`f(x)=abs(x-1)+abs(x-2), -1 le x le 3`. Find the range of f(x).

Answer

To find the range of the function \( f(x) = |x - 1| + |x - 2| \) for \( -1 \leq x \leq 3 \), we will analyze the function by breaking it down into intervals based on the critical points where the expressions inside the absolute values change sign.

### Step 1: Identify critical points
The critical points occur where the expressions inside the absolute values are zero:
- \( x - 1 = 0 \) gives \( x = 1 \)
- \( x - 2 = 0 \) gives \( x = 2 \)

Thus, we will consider the intervals:
1. \( -1 \leq x < 1 \)
2. \( 1 \leq x < 2 \)
3. \( 2 \leq x \leq 3 \)

### Step 2: Evaluate \( f(x) \) in each interval

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

#### Interval 2: \( 1 \leq x < 2 \)
Here, \( x - 1 \) is non-negative and \( x - 2 \) is negative:
\[ f(x) = (x - 1) - (x - 2) = x - 1 - x + 2 = 1 \]

#### Interval 3: \( 2 \leq x \leq 3 \)
In this interval, both \( x - 1 \) and \( x - 2 \) are non-negative:
\[ f(x) = (x - 1) + (x - 2) = x - 1 + x - 2 = 2x - 3 \]

### Step 3: Determine the values at the boundaries and critical points
Now we will evaluate \( f(x) \) at the boundaries and critical points:
- At \( x = -1 \):
\[ f(-1) = -2(-1) + 3 = 2 + 3 = 5 \]
- At \( x = 1 \):
\[ f(1) = 1 \]
- At \( x = 2 \):
\[ f(2) = 1 \]
- At \( x = 3 \):
\[ f(3) = 2(3) - 3 = 6 - 3 = 3 \]

### Step 4: Collect all values
From the evaluations:
- In the interval \( -1 \leq x < 1 \), \( f(x) \) varies from \( 5 \) to \( 1 \).
- In the interval \( 1 \leq x < 2 \), \( f(x) = 1 \).
- In the interval \( 2 \leq x \leq 3 \), \( f(x) \) varies from \( 1 \) to \( 3 \).

### Step 5: Determine the range
Combining all the values, the minimum value of \( f(x) \) is \( 1 \) and the maximum value is \( 5 \). Thus, the range of \( f(x) \) is:
\[ \text{Range of } f(x) = [1, 5] \]