If f is any function, then `(1)/(2) [ f (x) + f(-x) ] ` is always

Options

A. even
B. odd
C. neither even nor odd
D. None of these

Answer

To solve the problem, we need to analyze the expression \( \frac{1}{2} [ f(x) + f(-x) ] \) and determine its nature (whether it is an even function, odd function, or neither).

### Step-by-Step Solution:

1. **Understanding the Function**:
We start with the expression \( g(x) = \frac{1}{2} [ f(x) + f(-x) ] \).

2. **Substituting \(-x\)**:
To check if \( g(x) \) is even or odd, we need to find \( g(-x) \):
\[ g(-x) = \frac{1}{2} [ f(-x) + f(-(-x)) ] = \frac{1}{2} [ f(-x) + f(x) ] \]

3. **Rearranging the Expression**:
Notice that \( f(-x) + f(x) \) is the same as \( f(x) + f(-x) \). Therefore, we can write:
\[ g(-x) = \frac{1}{2} [ f(-x) + f(x) ] = \frac{1}{2} [ f(x) + f(-x) ] = g(x) \]

4. **Conclusion about the Function**:
Since \( g(-x) = g(x) \), this means that \( g(x) \) is an even function.

### Final Answer:
Thus, \( \frac{1}{2} [ f(x) + f(-x) ] \) is always an even function.

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