`f(x) = |x| + |x-1|` at `x = 1`.

Explanation

To determine the continuity of the function \( f(x) = |x| + |x-1| \) at \( x = 1 \), we will follow these steps:

### Step 1: Find the value of the function at \( x = 1 \)
We start by calculating \( f(1) \):
\[ f(1) = |1| + |1 - 1| = 1 + 0 = 1 \]

### Step 2: Find the left-hand limit as \( x \) approaches 1
We need to calculate the left-hand limit:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (|x| + |x-1|) \]
For \( x < 1 \):
- \( |x| = x \) (since \( x \) is non-negative)
- \( |x-1| = -(x-1) = 1 - x \)

Thus, we have:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + (1 - x)) = \lim_{x \to 1^-} 1 = 1 \]

### Step 3: Find the right-hand limit as \( x \) approaches 1
Next, we calculate the right-hand limit:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (|x| + |x-1|) \]
For \( x \geq 1 \):
- \( |x| = x \)
- \( |x-1| = x - 1 \)

Thus, we have:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x + (x - 1)) = \lim_{x \to 1^+} (2x - 1) = 2(1) - 1 = 1 \]

### Step 4: Compare the limits and the function value
Now we compare the left-hand limit, right-hand limit, and the function value at \( x = 1 \):
- \( f(1) = 1 \)
- \( \lim_{x \to 1^-} f(x) = 1 \)
- \( \lim_{x \to 1^+} f(x) = 1 \)

Since both the left-hand limit and right-hand limit are equal to the function value at \( x = 1 \), we conclude that:
\[ \text{The function } f(x) \text{ is continuous at } x = 1. \]