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Math Help - Stating the largest value of a constant. Difficulty with trigno in functions.

  1. #1
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    Stating the largest value of a constant. Difficulty with trigno in functions.

    Greetings all.

    I have a simple question but which I am unable to do. It involves both functions and trignometry. The question is in 2 parts. I'll post the data, then the question.

    Data:
    function: g(x) = 3-2sinx
    condition: 0 degrees <or= x <or= A degrees.
    A is a constant.

    The question (p1):
    I am required to state the largest value of A for which g has an inverse.

    Normally, this wouldn't be a problem if it weren't for that sin value. I'm not sure we can apply the same inverse method of replacing x and y that we normally do, so I don't really know what to do here.

    (p2):
    Obtain an expression in terms of x, for g^-1(x).

    I can only do this once the value of A has been obtained.

    Any help would be appreciated.
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  2. #2
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    Firstly, you need to restrict the function for where 3-2\sin x is one to one. This will give you 'A'. Do you know the shape of this function?
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  3. #3
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    Thanks for the reply, pickslides.

    I'm not sure what you mean by restrict.

    I used degrees mode for the sin values and obtained the following values after plugging them in the equation:

    Sin(0) : 3-2(0) = 3
    Sin(90): 3-2(1) = 1
    Sin(180): 3-2(0) = 3
    Sin(270): 3-2(-1) = 5
    Sin(360): 3-2(0) = 3

    The least value being 1 and the highest being 5.
    Is this the restriction you meant: 1<=x<=5 ?

    Also, the shape coming from this is:



    The answer is pi/2 which is the largest value of A. Problem is, I don't know why it in particular is the largest value of A. Could you explain this to me?

    Thanks.
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  4. #4
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    Quote Originally Posted by SolCon View Post
    The answer is pi/2 which is the largest value of A. Problem is, I don't know why it in particular is the largest value of A. Could you explain this to me?
    I agree that x=\frac{\pi}{2} is the largest value. The reason for this being as you follow the function after this value it is no longer one-to-one.

    A function only has an inverse on the interval that is one-to-one.
    Last edited by pickslides; January 11th 2011 at 12:53 PM. Reason: No tags.
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  5. #5
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    Thanks. That clears a lot of things up.

    Just one last thing though. You have stated that after sin90, it is no longer one-to-one hence, this is the largest value for A. But what about sin270? Isn't it one-to-one also? Why do we take sin90 specifically but not sin270? Also, if the shape was such that sin0 were to give a one-to-one point, would that then have been the largest value for A?

    Just need to get this bit cleared up, otherwise you've helped me understand this much better now.
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  6. #6
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    A function is one-to-one, in a rough-and-ready way, if it passes a horizontal line test. That is, if you can draw any horizontal line whatsoever, and it intersects your function in at most one location, then you've got a one-to-one function. Note that in order to be a function, it had to pass the vertical line test (same as the horizontal line test, only with, oddly enough, vertical lines). If you look at your graph in post # 3, you might be able to see how this could determine your A.
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  7. #7
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    Ackbeet, thanks for the tip about the vertical line test.

    However, I still do not know why sin90 has been chosen as A and the largest value while sin270, which is also one-to-one like sin90, has not been selected. Also, how do I determine A if the vertical line test is being passed by all the sinx positions? A vertical line being drawn at any point will intersect the curve only once, making them all pass the vertical line test.
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  8. #8
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    The vertical line test is for determining if a given relation is a function. sin is already a function, so you don't need to worry about the vertical line test.

    A function is one-to-one if it passes the horizontal line test. If you look at your graph in post # 3, it obviously fails the horizontal line test. So you're asked to artificially restrict the domain of your function so that, on this new restricted domain, it is one-to-one (passes the horizontal line test). And, you're told that the left endpoint of this restricted domain must be 0. So you have to look at intervals of the form [0,A].

    Here's an example: on the interval from [180,360], your function fails the horizontal line test, because the line y = 3 intersects the function at both endpoints. But suppose you bring in the right endpoint a bit? You can see that [180,300] still fails, [180,280] fails. But finally, [180,270] succeeds. So, on that interval, your function is one-to-one.

    why sin90 has been chosen as A
    Technically, A=90, not sin(A)=90.

    sin270, which is also one-to-one like sin90
    This is nonsense. If you're talking about sin(270) as a value, it makes no sense to call it one-to-one, because the property of "one-to-one"-ness is only something you can say about functions. If you're talking about y = sin(270), then it's not one-to-one, because no horizontal line is one-to-one (it massively fails the horizontal line test!!)

    The reason you have to have A = 90, is because the problem already constrains you to have the left-hand endpoint at 0. If the left hand endpoint could be at, say, 90, then 270 would be A.

    Does that clear things up?
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  9. #9
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    Thanks for the reply.

