A function is defined as `f(x) = {{:(e^(x)",",x le 0),(|x-1|",",x gt 0):}`, then f(x) is

Options

A. A. continuous at x = 0
B. B. continuous at x = 1
C. C. differentiable at x = 0
D. D. differentiable at x = 1

Answer

To determine the continuity and differentiability of the function defined as:

\[ f(x) = \begin{cases} e^x & \text{if } x \leq 0 \\ |x - 1| & \text{if } x > 0 \end{cases} \]

we will analyze the function at the points of interest, specifically \(x = 0\) and \(x = 1\).

### Step 1: Check Continuity at \(x = 0\)

**Left-hand limit as \(x\) approaches 0:**
\[ f(0^-) = e^0 = 1 \]

**Right-hand limit as \(x\) approaches 0:**
\[ f(0^+) = |0 - 1| = 1 \]

Since both the left-hand limit and right-hand limit at \(x = 0\) are equal to \(1\), we have:
\[ f(0) = 1 \]
Thus, \(f(x)\) is continuous at \(x = 0\).

### Step 2: Check Continuity at \(x = 1\)

**Left-hand limit as \(x\) approaches 1:**
\[ f(1^-) = |1 - 1| = 0 \]

**Right-hand limit as \(x\) approaches 1:**
\[ f(1^+) = 1 - 1 = 0 \]

Since both the left-hand limit and right-hand limit at \(x = 1\) are equal to \(0\), we have:
\[ f(1) = 0 \]
Thus, \(f(x)\) is continuous at \(x = 1\).

### Step 3: Check Differentiability at \(x = 0\)

**Left-hand derivative at \(x = 0\):**
\[ \lim_{h \to 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h \to 0} \frac{e^{-h} - 1}{-h} \]
Using L'Hôpital's Rule:
\[ = \lim_{h \to 0} \frac{-e^{-h}}{-1} = e^0 = 1 \]

**Right-hand derivative at \(x = 0\):**
\[ \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{e^h - 1}{h} \]
Using L'Hôpital's Rule:
\[ = \lim_{h \to 0} \frac{e^h}{1} = e^0 = 1 \]

Since both derivatives are equal, \(f(x)\) is differentiable at \(x = 0\).

### Step 4: Check Differentiability at \(x = 1\)

**Left-hand derivative at \(x = 1\):**
\[ \lim_{h \to 0} \frac{f(1 - h) - f(1)}{-h} = \lim_{h \to 0} \frac{1 - h - 0}{-h} = \lim_{h \to 0} \frac{1 - h}{-h} = -1 \]

**Right-hand derivative at \(x = 1\):**
\[ \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0} \frac{(1 + h) - 1 - 0}{h} = \lim_{h \to 0} \frac{h}{h} = 1 \]

Since the left-hand derivative is \(-1\) and the right-hand derivative is \(1\), they are not equal. Thus, \(f(x)\) is not differentiable at \(x = 1\).

### Conclusion

The function \(f(x)\) is continuous at \(x = 0\) and \(x = 1\), but it is only differentiable at \(x = 0\). Therefore, the final answer is:

- **Continuous at \(x = 0\) and \(x = 1\)**
- **Differentiable at \(x = 0\)**
- **Not differentiable at \(x = 1\)**