Sure: if you draw one horizontal line consisting of y=g(270), it will only intersect the function once. However, this will not tell you whether g passes the HLT on the entire interval [0,360], because you would need to test y = g(269), y = g(268), y = g(267), etc., AND all the numbers in-between. You'd have to test the entire continuum of horizontal lines in-between y = 1 and y = 5, since the range corresponding to domain [0,360] is [1,5].So for [0,360], at the x-point of 270, it will pass the HLT, yes?

You keep using the language "45 passes the HLT". That doesn't parse mathematically. It would be like saying, "You rightly bird blue fly." The correct language is this:For the question that I posted and the figure in post#3, 90 would be better because it passes the HLT whereas 45 does not pass the HLT at all let alone being compared to 90.

Function g passes the HLT on the interval [0,45] (which it does, incidentally!) as well as on the interval [0,90] (which it does, incidentally!).

Are you confused by the difference between the notation for aninterval,which looks like [0,90], and consists of all the real numbers between and including 0 and 90; versus the notation for apointin the xy plane, which looks like (4,5), and indicates the point where x = 4 and y = 5? I've tried to be consistent, and in this thread, I've been using the interval notation much more than the point notation. It is confusing, because the open interval (0,90) (all the numbers between 0 and 90 but NOT including the endpoints) looks like the notation for a point. You have to use context to know whether someone is talking about the point (4,5) or the interval (4,5). Of course, if someone is talking about the point (5,4), then it's impossible to be confused, because you generally don't write intervals with the right-hand endpoint on the left! That would be too confusing.

Getting back to the problem: g is one-to-one on the interval [0,45], and g is one-to-one on the interval [0,90]. So why is A = 90 the best answer to your problem?