    This is nonsense. If you're talking about sin(270) as a value, it makes no sense to call it one-to-one, because the property of "one-to-one"-ness is only something you can say about functions.
    My apologies, I think I should have been more clear. I was not talking about them being values but merely the x-points. I should have probably used 0pi, pi/2, pi, 3/2pi and 2pi instead of the sin 0-360 values for x.

    If you're talking about y = sin(270), then it's not one-to-one, because no horizontal line is one-to-one (it massively fails the horizontal line test!!)
    Okay, but how about simply being 270 within the restriction? Like you have said for A=90. Isn't the line one-to-one on both 90 and 270, seeing as, if we were to draw a horizontal line it would meet at only one point at both these intervals? Please just confirm this that whether both the 90 and 270 points are one-to-one or if only 90 is one-to-one. And if 90 is one-to-one, isn't 270 one-to-one also?

    The reason you have to have A = 90, is because the problem already constrains you to have the left-hand endpoint at 0. If the left hand endpoint could be at, say, 90, then 270 would be A.
    From what I understand, we have chosen 90 simply because it is the first one-to-one function in the restriction. Is that right?

    Does that clear things up?
    heh, I've always been weak in maths but it has to be done. It's getting more clear now and if you could just confirm my assumptions, all will be clear.
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  10. #10
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    Isn't the line one-to-one on both 90 and 270?
    Any function that doesn't have the same values at both 90 and 270 would be one-to-one "on both 90 and 270". But your question does not ask about one-to-one-ness at two points, but on an interval, which is a continuum of points: ALL the points from 0 to 90, say. So, when I claim that your function is one-to-one on the interval [0,90], I mean, using the horizontal line test, that if I lop off any portion of the function outside that interval, and only consider the function inside that interval, it passes the horizontal line test.

    Let me ask you these questions:

    1. Do you see why your function is one-to-one on [0,90]?
    2. Do you see why it is not one-to-one on [0,180]?
    3. Is it one-to-one on [0,45]?
    4. Why is [0,90] a better answer to your problem than [0,45]?
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  11. #11
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    Let me ask you these questions:

    1. Do you see why your function is one-to-one on [0,90]?
    2. Do you see why it is not one-to-one on [0,180]?
    3. Is it one-to-one on [0,45]?
    4. Why is [0,90] a better answer to your problem than [0,45]?
    1. It is one-to-one on [0,90] because it passes the horizontal line test within the restriction, i.e a horizontal line drawn at this point will only meet the curve once.

    2. It is not one-to-one on [0,180] because it does not pass the horizontal line test within the restriction, i.e a horizontal line drawn at this point meets the curve more than once.

    3. <Same as 2> for [0,45]

    4. It is better because a function only has an inverse on the interval that is one-to-one and the question states the 'A' must be a value for which there is an inverse. At [0,90], it is one-to-one while at [0,45] it is not.

    Any function that doesn't have the same values at both 90 and 270 would be one-to-one "on both 90 and 270". But your question does not ask about one-to-one-ness at two points, but on an interval, which is a continuum of points: ALL the points from 0 to 90, say. So, when I claim that your function is one-to-one on the interval [0,90], I mean, using the horizontal line test, that if I lop off any portion of the function outside that interval, and only consider the function inside that interval, it passes the horizontal line test.
    So if we apply the same thing to the interval [0,270] and if any part of the function outside that interval is removed, and we only consider the function inside that interval, will it not also pass the horizontal line test?

    Here's another figure:

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  12. #12
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    1. Correct.
    2. Correct. What's an example of a horizontal line drawn that intersects the graph twice?
    3. Incorrect. If a function passes the horizontal line test (hereby abbreviated HLT) on one interval, like [0,90], then it will pass the HLT on a subset, like [0,45].
    4. Since your answer for this question depended on your incorrect answer for 3, this answer is also incorrect.

    I'm going to include five plots for you. I've plotted your function on different intervals so you can see, perhaps, what's going on. In each case, I've drawn a representative horizontal line (y = 2 in each case), so you can see whether the function passes or does not pass the HLT. Hopefully this will clear things up for you. I've labeled each file by the interval over which I plotted the function.

    Stating the largest value of a constant. Difficulty with trigno in functions.-0-45.jpg
    Stating the largest value of a constant. Difficulty with trigno in functions.-0-90.jpg
    Stating the largest value of a constant. Difficulty with trigno in functions.-0-180.jpg
    Stating the largest value of a constant. Difficulty with trigno in functions.-0-270.jpg
    Stating the largest value of a constant. Difficulty with trigno in functions.-90-270.jpg

    So if we apply the same thing to the interval [0,270] and if any part of the function outside that interval is removed, and we only consider the function inside that interval, will it not also pass the HLT?
    No, it will not. It didn't pass the HLT on the interval [0,180], so why should it pass on an interval containing [0,180]? You can generalize here and say that if it doesn't pass the HLT on an interval, it will not pass the HLT on any interval containing the first interval.

