Find the number of critical points of the function
`f(x) = |x - 2| + |x +1 | , x in R`

Answer

To find the number of critical points of the function \( f(x) = |x - 2| + |x + 1| \), we will analyze the function by breaking it down based on the values of \( x \) where the absolute value expressions change.

### Step 1: Identify the points where the function changes
The function \( f(x) \) consists of two absolute value terms: \( |x - 2| \) and \( |x + 1| \). The points where these expressions change are:
- \( x = 2 \) (where \( |x - 2| \) changes)
- \( x = -1 \) (where \( |x + 1| \) changes)

### Step 2: Define the function in intervals
We will analyze the function in the following intervals:
1. \( x < -1 \)
2. \( -1 \leq x < 2 \)
3. \( x \geq 2 \)

#### Interval 1: \( x < -1 \)
In this interval, both expressions are negative:
\[ f(x) = -(x - 2) - (x + 1) = -x + 2 - x - 1 = -2x + 1 \]

#### Interval 2: \( -1 \leq x < 2 \)
In this interval, \( |x - 2| \) is negative and \( |x + 1| \) is positive:
\[ f(x) = -(x - 2) + (x + 1) = -x + 2 + x + 1 = 3 \]

#### Interval 3: \( x \geq 2 \)
In this interval, both expressions are positive:
\[ f(x) = (x - 2) + (x + 1) = x - 2 + x + 1 = 2x - 1 \]

### Step 3: Find the derivative in each interval
To find critical points, we need to find the derivative of \( f(x) \) in each interval.

1. **For \( x < -1 \)**:
\[ f'(x) = -2 \]

2. **For \( -1 \leq x < 2 \)**:
\[ f'(x) = 0 \]

3. **For \( x \geq 2 \)**:
\[ f'(x) = 2 \]

### Step 4: Identify critical points
- In the interval \( x < -1 \), \( f'(x) = -2 \) (no critical points).
- In the interval \( -1 \leq x < 2 \), \( f'(x) = 0 \) (this means the function is constant, and thus every point in this interval is a critical point).
- In the interval \( x \geq 2 \), \( f'(x) = 2 \) (no critical points).

### Conclusion
The critical points occur in the interval \( -1 \leq x < 2 \), which contains infinitely many points since the derivative is zero throughout this interval.

Thus, the number of critical points of the function \( f(x) = |x - 2| + |x + 1| \) is **infinite**.

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