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

Explanation

To determine whether the function \( f(x) = |x| + |x - 1| \) is continuous at \( x = 1 \), we need to check the following three conditions:

1. \( f(1) \)
2. \( \lim_{x \to 1^-} f(x) \) (left-hand limit)
3. \( \lim_{x \to 1^+} f(x) \) (right-hand limit)

The function is continuous at \( x = 1 \) if all three values are equal.

### Step 1: Calculate \( f(1) \)

\[ f(1) = |1| + |1 - 1| = 1 + 0 = 1 \]

### Step 2: Calculate \( \lim_{x \to 1^-} f(x) \)

For \( x \) approaching 1 from the left (\( x < 1 \)), we have:

\[ f(x) = |x| + |x - 1| \]

Since \( x < 1 \), \( |x - 1| = 1 - x \). Thus,

\[ f(x) = |x| + (1 - x) \]

For \( x \) in this range, \( |x| = x \) (since \( x \) is non-negative when \( x < 1 \)). Therefore,

\[ f(x) = x + (1 - x) = 1 \]

Taking the limit as \( x \) approaches 1 from the left:

\[ \lim_{x \to 1^-} f(x) = 1 \]

### Step 3: Calculate \( \lim_{x \to 1^+} f(x) \)

For \( x \) approaching 1 from the right (\( x > 1 \)), we have:

\[ f(x) = |x| + |x - 1| \]

Since \( x > 1 \), \( |x - 1| = x - 1 \). Thus,

\[ f(x) = |x| + (x - 1) \]

For \( x \) in this range, \( |x| = x \). Therefore,

\[ f(x) = x + (x - 1) = 2x - 1 \]

Taking the limit as \( x \) approaches 1 from the right:

\[ \lim_{x \to 1^+} f(x) = 2(1) - 1 = 1 \]

### Conclusion

Now we have:

- \( f(1) = 1 \)
- \( \lim_{x \to 1^-} f(x) = 1 \)
- \( \lim_{x \to 1^+} f(x) = 1 \)

Since all three values are equal, we conclude that the function \( f(x) = |x| + |x - 1| \) is continuous at \( x = 1 \).