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

Explanation

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

### Step 1: Define the function in piecewise form
The function \( f(x) \) can be expressed in different forms depending on the value of \( x \):

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

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

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

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

### Step 2: Check continuity at \( x = 1 \)
To check for continuity at \( x = 1 \), we need to verify three conditions:
1. \( f(1) \) is defined.
2. The limit \( \lim_{x \to 1} f(x) \) exists.
3. \( \lim_{x \to 1} f(x) = f(1) \).

#### Calculate \( f(1) \):
Using the piecewise definition for \( x \geq 1 \):
\[ f(1) = 2(1) - 1 = 1 \]

#### Calculate \( \lim_{x \to 1^-} f(x) \):
For \( x \) approaching 1 from the left (i.e., \( 0 \leq x < 1 \)):
\[ \lim_{x \to 1^-} f(x) = 1 \]

#### Calculate \( \lim_{x \to 1^+} f(x) \):
For \( x \) approaching 1 from the right (i.e., \( x \geq 1 \)):
\[ \lim_{x \to 1^+} f(x) = 2(1) - 1 = 1 \]

#### Conclusion on continuity:
Since:
- \( f(1) = 1 \)
- \( \lim_{x \to 1^-} f(x) = 1 \)
- \( \lim_{x \to 1^+} f(x) = 1 \)

We have:
\[ \lim_{x \to 1} f(x) = f(1) \]

Thus, \( f(x) \) is continuous at \( x = 1 \).

### Step 3: Graph the function
To visualize the function, we can sketch the graph based on the piecewise definitions:
- For \( x < 0 \), the line \( -2x + 1 \) is decreasing.
- For \( 0 \leq x < 1 \), the function is constant at \( y = 1 \).
- For \( x \geq 1 \), the line \( 2x - 1 \) is increasing.

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

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