If `|x-1|+|x| + |x+1| ge 6` , then x belongs to

Options

A. `(-oo, 2)`
B. `(-oo, -2] cup [2, oo)`
C. R
D. `phi`

Answer

To solve the inequality \( |x-1| + |x| + |x+1| \geq 6 \), we will analyze the expression by considering different cases based on the values of \( x \).

### Step 1: Identify critical points
The critical points occur where the expressions inside the absolute values change sign. These points are:
- \( x - 1 = 0 \) → \( x = 1 \)
- \( x = 0 \)
- \( x + 1 = 0 \) → \( x = -1 \)

Thus, the critical points are \( x = -1, 0, 1 \). We will consider the intervals defined by these points:
1. \( (-\infty, -1) \)
2. \( [-1, 0) \)
3. \( [0, 1) \)
4. \( [1, \infty) \)

### Step 2: Analyze each case

#### Case 1: \( x < -1 \)
In this case, all terms inside the absolute values are negative:
\[ |x-1| = -(x-1) = -x + 1, \quad |x| = -x, \quad |x+1| = -(x+1) = -x - 1 \]
Thus, the inequality becomes:
\[ (-x + 1) + (-x) + (-x - 1) \geq 6 \]
Simplifying:
\[ -3x \geq 6 \implies x \leq -2 \]
Since \( x < -1 \), the solution for this case is:
\[ x \in (-\infty, -2] \]

#### Case 2: \( -1 \leq x < 0 \)
In this interval, we have:
\[ |x-1| = -(x-1) = -x + 1, \quad |x| = -x, \quad |x+1| = x + 1 \]
The inequality becomes:
\[ (-x + 1) + (-x) + (x + 1) \geq 6 \]
Simplifying:
\[ 1 \geq 6 \]
This is a contradiction, so there are no solutions in this case.

#### Case 3: \( 0 \leq x < 1 \)
In this case:
\[ |x-1| = -(x-1) = -x + 1, \quad |x| = x, \quad |x+1| = x + 1 \]
The inequality becomes:
\[ (-x + 1) + x + (x + 1) \geq 6 \]
Simplifying:
\[ 2 \geq 6 \]
This is also a contradiction, so there are no solutions in this case.

#### Case 4: \( x \geq 1 \)
In this case, all terms are positive:
\[ |x-1| = x - 1, \quad |x| = x, \quad |x+1| = x + 1 \]
The inequality becomes:
\[ (x - 1) + x + (x + 1) \geq 6 \]
Simplifying:
\[ 3x \geq 6 \implies x \geq 2 \]
Thus, the solution for this case is:
\[ x \in [2, \infty) \]

### Step 3: Combine the solutions
From all cases, we have:
- From Case 1: \( x \in (-\infty, -2] \)
- From Case 2: No solutions
- From Case 3: No solutions
- From Case 4: \( x \in [2, \infty) \)

Combining these results, we find:
\[ x \in (-\infty, -2] \cup [2, \infty) \]

### Final Answer
Thus, the solution to the inequality \( |x-1| + |x| + |x+1| \geq 6 \) is:
\[ x \in (-\infty, -2] \cup [2, \infty) \]