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

Options

A. f(x) is continuous at x=0 as well as at x=1
B. f(x) is continuous at x=0, but not at x=1
C. f(x) is continuous at x=1, but not at x=0
D. None of these

Answer

To determine the continuity of the function \( f(x) = |x| + |x-1| \) at the points \( x = 0 \) and \( x = 1 \), we need to analyze the function in different intervals defined by the points where the absolute values change.

### Step 1: Identify intervals
The function \( f(x) \) can be expressed differently based on the values of \( x \):
1. For \( x < 0 \)
2. For \( 0 \leq x < 1 \)
3. For \( x \geq 1 \)

### Step 2: Define the function in each interval
- **For \( x < 0 \)**:
\[ f(x) = -x + (-(x-1)) = -x + 1 + x = 1 \]

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

- **For \( x \geq 1 \)**:
\[ f(x) = x + (x-1) = 2x - 1 \]

### Step 3: Check continuity at \( x = 0 \)
To check continuity at \( x = 0 \), we need to find:
- \( f(0) \)
- Left-hand limit as \( x \to 0^- \)
- Right-hand limit as \( x \to 0^+ \)

1. **Value at \( x = 0 \)**:
\[ f(0) = 1 \]

2. **Left-hand limit**:
\[ \lim_{x \to 0^-} f(x) = 1 \]

3. **Right-hand limit**:
\[ \lim_{x \to 0^+} f(x) = 1 \]

Since the left-hand limit, right-hand limit, and the function value at \( x = 0 \) are all equal, \( f(x) \) is continuous at \( x = 0 \).

### Step 4: Check continuity at \( x = 1 \)
To check continuity at \( x = 1 \), we need to find:
- \( f(1) \)
- Left-hand limit as \( x \to 1^- \)
- Right-hand limit as \( x \to 1^+ \)

1. **Value at \( x = 1 \)**:
\[ f(1) = 1 \]

2. **Left-hand limit**:
\[ \lim_{x \to 1^-} f(x) = 1 \]

3. **Right-hand limit**:
\[ \lim_{x \to 1^+} f(x) = 2 \cdot 1 - 1 = 1 \]

Again, since the left-hand limit, right-hand limit, and the function value at \( x = 1 \) are all equal, \( f(x) \) is continuous at \( x = 1 \).

### Conclusion
The function \( f(x) = |x| + |x-1| \) is continuous at both \( x = 0 \) and \( x = 1 \).