suppose $f(x)$ is a function such that if $p < q$, $f(p) < f(q)$. which statement best describes…

Brief Explanations:

First, define the given function: the condition $p<q \implies f(p)<f(q)$ means $f(x)$ is a strictly increasing function.

• An even function satisfies $f(-x)=f(x)$. For a strictly increasing function, $f(-x)=f(x)$ would require $-x=x$ (since if $x>0$, $-x<x$ so $f(-x)<f(x)$, contradicting $f(-x)=f(x)$), which only holds for $x=0$, not all $x$.
• An odd function satisfies $f(-x)=-f(x)$. For a strictly increasing function, take $x>0$, so $-x<x$, then $f(-x)<f(x)$. For an odd function, this becomes $-f(x)<f(x)$, so $2f(x)>0 \implies f(x)>0$ for $x>0$, and $f(x)<0$ for $x<0$. But a strictly increasing odd function cannot exist globally: as $x\to-\infty$, $f(x)\to-\infty$, and as $x\to+\infty$, $f(x)\to+\infty$, but at $x=0$, $f(0)=0$ (from odd function property). However, for any $a>0$, $-a<0<a$, so $f(-a)<f(0)<f(a)$, which fits, but the core issue is that a strictly increasing function cannot be even, and cannot be odd because the symmetry of odd functions conflicts with strict monotonicity across all real numbers (the only function that is both strictly increasing and odd would have to violate the strict increase for negative and positive pairs consistently). Thus, $f(x)$ cannot be odd or even.

Answer:

f(x) cannot be odd or even.