    However, as you can see, I've included the plot of the function on the interval [90,270]. It does pass the HLT on that interval.

    So, let me repeat my question # 4:

    4. Why is [0,90] a better answer to your problem than [0,45]?
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  13. #13
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    Apologies for the late reply. Had some things to take care of.

    2. Correct. What's an example of a horizontal line drawn that intersects the graph twice?
    [0,225]. If we draw a h-line from, say, y=2, it will meet the curve at 2 points.

    3. Incorrect. If a function passes the horizontal line test (hereby abbreviated HLT) on one interval, like [0,90], then it will pass the HLT on a subset, like [0,45].
    Firstly, I thank you for the figures. They are quite useful here.
    I agree that from the first figure you've provided, [0,45] will pass the HLT. But what about the entire restriction 0-360? If we consider the 45 point of x in it, and draw a vertical line from there and draw a h-line where it meets the curve, it won't pass the HLT, will it seeing as there isn't any h-line there already? Below is a figure of what I mean:



    No, it will not. It didn't pass the HLT on the interval [0,180], so why should it pass on an interval containing [0,180]? You can generalize here and say that if it doesn't pass the HLT on an interval, it will not pass the HLT on any interval containing the first interval.

    However, as you can see, I've included the plot of the function on the interval [90,270]. It does pass the HLT on that interval.
    Alright, I think I'm getting it. But for the figures you've provided, there is already a h-line which is y=2. But what about the question that I've provided where there isn't any h-line to begin with? What do we do there?

    So, let me repeat my question # 4:

    4. Why is [0,90] a better answer to your problem than [0,45]?
    Well, my first guess would be that 3-2sin90 gives us a whole value, i.e 1 whereas 3-2sin45 gives us a decimal value, i.e 1.58..... Is this the reason? Because I can't think of any other.
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  14. #14
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    But what about the entire restriction 0-360? If we consider the 45 point of x in it, and draw a vertical line from there and draw a h-line where it meets the curve, it won't pass the HLT, will it seeing as there isn't any h-line there already?
    You are correct: it will not pass the HLT. Whether there is a horizontal line there already or not is completely irrelevant to whether the original function passes or does not pass the HLT. The graph you drew is a visual confirmation that the function fails the HLT on [0,360].

    But for the figures you've provided, there is already a h-line which is y=2. But what about the question that I've provided where there isn't any h-line to begin with? What do we do there?
    The idea behind my representative y = 2 line is that you should mentally move that horizontal line up and down the entire range of the function on each restriction of the domain. If it only ever intersects the function once over the entire range, then the function passes the HLT for that domain restriction. If, on the other hand, it ever intersects more than once, the function fails the HLT for that domain restriction.

    Well, my first guess would be that 3-2sin90 gives us a whole value, i.e 1 whereas 3-2sin45 gives us a decimal value, i.e 1.58..... Is this the reason? Because I can't think of any other.
    Unfortunately, this is entirely irrelevant. If you look back at the wording of the original problem, it says this, "I am required to state the largest value of A for which g has an inverse." - emphasis added. Does this trigger something in your mind concerning my question # 4?
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  15. #15
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    You are correct: it will not pass the HLT. Whether there is a horizontal line there already or not is completely irrelevant to whether the original function passes or does not pass the HLT. The graph you drew is a visual confirmation that the function fails the HLT on [0,360].
    So for [0,360], at the x-point of 270, it will pass the HLT, yes?

    The idea behind my representative y = 2 line is that you should mentally move that horizontal line up and down the entire range of the function on each restriction of the domain. If it only ever intersects the function once over the entire range, then the function passes the HLT for that domain restriction. If, on the other hand, it ever intersects more than once, the function fails the HLT for that domain restriction.
    Like I said, those figures helped clear things up.

    Unfortunately, this is entirely irrelevant. If you look back at the wording of the original problem, it says this, "I am required to state the largest value of A for which g has an inverse." - emphasis added. Does this trigger something in your mind concerning my question # 4?
    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've just stated this and have said that 45 will not pass the HLT, yet you said I was incorrect earlier. I'm thinking this was so because I had written [0,45] instead of just the x-position within the entire interval [0,360] (as I've done now).

    However, if you are talking about the figures you've posted, that is, [0,90] and [0,45], then the larger interval which is [0,90] would be taken. But my question was, in the first place, about the individual x-positions within the [0,360] restriction, not the intervals. Somewhere above I apologised for this mishap and stated that it would have been better if I had used pi values instead of what I did not initially recognize as intervals.

    In any case, I thank you for taking time into helping me with this. Your explanations have been invaluable.
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