Let `f(x) = |x-1|+|x+1|` Discuss the continuity and differentiability of the function.

Answer

To discuss the continuity and differentiability of the function \( f(x) = |x-1| + |x+1| \), we will analyze the function by breaking it down into different intervals based on the points where the expressions inside the absolute values change sign.

### Step 1: Identify critical points
The critical points occur where the expressions inside the absolute values equal zero:
- \( x - 1 = 0 \) → \( x = 1 \)
- \( x + 1 = 0 \) → \( x = -1 \)

Thus, we will analyze the function in three intervals:
1. \( x < -1 \)
2. \( -1 \leq x < 1 \)
3. \( x \geq 1 \)

### Step 2: Define the function in each interval

**Interval 1: \( x < -1 \)**
- Here, both expressions inside the absolute values are negative:
\[ f(x) = -(x - 1) - (x + 1) = -x + 1 - x - 1 = -2x \]

**Interval 2: \( -1 \leq x < 1 \)**
- In this interval, the first expression is negative and the second is positive:
\[ f(x) = -(x - 1) + (x + 1) = -x + 1 + x + 1 = 2 \]

**Interval 3: \( x \geq 1 \)**
- Here, both expressions are positive:
\[ f(x) = (x - 1) + (x + 1) = x - 1 + x + 1 = 2x \]

### Step 3: Summary of the function
We can summarize the function as:
\[ f(x) = \begin{cases} -2x & \text{if } x < -1 \\ 2 & \text{if } -1 \leq x < 1 \\ 2x & \text{if } x \geq 1 \end{cases} \]

### Step 4: Check continuity
To check for continuity at the critical points \( x = -1 \) and \( x = 1 \), we need to verify if:
\[ \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) \]

**At \( x = -1 \):**
- \( \lim_{x \to -1^-} f(x) = -2(-1) = 2 \)
- \( \lim_{x \to -1^+} f(x) = 2 \)
- \( f(-1) = 2 \)
- Thus, \( f(x) \) is continuous at \( x = -1 \).

**At \( x = 1 \):**
- \( \lim_{x \to 1^-} f(x) = 2 \)
- \( \lim_{x \to 1^+} f(x) = 2(1) = 2 \)
- \( f(1) = 2 \)
- Thus, \( f(x) \) is continuous at \( x = 1 \).

Since \( f(x) \) is continuous for all \( x \), we conclude that \( f(x) \) is continuous everywhere.

### Step 5: Check differentiability
To check for differentiability, we need to find the derivative in each interval and check the limits at the critical points.

**Interval 1: \( x < -1 \)**
- \( f'(x) = -2 \)

**Interval 2: \( -1 < x < 1 \)**
- \( f'(x) = 0 \)

**Interval 3: \( x > 1 \)**
- \( f'(x) = 2 \)

**At \( x = -1 \):**
- \( \lim_{x \to -1^-} f'(x) = -2 \)
- \( \lim_{x \to -1^+} f'(x) = 0 \)
- Since the left-hand limit and right-hand limit are not equal, \( f(x) \) is not differentiable at \( x = -1 \).

**At \( x = 1 \):**
- \( \lim_{x \to 1^-} f'(x) = 0 \)
- \( \lim_{x \to 1^+} f'(x) = 2 \)
- Since the left-hand limit and right-hand limit are not equal, \( f(x) \) is not differentiable at \( x = 1 \).

### Conclusion
- The function \( f(x) \) is continuous everywhere.
- The function \( f(x) \) is not differentiable at \( x = -1 \) and \( x = 1 \).