Solve : `| x - 1| + |x - 2| + | x - 3 | gt 6 `

Options

A. `x in (-oo, 1)`
B. `x in (-oo, 0)`
C. `x in (4, oo)`
D. ` (2, oo)`

Answer

To solve the inequality \( |x - 1| + |x - 2| + |x - 3| > 6 \), we will consider different cases based on the values of \( x \).

### Step 1: Identify critical points
The critical points where the expressions inside the absolute values change are \( x = 1, 2, 3 \). We will consider intervals based on these points:
1. \( (-\infty, 1) \)
2. \( [1, 2) \)
3. \( [2, 3) \)
4. \( [3, \infty) \)

### Step 2: Solve for each interval

**Case 1: \( x < 1 \)**
In this interval, all expressions are negative:
\[ |x - 1| = 1 - x, \quad |x - 2| = 2 - x, \quad |x - 3| = 3 - x \]
Thus, the inequality becomes:
\[ (1 - x) + (2 - x) + (3 - x) > 6 \]
Simplifying this:
\[ 6 - 3x > 6 \]
Subtracting 6 from both sides:
\[ -3x > 0 \implies x < 0 \]
So, for this case, the solution is \( x < 0 \).

**Case 2: \( 1 \leq x < 2 \)**
In this interval, \( |x - 1| = x - 1 \) and the others are negative:
\[ |x - 1| = x - 1, \quad |x - 2| = 2 - x, \quad |x - 3| = 3 - x \]
The inequality becomes:
\[ (x - 1) + (2 - x) + (3 - x) > 6 \]
Simplifying this:
\[ 4 - x > 6 \]
Subtracting 4 from both sides:
\[ -x > 2 \implies x < -2 \]
Since this contradicts our assumption \( 1 \leq x < 2 \), there are no solutions in this interval.

**Case 3: \( 2 \leq x < 3 \)**
In this interval, \( |x - 1| = x - 1 \), \( |x - 2| = x - 2 \), and \( |x - 3| = 3 - x \):
\[ (x - 1) + (x - 2) + (3 - x) > 6 \]
Simplifying this:
\[ x > 6 \]
This is impossible since \( x < 3 \). Thus, there are no solutions in this interval.

**Case 4: \( x \geq 3 \)**
In this interval, all expressions are positive:
\[ |x - 1| = x - 1, \quad |x - 2| = x - 2, \quad |x - 3| = x - 3 \]
The inequality becomes:
\[ (x - 1) + (x - 2) + (x - 3) > 6 \]
Simplifying this:
\[ 3x - 6 > 6 \]
Adding 6 to both sides:
\[ 3x > 12 \implies x > 4 \]

### Step 3: Combine the solutions
From our cases, we have:
- From Case 1: \( x < 0 \)
- From Case 2: No solutions
- From Case 3: No solutions
- From Case 4: \( x > 4 \)

Thus, the final solution is:
\[ x \in (-\infty, 0) \cup (4, \infty) \]