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Quertior

Newbie Joined: 30 Aug 2009
Posts: 12 Posted: 24 Sep 2009 08:04:23 pm    Post subject: My math teacher and I disagree about the domain of f(x) = (√(2-x))^2 . He says that it is (-∞, 2]. I say that it is all real numbers. His reasoning goes as follows: Think of it as a composition f(g(x)) of two functions, f(x) = x^2 and g(x) = √(2-x) . The domain of two functions' composition is limited by the most restrictive single function's domain. Therefore, f(g(x)) should retain the domain restriction of (-∞, 2] placed on g(x) . Also, think of a function such as (x-2)/(x-2). This function's domain is (real numbers) \ {2}, because that value makes the function undefined. The same logic applies to f(g(x)), because g(x) is undefined for any x > 2 . Additionally, TI-83, 84, and 86 calculators only graph the function up to x=2 . ---------------- Now my argument: Plug in an x-value outside the supposed (-∞, 2] domain, such as 3. (√(2-3))^2 = (√(-1))^2 = -1. This is a perfectly valid real numbered answer. The bottom line is, for any real or imaginary z, z^2 (is an element of) the real number set. √(2-x) can (and will) cross into the imaginary numbers, but when squared will become real again. Any real x plugged in and fully evaluated gives a real valued answer. Additionally, my TI-89 graphs the function across all real numbers. ----------------- So what ought we to make of this?
Graphmastur Joined: 25 Mar 2009
Posts: 360 Posted: 24 Sep 2009 08:26:25 pm    Post subject: Your teacher is right, sort of... Your are probably thinking this: if it is k(X) = (√(2-x))^2 Then the square root and the square cancel, which means your function is the same is 2-X. Interestingly enough when you graph it, it takes the square root first, and then the square. But you know mathematically, that it is the same thing as (X-2)^(1/2) power to the 2 power. So in other words, your graphing calculator says the domain is (-∞, 2], but mathematically, I think it should be all reals. Interesting. It just depends on which way you work it. But seeing that the order of operations has parenthesis, and then powers, I would say that the square would get rid of the square root first. Then you have X-2. But all in all, the domain is (-∞, 2]. Simple because If you where to do X=1, that would mean that it would be the √-1, which is i, which is squared, so it is equal to -1. so my guess is that the domain is (-∞, 1]. Oh the joy of math... EDIT: btw, your calculator probably uses limits, or figured it out, like I did.Last edited by Guest on 24 Sep 2009 08:27:24 pm; edited 1 time in total
Quertior

Newbie Joined: 30 Aug 2009
Posts: 12 Posted: 24 Sep 2009 08:37:11 pm    Post subject: Graphmastur wrote:So in other words, your graphing calculator says the domain is (-∞, 2], but mathematically, I think it should be all reals. Quote:But all in all, the domain is (-∞, 2]. Simple because If you where to do X=1, that would mean that it would be the √-1, which is i, which is squared, so it is equal to -1. so my guess is that the domain is (-∞, 1]. Sorry, I can't quite make sense of your wording. Can you please clarify? Thanks. Graphmastur Joined: 25 Mar 2009
Posts: 360 Posted: 24 Sep 2009 09:49:19 pm    Post subject: In other words, the domain is all reals. You are correct.
Builderboy2005 Joined: 19 Apr 2009
Posts: 51 Posted: 24 Sep 2009 10:56:12 pm    Post subject: You can actually get the calculator to graph it through a little trickery. Put this into Y= instead: real(√(2-x)^2 . and make sure you are in a+bi mode, or else the calculator treats taking the square root of a negative number as an error, which results in not graphing. The real( is a bit harder to explain. I think it might have something to do with the calculator not knowing that i^2 is not imaginary. Try storing i^2 into A, you will notice that A appears in the complex menu, under the memory menu, even though it has no complex portion. So a glitch and a mode change later, the calculator is agreeing with you! And i belive you are correct. Even the simple function of √(X) has a domain of All Real Numbers, as taking the square root of a negative number is not illigal, it just results in an imaginary number Although your teacher might be considering this illegal for your class, and since its his class, he would be right. If you ignore imaginary numbers, taking the square root of a negative number is just as bad as dividing by zero polarBody

Newbie Joined: 02 Sep 2009
Posts: 30 Posted: 25 Sep 2009 12:04:55 am    Post subject: I think they can both be right simultaneously since both explanations are based on legitimate assumptions, and choosing one assumption over the other is quite arbitrary. On the one hand, Quertior's teacher assumes that the range of sqrt(x) is only nonnegative reals and that the domain of x² is all reals. On the other hand, Quertior assumes that the range of sqrt(x) includes all imaginary numbers and that the domain of x² includes all imaginary numbers as well. These are both valid and self-consistent assumptions. I think this situation is similar to the difference between Euclidean and Non-Euclidean Geometries. In Euclidean Geometry, it is assumed that two straight, parallel lines remain the same distance apart at all points. In Non-Euclidean Geometries (there are more than one), it is assumed that the distance between these two lines will change over time. Obviously, Euclidean Geometry makes more intuitive sense, but nevertheless, these two different assumptions lead to two separate but self-consistent branches of mathematics.
Mapar007 Joined: 04 Oct 2008
Posts: 365 Posted: 25 Sep 2009 11:14:45 am    Post subject: I second you, Quertior. The final outcome will always be real, so it seems. Quote:√(2-x) can (and will) cross into the imaginary numbers, but when squared will become real again. Indeed, this kills your teacher's reasoning.Last edited by Guest on 25 Sep 2009 11:15:45 am; edited 1 time in total
polarBody

