# Thread: Stating the largest value of a constant. Difficulty with trigno in functions.

1. So for [0,360], at the x-point of 270, it will pass the HLT, yes?
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].

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.
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:

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 an interval, which looks like [0,90], and consists of all the real numbers between and including 0 and 90; versus the notation for a point in 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?

2. Are you confused by the difference between the notation for an interval, which looks like [0,90], and consists of all the real numbers between and including 0 and 90; versus the notation for a point in the xy plane, which looks like (4,5), and indicates the point where x = 4 and y = 5?
Yes, this was what confused me initially. I did not know you were talking about intervals and assumed them to be points on the xy plane, not in the manner (x,y) because that would mean y was 90, but in the way that you were considering 0-90 or simply 90 and within the entire restriction [0,360]. My bad.

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?
Well, you just said earlier that we have to state the largest value of A, and I forgot that the question also asked this so I would have to say that [0,90] is better simply because it is a larger interval than [0,45] and has a larger continuum. Or that 90 is also written as pi/2 and 45 as pi/4 so pi/2, being the larger value of the two, is chosen.

Could this be it?

3. Well, you just said earlier that we have to state the largest value of A, and I forgot that the question also asked this so I would have to say that [0,90] is better simply because it is a larger interval than [0,45] and has a larger continuum.
There you go. That's the reason. [0,90] is the largest interval that looks like [0,A], over which g has an inverse.

4. There you go. That's the reason. [0,90] is the largest interval that looks like [0,A], over which g has an inverse.

Well, that's good to hear.
Didn't realize it was that simple but wouldn't have understood it if it weren't for your explanation about the interval and continuum.

I see you haven't overlooked the second part.

(p2):
Obtain an expression in terms of x, for g^-1(x).
Turns out that A isn't really needed for this as I was able to obtain the answer without having the need of A being pi/2. The answer to this question is: y=Sin^-1[(3-x)/2]

I'm certain that this is the answer but what do you think?

5. I'm certain that this is the answer but what do you think?
Looks good to me! I'd say you're done with this problem.

6. Looks good to me! I'd say you're done with this problem.
Great.

Ackbeet, thank you for your helping me understand this problem and not giving up at any time. Hopefully, you'll be able to help me solve problems I may (likely) have in the future.

7. You're very welcome. Have a good one!

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