Question

A function does not have any x-intercepts. What might be true about its domain and range?

Answer 1

Answer:

Here's my two cents worth about this one:

This is a discontinuous function - and it passes the vetical line test because any vertical line would only cut the "graph" in one place. The graph would have an infinite number of disconnected "peaks" and "valleys" because the x values along the number line are continuous. In other words, all the rationals would have y values of 1 and all the irrationals would have values of 0. The graph is definitely non-continuous because there aren't any "intermediate"  y values - i.e., "y" is either 0 or 1 - and nothing in between !!   Basically, the graph would be just  "dots" at either y= 0 or at y =1. In a strange way though, this is an "even" function, because the negative of any positive rational would have a corresponding y value of 1 and the negative of any positive irrational would have a corresponding y value of 0. Thus, f(x) = f(-x). The y intercept would be  the "dot" at (0, 1) - because 0 is rational. The "x" intercepts would just be the "dots" where the irrational numbers are located on the number line because "f(x)" would be 0 at those points!!!  The domain is all real numbers, but there are onl two values for the "range" .....0 and 1 !!!

That's my "take" on this one !!!

Step-by-step explanation: