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: Understand the function
The function \( f(x) = e^{|x|} \) involves the absolute value of \( x \). The absolute value function \( |x| \) is defined as:
\[ |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} \]
Thus, we can express \( f(x) \) as:
\[ f(x) = \begin{cases} e^x & \text{if } x \geq 0 \\ e^{-x} & \text{if } x < 0 \end{cases} \]

### Step 2: Check for continuity
To check if \( f(x) \) is continuous, we need to evaluate the limits from both sides at \( x = 0 \):
- For \( x \geq 0 \):
\[ \lim_{x \to 0^+} f(x) = f(0) = e^0 = 1 \]
- For \( x < 0 \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} e^{-x} = e^0 = 1 \]

Since both limits equal \( f(0) \), we conclude that:
\[ \lim_{x \to 0} f(x) = f(0) = 1 \]
Thus, \( f(x) \) is continuous everywhere.

### Step 3: Check for differentiability
Next, we check the differentiability of \( f(x) \) at \( x = 0 \). We need to find the derivative from both sides:
- For \( x \geq 0 \):
\[ f'(x) = \frac{d}{dx}(e^x) = e^x \]
Thus,
\[ f'(0^+) = e^0 = 1 \]
- For \( x < 0 \):
\[ f'(x) = \frac{d}{dx}(e^{-x}) = -e^{-x} \]
Thus,
\[ f'(0^-) = -e^0 = -1 \]

Since \( f'(0^+) \neq f'(0^-) \) (1 is not equal to -1), the function \( f(x) \) is not differentiable at \( x = 0 \).

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

### Final Answer
The function \( f(x) = e^{|x|} \) is continuous everywhere but not differentiable at \( x = 0 \).

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