If `|x-1|+|x+1|=2`, then find x.

Options

A. `-2 lexle1`
B. `x= -1, 1`
C. `-1 lexle1`
D. `-1 lexle2`

Explanation

To solve the equation \( |x-1| + |x+1| = 2 \), we will consider different cases based on the values of \( x \).

### Step 1: Identify the critical points
The critical points for the absolute values are \( x = -1 \) and \( x = 1 \). We will analyze the equation in three intervals:
1. \( x < -1 \)
2. \( -1 \leq x < 1 \)
3. \( x \geq 1 \)

### Step 2: Case 1: \( x < -1 \)
In this case, both expressions inside the absolute values are negative:
\[ |x-1| = -(x-1) = -x + 1 \]
\[ |x+1| = -(x+1) = -x - 1 \]
Substituting these into the equation gives:
\[ (-x + 1) + (-x - 1) = 2 \]
Simplifying this:
\[ -2x = 2 \]
\[ x = -1 \]
Since \( x = -1 \) is not less than \(-1\), there are no solutions in this case.

### Step 3: Case 2: \( -1 \leq x < 1 \)
Here, \( |x-1| \) is negative and \( |x+1| \) is positive:
\[ |x-1| = -(x-1) = -x + 1 \]
\[ |x+1| = x + 1 \]
Substituting into the equation gives:
\[ (-x + 1) + (x + 1) = 2 \]
Simplifying this:
\[ 2 = 2 \]
This is always true, meaning any \( x \) in the interval \([-1, 1)\) is a solution.

### Step 4: Case 3: \( x \geq 1 \)
In this case, both expressions inside the absolute values are positive:
\[ |x-1| = x - 1 \]
\[ |x+1| = x + 1 \]
Substituting into the equation gives:
\[ (x - 1) + (x + 1) = 2 \]
Simplifying this:
\[ 2x = 2 \]
\[ x = 1 \]
Since \( x = 1 \) is included in this case, it is a valid solution.

### Conclusion
Combining the results from all cases, we find that the solution set is:
\[ x \in [-1, 1] \]