At x = 0 , the function ` y = e^(-|x|)` is

Options

A. Continuous
B. Continuous and differentiable
C. Differentiable with derivative = 1
D. Differentiable with derivative =- 1

Answer

To determine the properties of the function \( y = e^{-|x|} \) at \( x = 0 \), we will check for continuity and differentiability step by step.

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

A function is continuous at a point if:
1. The function is defined at that point.
2. The limit of the function as \( x \) approaches that point exists.
3. The limit equals the function value at that point.

**1. Function Value at \( x = 0 \)**:
\[ y(0) = e^{-|0|} = e^{0} = 1 \]

**2. Limit as \( x \) approaches 0**:
We need to find:
\[ \lim_{x \to 0} e^{-|x|} \]
As \( x \) approaches 0, \( |x| \) approaches 0 as well. Therefore:
\[ \lim_{x \to 0} e^{-|x|} = e^{0} = 1 \]

**3. Compare Limit and Function Value**:
Since:
\[ \lim_{x \to 0} e^{-|x|} = 1 = y(0) \]
The function is continuous at \( x = 0 \).

### Step 2: Check Differentiability at \( x = 0 \)

A function is differentiable at a point if the left-hand derivative and right-hand derivative at that point exist and are equal.

**1. Left-hand Derivative**:
The left-hand derivative at \( x = 0 \) is given by:
\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{e^{-|h|} - 1}{h} \]
Since \( h \) is negative, \( |h| = -h \):
\[ f'(0^-) = \lim_{h \to 0^-} \frac{e^{h} - 1}{h} \]
Using the limit definition of the derivative:
\[ f'(0^-) = 1 \quad \text{(as } h \to 0\text{)} \]

**2. Right-hand Derivative**:
The right-hand derivative at \( x = 0 \) is given by:
\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{e^{-|h|} - 1}{h} \]
Since \( h \) is positive, \( |h| = h \):
\[ f'(0^+) = \lim_{h \to 0^+} \frac{e^{-h} - 1}{h} \]
Using the limit definition of the derivative:
\[ f'(0^+) = -1 \quad \text{(as } h \to 0\text{)} \]

**3. Compare Left-hand and Right-hand Derivatives**:
Since:
\[ f'(0^-) = 1 \quad \text{and} \quad f'(0^+) = -1 \]
The left-hand and right-hand derivatives are not equal, thus the function is not differentiable at \( x = 0 \).

### Conclusion
- The function \( y = e^{-|x|} \) is continuous at \( x = 0 \).
- The function is not differentiable at \( x = 0 \).