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 \( f(x) = |x| + |x - 1| \), we will analyze the function in different intervals based on the properties of the absolute value function.

### Step 1: Identify intervals based on the absolute values
The function \( f(x) \) involves two absolute value expressions: \( |x| \) and \( |x - 1| \). We need to identify the points where the expressions inside the absolute values change sign. These points are \( x = 0 \) and \( x = 1 \).

Thus, we will consider three intervals:
1. \( x < 0 \)
2. \( 0 \leq x < 1 \)
3. \( x \geq 1 \)

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

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

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

3. **For \( x \geq 1 \)**:
- Here, \( |x| = x \) and \( |x - 1| = x - 1 \).
- Therefore,
\[ f(x) = x + (x - 1) = 2x - 1. \]

### Step 3: Write the piecewise function
Combining the results from the three intervals, we can express \( f(x) \) as:
\[ f(x) = \begin{cases} -2x + 1 & \text{if } x < 0 \\ 1 & \text{if } 0 \leq x < 1 \\ 2x - 1 & \text{if } x \geq 1 \end{cases} \]

### Step 4: Check continuity at the points \( x = 0 \) and \( x = 1 \)

1. **At \( x = 0 \)**:
- Left-hand limit (LHL):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-2x + 1) = -2(0) + 1 = 1. \]
- Right-hand limit (RHL):
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (1) = 1. \]
- Value of the function:
\[ f(0) = 1. \]
- Since LHL = RHL = \( f(0) \), the function is continuous at \( x = 0 \).

2. **At \( x = 1 \)**:
- Left-hand limit (LHL):
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (1) = 1. \]
- Right-hand limit (RHL):
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - 1) = 2(1) - 1 = 1. \]
- Value of the function:
\[ f(1) = 1. \]
- Since LHL = RHL = \( f(1) \), the function is continuous at \( x = 1 \).

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

---