Newbie Joined: 02 Sep 2009
Posts: 30 Posted: 25 Sep 2009 11:37:16 am    Post subject: Mapar007 wrote: Quote:√(2-x) can (and will) cross into the imaginary numbers, but when squared will become real again. Indeed, this kills your teacher's reasoning. I disagree. The teacher makes the implicit assumption that the square root of a negative number is undefined, so the domain of sqrt(x) will only be nonnegative reals, and the range of sqrt(x) will not include imaginary numbers. Thus, √(2-x) will not cross into the imaginary numbers, by the teacher's reasoning. I think it's perfectly legitimate to assume that the square root of a negative number is either defined or undefined, as long as you are consistent through your chain of reasoning. Am I wrong? Is it incorrect to assume that the square root of a negative number is undefined?Last edited by Guest on 25 Sep 2009 11:40:37 am; edited 1 time in total
Mapar007 Joined: 04 Oct 2008
Posts: 365 Posted: 25 Sep 2009 11:56:47 am    Post subject: Sqrt and squaring neutralize each other.
polarBody

Newbie Joined: 02 Sep 2009
Posts: 30 Posted: 25 Sep 2009 12:18:24 pm    Post subject: True. But if you decide to cancel the two exponents of f(x) = (√(2-x))^2, then you are dealing with f(x)=2-x, which is no longer a composite function. In that case, the domain of f(x)=2-x is simply all reals. But if you decide not to cancel the exponents of f(x) = (√(2-x))^2, then in order to figure out the domain of this composite function, you must first make the decision of whether or not to include imaginary numbers in the range of f(x)=√(2-x). The decision to either include imaginary numbers or not is what produced the two different arguments listed in the original post, and I'm making the argument that this decision is arbitrary.Last edited by Guest on 25 Sep 2009 12:20:43 pm; edited 1 time in total
Mapar007 Joined: 04 Oct 2008
Posts: 365 Posted: 25 Sep 2009 12:23:11 pm    Post subject: Yeah, I see how you reason. "Common Sense" is my main argument polarBody

Newbie Joined: 02 Sep 2009
Posts: 30 Posted: 25 Sep 2009 01:35:52 pm    Post subject: I hear you. Unfortunately, sometimes mathematics is the art of splitting infinitely small hairs! FloppusMaximus Joined: 22 Aug 2008
Posts: 472 Posted: 25 Sep 2009 05:11:39 pm    Post subject: You can't just go around arbitrarily cancelling things. Saying you can "cancel" the square and the square root is tantamount to saying that the square root is defined for negative numbers - but according to the teacher's definition, it isn't. It really does come down to definitions. If you're using the "real square root" function (whose domain and range are each the set of non-negative reals), then your function f is a function from (-∞, 2] to [0, ∞). But if you're using the "complex square root" function - which is what I would assume in the absence of any other information - then f is a function from C to C.
Graphmastur Joined: 25 Mar 2009
Posts: 360 Posted: 25 Sep 2009 06:30:18 pm    Post subject: polarBody wrote:But if you decide not to cancel the exponents of f(x) = (√(2-x))^2, then in order to figure out the domain of this composite function, you must first make the decision of whether or not to include imaginary numbers in the range of f(x)=√(2-x). Yes, but the range is all the outputs of f(x). No one said you could stop half way, just because you see i. The range would be all reals, as would the domain. Not because of canceling, well that, but because of this: f(x) = (√(2-X))^2 f(5) = (√(2-5))^2 f(5) = (√(-3))^2 f(5) = (i√3)^2 f(5) = (i^2)((√3)^2) f(5) = -1(3) f(5) = -3 The domain and range are all real numbers.
FloppusMaximus Joined: 22 Aug 2008
Posts: 472 Posted: 25 Sep 2009 06:57:42 pm    Post subject: You know what, you're all wrong. f is clearly a function of the integers mod 17. The domain is the set {0, 1, 2, 3, 4, 6, 10, 11, 15}.
polarBody

Newbie Joined: 02 Sep 2009
Posts: 30 Posted: 25 Sep 2009 07:15:34 pm    Post subject: Graphmastur wrote: polarBody wrote:But if you decide not to cancel the exponents of f(x) = (√(2-x))^2, then in order to figure out the domain of this composite function, you must first make the decision of whether or not to include imaginary numbers in the range of f(x)=√(2-x). Yes, but the range is all the outputs of f(x). No one said you could stop half way, just because you see i. Who says that you have to recognize the existence of i? What law of mathematics states that you must define √(-1) as i? I am aware that you can define √(-1) as i, and there's some very interesting mathematics that stem from this definition, but that doesn't mean that you have to make that definition in all cases. You're right, Floppus (or is it Mr. Maximus); this is just a case of using different definitions of the square root function.Last edited by Guest on 25 Sep 2009 07:17:08 pm; edited 1 time in total
Graphmastur Joined: 25 Mar 2009
Posts: 360 Posted: 25 Sep 2009 09:02:29 pm    Post subject: You don't have to use I at all. You could just do it with square root of a negative, square it, and move on. The domain is all real numbers.
FloppusMaximus Joined: 22 Aug 2008
Posts: 472 Posted: 25 Sep 2009 10:14:11 pm    Post subject: If you're working in the real numbers, negative numbers have no square roots. Whether you call it i, j, or a suffusion of yellow, it is not a real number. And why do you keep saying the domain is all real numbers? Clearly, if you're allowing i, the domain includes all complex numbers.
polarBody

Newbie Joined: 02 Sep 2009
Posts: 30 Posted: 25 Sep 2009 10:44:53 pm    Post subject: Wow, FloppusMaximus. You're taking us ALL to school.
FloppusMaximus Joined: 22 Aug 2008
Posts: 472 Posted: 25 Sep 2009 11:13:07 pm    Post subject: Sorry, my brain got stuck in math geek mode. Display posts from previous: All Posts Oldest FirstNewest First
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