Consider the following statement
1. `f(x)=e^(x)`, where `xgt0`
2. `g(x)=|x-3|`
which of the above function is/are continuous?

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Answer

To determine the continuity of the functions \( f(x) = e^x \) and \( g(x) = |x - 3| \), let's analyze each function step by step.

### Step 1: Analyze the function \( f(x) = e^x \)

1. **Definition of Continuity**: A function is continuous at a point \( c \) if:
- \( f(c) \) is defined.
- \( \lim_{x \to c} f(x) \) exists.
- \( \lim_{x \to c} f(x) = f(c) \).

2. **Exponential Function**: The function \( f(x) = e^x \) is an exponential function. Exponential functions are continuous everywhere on their domain, which is all real numbers.

3. **Conclusion for \( f(x) \)**: Since \( e^x \) is continuous for all \( x \), it is certainly continuous for \( x > 0 \).

### Step 2: Analyze the function \( g(x) = |x - 3| \)

1. **Definition of Absolute Value**: The function \( g(x) = |x - 3| \) represents the absolute value of \( x - 3 \). This function is piecewise defined:
- For \( x < 3 \), \( g(x) = 3 - x \).
- For \( x \geq 3 \), \( g(x) = x - 3 \).

2. **Check Continuity at \( x = 3 \)**:
- **Left-hand limit**: \( \lim_{x \to 3^-} g(x) = \lim_{x \to 3^-} (3 - x) = 3 - 3 = 0 \).
- **Right-hand limit**: \( \lim_{x \to 3^+} g(x) = \lim_{x \to 3^+} (x - 3) = 3 - 3 = 0 \).
- **Value at the point**: \( g(3) = |3 - 3| = 0 \).

3. **Conclusion for \( g(x) \)**: Since both the left-hand limit and right-hand limit exist and are equal to \( g(3) \), the function \( g(x) = |x - 3| \) is continuous at \( x = 3 \) and is also continuous for all other values of \( x \).

### Final Conclusion

Both functions \( f(x) = e^x \) and \( g(x) = |x - 3| \) are continuous.

### Answer: Both functions are continuous.

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