The set of points where the function `f(x)=|x-1|e^(x)` is differentiable, is

Options

A. `R`
B. `R-[1]`
C. `R-[-1]`
D. `R-(0)`

Answer

To determine the set of points where the function \( f(x) = |x - 1| e^x \) is differentiable, we need to analyze the function and its behavior at the point where the absolute value expression changes, which is at \( x = 1 \).

### Step-by-Step Solution:

1. **Identify the Function**:
The function is given as:
\[ f(x) = |x - 1| e^x \]

2. **Break Down the Absolute Value**:
The absolute value function \( |x - 1| \) can be expressed in piecewise form:
- For \( x \geq 1 \): \( |x - 1| = x - 1 \)
- For \( x < 1 \): \( |x - 1| = -(x - 1) = 1 - x \)

Therefore, we can write \( f(x) \) as:
\[ f(x) = \begin{cases} (x - 1)e^x & \text{if } x \geq 1 \\ (1 - x)e^x & \text{if } x < 1 \end{cases} \]

3. **Differentiate the Function**:
Now, we will differentiate \( f(x) \) in both cases.

- For \( x \geq 1 \):
\[ f'(x) = \frac{d}{dx}((x - 1)e^x) = (x - 1)e^x + e^x = x e^x \]

- For \( x < 1 \):
\[ f'(x) = \frac{d}{dx}((1 - x)e^x) = (1 - x)e^x - e^x = (1 - x - 1)e^x = -x e^x \]

Thus, we have:
\[ f'(x) = \begin{cases} x e^x & \text{if } x \geq 1 \\ -x e^x & \text{if } x < 1 \end{cases} \]

4. **Check Differentiability at \( x = 1 \)**:
To check if \( f(x) \) is differentiable at \( x = 1 \), we need to see if the left-hand derivative and the right-hand derivative at that point are equal.

- Right-hand derivative at \( x = 1 \):
\[ f'(1^+) = 1 \cdot e^1 = e \]

- Left-hand derivative at \( x = 1 \):
\[ f'(1^-) = -1 \cdot e^1 = -e \]

Since \( f'(1^+) = e \) and \( f'(1^-) = -e \), we see that:
\[ f'(1^+) \neq f'(1^-) \]

5. **Conclusion**:
The function \( f(x) \) is not differentiable at \( x = 1 \) because the left-hand and right-hand derivatives do not match. However, it is differentiable everywhere else.

Therefore, the set of points where the function is differentiable is:
\[ \text{All real numbers except } x = 1 \]

### Final Answer:
The set of points where the function \( f(x) = |x - 1| e^x \) is differentiable is \( \mathbb{R} \setminus \{1\} \).