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

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

### Step 2: Checking Continuity
To check if \( f(x) \) is continuous, we need to verify if the limit of \( f(x) \) as \( x \) approaches 0 from both sides is equal to \( f(0) \).

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

2. Calculate the left-hand limit as \( x \) approaches 0:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} e^{-x} = e^0 = 1 \]

3. Calculate the right-hand limit as \( x \) approaches 0:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^x = e^0 = 1 \]

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

### Step 3: Checking Differentiability
To check if \( f(x) \) is differentiable at \( x = 0 \), we need to find the left-hand derivative and the right-hand derivative at that point.

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

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

Since the left-hand derivative and the right-hand derivative at \( x = 0 \) are not equal:
\[ \lim_{x \to 0^-} f'(x) = -1 \quad \text{and} \quad \lim_{x \to 0^+} f'(x) = 1 \]
Thus, \( f(x) \) 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 for all \( x \in \mathbb{R} \) and not differentiable at \( x = 0 \).

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