To solve the problem, we need to find the values of \( x \) in the interval \([-2, 2]\) for which \( f(x-3) \), \( f(x-1) \), and \( f(x+1) \) are in arithmetic progression.
### Step 1: Define the function
The function is given as:
\[
f(x) = |x + 1|
\]
### Step 2: Calculate \( f(x-3) \), \( f(x-1) \), and \( f(x+1) \)
1. **Calculate \( f(x-3) \)**:
\[
f(x-3) = |(x-3) + 1| = |x - 2|
\]
2. **Calculate \( f(x-1) \)**:
\[
f(x-1) = |(x-1) + 1| = |x|
\]
3. **Calculate \( f(x+1) \)**:
\[
f(x+1) = |(x+1) + 1| = |x + 2|
\]
### Step 3: Set up the arithmetic progression condition
For three numbers \( a, b, c \) to be in arithmetic progression, the middle term must equal the average of the other two:
\[
f(x-1) = \frac{f(x-3) + f(x+1)}{2}
\]
Substituting the values we calculated:
\[
|x| = \frac{|x - 2| + |x + 2|}{2}
\]
Multiplying both sides by 2:
\[
2|x| = |x - 2| + |x + 2|
\]
### Step 4: Analyze the absolute values
The absolute values will change depending on the value of \( x \). We will consider different cases based on the critical points where the expressions inside the absolute values change sign: \( x = -2, 0, 2 \).
#### Case 1: \( x \in [-2, 0) \)
- Here, \( |x| = -x \), \( |x - 2| = -x + 2 \), and \( |x + 2| = x + 2 \).
\[
2(-x) = (-x + 2) + (x + 2)
\]
\[
-2x = 4 \implies x = -2
\]
Since \( -2 \) is in the interval, it is a valid solution.
#### Case 2: \( x = 0 \)
- Check if \( x = 0 \) satisfies the condition:
\[
2|0| = |0 - 2| + |0 + 2| \implies 0 = 2 + 2 \implies 0 \neq 4
\]
Thus, \( x = 0 \) is not a solution.
#### Case 3: \( x \in (0, 2] \)
- Here, \( |x| = x \), \( |x - 2| = -x + 2 \), and \( |x + 2| = x + 2 \).
\[
2x = (-x + 2) + (x + 2)
\]
\[
2x = 4 \implies x = 2
\]
Since \( 2 \) is in the interval, it is also a valid solution.
### Step 5: Conclusion
The valid solutions for \( x \) in the interval \([-2, 2]\) are:
- \( x = -2 \)
- \( x = 2 \)
Thus, the total number of values of \( x \) for which \( f(x-3), f(x-1), f(x+1) \) are in arithmetic progression is **2**.
