The function f(x)=|x|+|x-1| is

Options

A. differentiable at x=0 but not at x=1
B. differentiable at x=1 bit not at x=0
C. neither differentiable at x=0 nor at x=1
D. differentiable at x=0 as well as at x=1

Answer

To determine the differentiability of the function \( f(x) = |x| + |x - 1| \) at the points \( x = 0 \) and \( x = 1 \), we will analyze the function by breaking it down into intervals based on the points where the absolute values change.

### Step 1: Identify critical points
The critical points occur where the expressions inside the absolute values are zero. Thus, we have:
- \( |x| = 0 \) at \( x = 0 \)
- \( |x - 1| = 0 \) at \( x = 1 \)

### Step 2: Break down the function into intervals
We will analyze the function in three intervals:
1. \( x < 0 \)
2. \( 0 \leq x < 1 \)
3. \( x \geq 1 \)

### Step 3: Evaluate \( f(x) \) in each interval

**Interval 1: \( x < 0 \)**
- Here, \( |x| = -x \) and \( |x - 1| = -(x - 1) = -x + 1 \).
- Therefore, \( f(x) = -x + (-x + 1) = -2x + 1 \).

**Interval 2: \( 0 \leq x < 1 \)**
- Here, \( |x| = x \) and \( |x - 1| = -(x - 1) = -x + 1 \).
- Therefore, \( f(x) = x + (-x + 1) = 1 \).

**Interval 3: \( x \geq 1 \)**
- Here, \( |x| = x \) and \( |x - 1| = x - 1 \).
- Therefore, \( f(x) = x + (x - 1) = 2x - 1 \).

### Step 4: Summary of the piecewise function
We can summarize the function as:
\[ f(x) = \begin{cases} -2x + 1 & \text{if } x < 0 \\ 1 & \text{if } 0 \leq x < 1 \\ 2x - 1 & \text{if } x \geq 1 \end{cases} \]

### Step 5: Check differentiability at \( x = 0 \)
To check differentiability at \( x = 0 \), we need to find the left-hand derivative and the right-hand derivative.

- **Left-hand derivative at \( x = 0 \)**:
\[ f'(x) = -2 \quad \text{(for } x < 0\text{)} \]

- **Right-hand derivative at \( x = 0 \)**:
\[ f'(x) = 0 \quad \text{(for } 0 \leq x < 1\text{)} \]

Since the left-hand derivative \(-2\) is not equal to the right-hand derivative \(0\), \( f(x) \) is not differentiable at \( x = 0 \).

### Step 6: Check differentiability at \( x = 1 \)
Now we check differentiability at \( x = 1 \).

- **Left-hand derivative at \( x = 1 \)**:
\[ f'(x) = 0 \quad \text{(for } 0 \leq x < 1\text{)} \]

- **Right-hand derivative at \( x = 1 \)**:
\[ f'(x) = 2 \quad \text{(for } x \geq 1\text{)} \]

Since the left-hand derivative \(0\) is not equal to the right-hand derivative \(2\), \( f(x) \) is not differentiable at \( x = 1 \).

### Conclusion
The function \( f(x) = |x| + |x - 1| \) is not differentiable at both \( x = 0 \) and \( x = 1 \).

---