Suppose [tex]$f(x)$[/tex] is a function such that if [tex]$p\ \textless \ q[/tex], [tex]f(p)\ \textless \ f(q)$[/tex]. Which statement best describes [tex][tex]$f(x)$[/tex][/tex]?

A. [tex]$f(x)$[/tex] can be odd or even.
B. [tex]$f(x)$[/tex] can be odd but cannot be even.
C. [tex][tex]$f(x)$[/tex][/tex] can be even but cannot be odd.
D. [tex]$f(x)$[/tex] cannot be odd or even.



Answer :

To analyze the given function [tex]\( f(x) \)[/tex], let's break down the information provided about the function and the properties of odd and even functions.

The statement says that if [tex]\( p < q \)[/tex], then [tex]\( f(p) < f(q) \)[/tex]. This suggests that the function [tex]\( f(x) \)[/tex] is strictly increasing. Now, let's examine what it means for a function to be odd or even:

1. Even Function:
A function [tex]\( f(x) \)[/tex] is even if [tex]\( f(-x) = f(x) \)[/tex] for all [tex]\( x \)[/tex] in its domain.

2. Odd Function:
A function [tex]\( f(x) \)[/tex] is odd if [tex]\( f(-x) = -f(x) \)[/tex] for all [tex]\( x \)[/tex] in its domain.

Let's consider how these properties interact with the given information about the function being strictly increasing.

### Analysis

- Strictly Increasing Function:
The property of being strictly increasing implies that the function continuously grows as the input [tex]\( x \)[/tex] increases. That is, for any [tex]\( p < q \)[/tex], [tex]\( f(p) < f(q) \)[/tex].

To understand whether such a function can be odd or even, let's consider possible scenarios:

1. Even Function:
- If [tex]\( f(x) \)[/tex] is even, then [tex]\( f(x) = f(-x) \)[/tex] for all [tex]\( x \)[/tex].
- However, if [tex]\( f(x) \)[/tex] is strictly increasing, [tex]\( f(x) \)[/tex] must be strictly different for different values of [tex]\( x \)[/tex]. Specifically, for any positive [tex]\( x \)[/tex], [tex]\( f(-x) \)[/tex] should be strictly less than [tex]\( f(x) \)[/tex] or strictly greater than [tex]\( f(x) \)[/tex], which contradicts [tex]\( f(x) = f(-x) \)[/tex].

2. Odd Function:
- If [tex]\( f(x) \)[/tex] is odd, then [tex]\( f(-x) = -f(x) \)[/tex] for all [tex]\( x \)[/tex].
- Imagine [tex]\( f(x) \)[/tex] has a strictly increasing nature. For negative values [tex]\( x \)[/tex] and [tex]\( -x \)[/tex], [tex]\( f(x) \)[/tex]'s strictly increasing property still holds, making [tex]\( f(-x) \)[/tex] incompatible with the nature of an increasing function if the function needs to cross zero at [tex]\( x = 0 \)[/tex].

### Conclusion

Neither of these properties (being even or odd) are compatible with a strictly increasing function because:

- An even function cannot maintain a strictly increasing nature across its entire domain.
- An odd function's symmetry around the origin disrupts a strict increasing relationship.

Therefore, the most fitting conclusion is that [tex]\( f(x) \)[/tex] cannot be odd or even.

### Answer

The best statement that describes [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{f(x) \text{ cannot be odd or even.}} \][/tex]