If h(x)=f(x)+f(-x), " then " h(x) has got and extreme value at a point where f'(x) is

Options

A. an even function
B. an odd function
C. zero
D. none of these

Explanation

To solve the problem, we need to analyze the function \( h(x) = f(x) + f(-x) \) and find the conditions under which it has an extreme value.

### Step-by-Step Solution:

1. **Define the Function:**
We start with the function given in the problem:
\[ h(x) = f(x) + f(-x) \]

2. **Differentiate \( h(x) \):**
To find the extreme values of \( h(x) \), we need to differentiate it with respect to \( x \):
\[ h'(x) = \frac{d}{dx}[f(x)] + \frac{d}{dx}[f(-x)] \]
Using the chain rule for the second term:
\[ h'(x) = f'(x) + f'(-x) \cdot (-1) = f'(x) - f'(-x) \]

3. **Set the Derivative to Zero:**
For \( h(x) \) to have an extreme value, we set the derivative equal to zero:
\[ h'(x) = f'(x) - f'(-x) = 0 \]
This implies:
\[ f'(x) = f'(-x) \]

4. **Interpret the Result:**
The equation \( f'(x) = f'(-x) \) indicates that the derivative \( f'(x) \) is equal when \( x \) is replaced by \( -x \). This is the definition of an even function. Therefore, we conclude that:
\[ f'(x) \text{ is an even function.} \]

### Conclusion:
Thus, \( h(x) \) has an extreme value at points where \( f'(x) \) is an even function.