Domain of a function

[Quertior]

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 .
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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.
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So what ought we to make of this?

[Graphmastur]

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.

[Quertior]

[Graphmastur]

In other words, the domain is all reals. You are correct.

[Builderboy2005]

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]

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]

I second you, Quertior. The final outcome will always be real, so it seems.

[polarBody]

[Mapar007]

Sqrt and squaring neutralize each other.

[polarBody]

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.

[Mapar007]

Yeah, I see how you reason.

"Common Sense" is my main argument

[polarBody]

I hear you. Unfortunately, sometimes mathematics is the art of splitting infinitely small hairs!

[FloppusMaximus]

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]

[FloppusMaximus]

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]

[Graphmastur]

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]

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]

Wow, FloppusMaximus. You're taking us ALL to school.

[FloppusMaximus]

Sorry, my brain got stuck in math geek mode.