or why not?
5. How does solving absolute value equations differ from solving absolute value inequalities?

Tip: Always check each interval's conditions when solving absolute value inequalities to ensure correct solutions.