Hi Tracy,

To answer this question you have to very carefully read the definitions of even and one-yo-one that appear in your textbook. The definitions I found on the web are below.

A real valued function f of a real variable is even if for each real number x, f(x) = f(-x).

A function f is one-to-one if for each a and b in the domain of f, if f(a) = f(b) then a = b.

Hence if f is an even function and for some number a, a and -a are both in the domain of f then f(a) = f(-a) and yet a ≠ -a and hence f is not one-to-one.

But what about the function f(x) = √x? The domain of f is the set of non-negative real numbers and from the definition above f(x) = √x is one-to-one.

Is it even? Suppose the definition of even read

A real valued function f of a real variable is even if for each real number x, if x and -x are in the domain of f then f(x) = f(-x).

In this case f(x) = √x is even since the only x for which x and -x are in the domain of f is x = 0.

Carefully read the definitions of even and one-yo-one that appear in your textbook. Is f(x) = √x?

Harley