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

Options

A. f(x) is continuous both at x = 0 and 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 at its breakpoints, which are determined by the points where the expressions inside the absolute values change their signs. The breakpoints here are at \( x = 0 \) and \( x = 1 \).

### Step 1: Identify Breakpoints
The breakpoints of the function are:
- \( x = 0 \) (from \( |x| \))
- \( x = 1 \) (from \( |x-1| \))

### Step 2: Define the Function in Different Intervals
We will evaluate the function in the intervals defined by these breakpoints:

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

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

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

Thus, we can summarize the function as:
\[ f(x) = \begin{cases} 1 - 2x & \text{if } x < 0 \\ 1 & \text{if } 0 \leq x < 1 \\ 2x - 1 & \text{if } x \geq 1 \end{cases} \]

### Step 3: Check Continuity at \( x = 0 \)
To check continuity at \( x = 0 \), we need to verify:
- \( \lim_{x \to 0^-} f(x) \)
- \( \lim_{x \to 0^+} f(x) \)
- \( f(0) \)

Calculating these:
- \( \lim_{x \to 0^-} f(x) = 1 - 2(0) = 1 \)
- \( \lim_{x \to 0^+} f(x) = 1 \)
- \( f(0) = 1 \)

Since \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 1 \), the function is continuous at \( x = 0 \).

### Step 4: Check Continuity at \( x = 1 \)
To check continuity at \( x = 1 \), we need to verify:
- \( \lim_{x \to 1^-} f(x) \)
- \( \lim_{x \to 1^+} f(x) \)
- \( f(1) \)

Calculating these:
- \( \lim_{x \to 1^-} f(x) = 1 \)
- \( \lim_{x \to 1^+} f(x) = 2(1) - 1 = 1 \)
- \( f(1) = 2(1) - 1 = 1 \)

Since \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 1 \), the function is continuous at \( x = 1 \).

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

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