The function `f(x) = e^(|x|)` is

Options

A. continuous everywhere but not differentiable at `x = 0`
B. continuous and differentiable everywhere
C. not continuous at `x = 0`
D. None of the above

Explanation

To determine the continuity and differentiability of the function \( f(x) = e^{|x|} \), we will analyze the function step by step.

### Step 1: Define the function
The function \( f(x) = e^{|x|} \) can be expressed in piecewise form:
- For \( x \geq 0 \), \( f(x) = e^x \)
- For \( x < 0 \), \( f(x) = e^{-x} \)

### Step 2: Check for continuity at \( x = 0 \)
To check for continuity at \( x = 0 \), we need to verify if:
\[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) \]

1. Calculate \( f(0) \):
\[ f(0) = e^{|0|} = e^0 = 1 \]

2. Calculate \( \lim_{x \to 0^-} f(x) \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} e^{-x} = e^0 = 1 \]

3. Calculate \( \lim_{x \to 0^+} f(x) \):
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{x} = e^0 = 1 \]

Since all three values are equal:
\[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) = 1 \]
Thus, \( f(x) \) is continuous at \( x = 0 \).

### Step 3: Check for differentiability at \( x = 0 \)
To check differentiability at \( x = 0 \), we need to find the left-hand derivative and the right-hand derivative.

1. Calculate the left-hand derivative \( f'(0^-) \):
\[ f'(x) = \frac{d}{dx}(e^{-x}) = -e^{-x} \quad \text{for } x < 0 \]
Thus,
\[ f'(0^-) = -e^{0} = -1 \]

2. Calculate the right-hand derivative \( f'(0^+) \):
\[ f'(x) = \frac{d}{dx}(e^{x}) = e^{x} \quad \text{for } x \geq 0 \]
Thus,
\[ f'(0^+) = e^{0} = 1 \]

Since \( f'(0^-) \neq f'(0^+) \):
\[ -1 \neq 1 \]
Thus, \( f(x) \) is not differentiable at \( x = 0 \).

### Conclusion
The function \( f(x) = e^{|x|} \) is:
- Continuous for all \( x \in \mathbb{R} \)
- Not differentiable at \( x = 0 \)