To determine the nature of the function \( f(x) = (x^m + 9)^2 \), let's analyze whether it is even, odd, or neither for different values of \( m \).
### Definitions:
1. Even Function: A function \( f(x) \) is even if \( f(x) = f(-x) \) for all \( x \) in its domain.
2. Odd Function: A function \( f(x) \) is odd if \( f(x) = -f(-x) \) for all \( x \) in its domain.
### Analysis:
First, we substitute \( -x \) into the function \( f(x) \) to see how \( f(-x) \) looks:
[tex]\[ f(-x) = \left((-x)^m + 9\right)^2 \][/tex]
Next, we consider two cases based on whether \( m \) is even or odd.
#### Case 1: \( m \) is even
If \( m \) is even, then \( (-x)^m = x^m \). Substituting this into \( f(-x) \), we get:
[tex]\[ f(-x) = \left(x^m + 9\right)^2 \][/tex]
Comparing this with the original function \( f(x) \):
[tex]\[ f(x) = \left(x^m + 9\right)^2 \][/tex]
We can see that:
[tex]\[ f(-x) = f(x) \][/tex]
Thus, if \( m \) is even, \( f(x) \) is an even function.
#### Case 2: \( m \) is odd
If \( m \) is odd, then \( (-x)^m = -x^m \). Substituting this into \( f(-x) \), we get:
[tex]\[ f(-x) = \left(-x^m + 9\right)^2 \][/tex]
To compare this with the original function \( f(x) \), let's expand both expressions:
For \( f(x) \):
[tex]\[ f(x) = (x^m + 9)^2 \][/tex]
And for \( f(-x) \):
[tex]\[ f(-x) = (-x^m + 9)^2 \][/tex]
Squaring removes the sign difference, hence \( f(-x) \neq -f(x) \) and it does not satisfy \( f(-x) = f(x) \) either. Therefore, \( f(x) \) can neither be odd nor even in this case.
### Conclusion:
- \( f(x) \) is an even function for all even values of \( m \).
- \( f(x) \) is neither even nor odd for odd values of \( m \).
The correct statement about \( f(x) \) is:
[tex]\[ f(x) \text{ is an even function for all even values of } m. \][/tex]
