If `f(x) = |x| + |x-1|-|x-2|`, then `f(x)`

Options

A. a) has minima at `x=1`
B. b) has maxima at `x=0`
C. c) has neither maxima nor minima at `x=3`
D. d) none of these

Answer

To solve the problem \( f(x) = |x| + |x-1| - |x-2| \), we will analyze the function by considering different cases based on the values of \( x \). The absolute value function changes behavior at the points where the expression inside the absolute value equals zero. Thus, we will consider the critical points \( x = 0 \), \( x = 1 \), and \( x = 2 \).

### Step 1: Identify the intervals
The critical points divide the number line into the following intervals:
1. \( x < 0 \)
2. \( 0 \leq x < 1 \)
3. \( 1 \leq x < 2 \)
4. \( x \geq 2 \)

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

**Case 1: \( x < 0 \)**
- Here, \( |x| = -x \), \( |x-1| = -(x-1) = -x + 1 \), and \( |x-2| = -(x-2) = -x + 2 \).
- Thus,
\[ f(x) = -x + (-x + 1) - (-x + 2) = -x - x + 1 + x - 2 = -x + 1. \]
- Therefore,
\[ f(x) = -3x + 3. \]

**Case 2: \( 0 \leq x < 1 \)**
- Here, \( |x| = x \), \( |x-1| = -(x-1) = -x + 1 \), and \( |x-2| = -(x-2) = -x + 2 \).
- Thus,
\[ f(x) = x + (-x + 1) - (-x + 2) = x - x + 1 + x - 2 = -x + 3. \]

**Case 3: \( 1 \leq x < 2 \)**
- Here, \( |x| = x \), \( |x-1| = x - 1 \), and \( |x-2| = -(x-2) = -x + 2 \).
- Thus,
\[ f(x) = x + (x - 1) - (-x + 2) = x + x - 1 + x - 2 = x + 1. \]

**Case 4: \( x \geq 2 \)**
- Here, \( |x| = x \), \( |x-1| = x - 1 \), and \( |x-2| = x - 2 \).
- Thus,
\[ f(x) = x + (x - 1) - (x - 2) = x + x - 1 - x + 2 = x + 1. \]

### Step 3: Combine the results
Now we can summarize the function \( f(x) \) in piecewise form:
\[ f(x) = \begin{cases} -3x + 3 & \text{if } x < 0 \\ -x + 3 & \text{if } 0 \leq x < 1 \\ x + 1 & \text{if } 1 \leq x < 2 \\ x + 1 & \text{if } x \geq 2 \end{cases} \]

### Step 4: Find the minimum and maximum
To find the minimum and maximum values, we can evaluate the function at the critical points \( x = 0, 1, 2 \):
- \( f(0) = 3 \)
- \( f(1) = 2 \)
- \( f(2) = 3 \)

From this, we can see that the minimum value occurs at \( x = 1 \) where \( f(1) = 2 \).

### Final Answer
The function \( f(x) \) has a minimum at \( x = 1 \) with \( f(1) = 2 \).