To solve the inequality \( |x - 1| + 2|x + 1| + |x - 2| \leq 8 \), we will break it down into intervals based on the critical points where the expressions inside the absolute values change sign. The critical points are \( x = -2, -1, 1, 2 \).
### Step 1: Identify the intervals
The critical points divide the number line into the following intervals:
1. \( (-\infty, -2) \)
2. \( [-2, -1) \)
3. \( [-1, 1) \)
4. \( [1, 2) \)
5. \( [2, \infty) \)
### Step 2: Analyze each interval
#### Interval 1: \( x < -2 \)
In this interval, all expressions are negative:
- \( |x - 1| = -(x - 1) = -x + 1 \)
- \( |x + 1| = -(x + 1) = -x - 1 \)
- \( |x - 2| = -(x - 2) = -x + 2 \)
Substituting these into the inequality:
\[
-x + 1 + 2(-x - 1) + (-x + 2) \leq 8
\]
This simplifies to:
\[
-x + 1 - 2x - 2 - x + 2 \leq 8
\]
\[
-4x + 1 \leq 8
\]
\[
-4x \leq 7 \quad \Rightarrow \quad x \geq -\frac{7}{4}
\]
Since this interval is \( (-\infty, -2) \), we take the intersection:
\[
-\frac{7}{4} \text{ is not in } (-\infty, -2) \quad \Rightarrow \quad \text{No solutions in this interval.}
\]
#### Interval 2: \( -2 \leq x < -1 \)
Here:
- \( |x - 1| = -x + 1 \)
- \( |x + 1| = -x - 1 \)
- \( |x - 2| = -x + 2 \)
Substituting:
\[
-x + 1 + 2(-x - 1) + (-x + 2) \leq 8
\]
This simplifies to:
\[
-x + 1 - 2x - 2 - x + 2 \leq 8
\]
\[
-4x + 1 \leq 8
\]
\[
-4x \leq 7 \quad \Rightarrow \quad x \geq -\frac{7}{4}
\]
Since \( -\frac{7}{4} \) is not in \( [-2, -1) \), we have:
\[
\text{No solutions in this interval.}
\]
#### Interval 3: \( -1 \leq x < 1 \)
Here:
- \( |x - 1| = -x + 1 \)
- \( |x + 1| = x + 1 \)
- \( |x - 2| = -x + 2 \)
Substituting:
\[
-x + 1 + 2(x + 1) + (-x + 2) \leq 8
\]
This simplifies to:
\[
-x + 1 + 2x + 2 - x + 2 \leq 8
\]
\[
0x + 5 \leq 8 \quad \Rightarrow \quad 5 \leq 8
\]
This is always true, so:
\[
x \in [-1, 1)
\]
#### Interval 4: \( 1 \leq x < 2 \)
Here:
- \( |x - 1| = x - 1 \)
- \( |x + 1| = x + 1 \)
- \( |x - 2| = -x + 2 \)
Substituting:
\[
(x - 1) + 2(x + 1) + (-x + 2) \leq 8
\]
This simplifies to:
\[
x - 1 + 2x + 2 - x + 2 \leq 8
\]
\[
2x + 3 \leq 8 \quad \Rightarrow \quad 2x \leq 5 \quad \Rightarrow \quad x \leq \frac{5}{2}
\]
Since \( \frac{5}{2} \) is greater than \( 2 \), we take the intersection:
\[
x \in [1, 2)
\]
#### Interval 5: \( x \geq 2 \)
Here:
- \( |x - 1| = x - 1 \)
- \( |x + 1| = x + 1 \)
- \( |x - 2| = x - 2 \)
Substituting:
\[
(x - 1) + 2(x + 1) + (x - 2) \leq 8
\]
This simplifies to:
\[
x - 1 + 2x + 2 + x - 2 \leq 8
\]
\[
4x - 1 \leq 8 \quad \Rightarrow \quad 4x \leq 9 \quad \Rightarrow \quad x \leq \frac{9}{4}
\]
Since \( \frac{9}{4} < 2 \), we take the intersection:
\[
\text{No solutions in this interval.}
\]
### Final Solution
Combining all valid intervals, we find:
\[
x \in [-1, 2)
\]
