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Author Topic: I suppose this would fall here...I need help. o-o  (Read 2562 times)

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Offline WolfyTopic starter

I suppose this would fall here...I need help. o-o
« on: October 11, 2010, 09:41:06 PM »
(If this doesn't belong here, please move it)

I need help..with math. >_< Functions and finding domains of said functions...if anyone could help, I'd be greatly appreciative.

Offline Jude

Re: I suppose this would fall here...I need help. o-o
« Reply #1 on: October 11, 2010, 09:59:06 PM »
A relation is a mathematical object that takes some sort of input and returns some sort of output.  The domain is what we call the input and the range is what we call the output.  Some relations:

Relation S takes in a letter and puts out the next letter in the alphabet.  For example S(q) = r.  The domain here is letters other than z, because there is no next alphabetical letter after z, therefore the function is undefined for z.  The range is all letters but a, and the reason a is included is because no alphabetical number put into the function can give us a because there is no letter before a.

Relation W takes in a number and puts out 1.  For example W(17000) = 1.  The domain is all numbers.  The range is simply 1 because for every number put into W we always get 1.

Relation Q takes in a number and outputs a number that is equal to the original number when squared.  For example, Q(4) = 2 or -2, because both of those squared become 4.  The domain becomes all numbers for which there is a square of (which is only positive numbers unless we're working with a set of numbers that allows for complex numbers), and the range is either all reals or complex numbers (depending on what we want to go with -- without complex numbers we can only make positive squares).

Generally speaking when you're working with a function that is complicated you'll want to specify sets the function is working between.  The domain and range will then be subsets of the sets the function moves between.  You'll notice I didn't really do that above, which made describing the domain and range of Q very problematic.  Q itself seems rather strange for another reason, which I will go into here:

You'll notice that in Q there are 2 values which can be chosen, whereas all of the others results in only 1 answer for each input.  This iswhat makes S and W a function where Q is only a relation.  Formally the definition of a function is:

f is a function if and only if for every f(x) there exists a unique y such that f(x)=y

This means essentially that you can't put in the same number twice and get a different answer.  When you're looking at a graph of a function, you can do a visual test of whether or not the graph represents a function or a relation called the vertical line test.  To do this, you simply hold a ruler to the graph vertically and then move the ruler from left to right (or right to left) over the graph.  If at any point the ruler touches the graph in two places, then it is not a function (because that would require there to be a 2 y's for one x).

Lets return to Q, the relation that isn't a function.  By applying the rigid definition we can see that because Q(4) = 2 or -2, it isn't a function (because we have 2 answers for the input of 4).  But what does this particular function actually appear to be when graphed?  Well, our value y squared (2 and -2) has to be equal to x, giving us the equation y*y = x, or two functional halves y = sqrt(x) and y = -sqrt(x).  So it's two graphs that are are mirrored by the x axis, meaning no matter where you touch the graph that it is defined, you're gonna hit two points.  So it fails the vertical line test.
« Last Edit: October 11, 2010, 10:05:29 PM by Jude »

Offline WolfyTopic starter

Re: I suppose this would fall here...I need help. o-o
« Reply #2 on: October 11, 2010, 10:02:32 PM »
Thanks Jude...but that doesn't really help. I need to find f+g, f-g, f times g, and f/g. o-o...So..a wiki answer isn't going to help.

Offline Jude

Re: I suppose this would fall here...I need help. o-o
« Reply #3 on: October 11, 2010, 10:06:04 PM »
Thanks Jude...but that doesn't really help. I need to find f+g, f-g, f times g, and f/g. o-o...So..a wiki answer isn't going to help.
Ahahahaha, that was me, not a wiki, but I can help you with this:

f+g, f-g, f times g, and f/g are questions of the compositions of functions.  That means that f and g are functions with definitions and we're utilizing them together.  For example:

Lets say f(x) = x - 3 and g(x) = 17x.  f + g is simply x - 3 + 17x = 18x - 3.
« Last Edit: October 11, 2010, 10:07:41 PM by Jude »

Offline WolfyTopic starter

Re: I suppose this would fall here...I need help. o-o
« Reply #4 on: October 11, 2010, 10:07:33 PM »
Exactly..o-o..now I need to know how to solve it. D:

the problem is f(x)=5x^2-2x+2 and g(x)=x^3

and I'm supposed to find all of those compositions, plus the domains. D:

Offline Jude

Re: I suppose this would fall here...I need help. o-o
« Reply #5 on: October 11, 2010, 10:13:31 PM »
the problem is f(x)=5x^2-2x+2 and g(x)=x^3

and I'm supposed to find all of those compositions, plus the domains. D:
I'll do the hardest for you.

f(x)=5x^2-2x+2 and g(x)=x^3

f/g = [5x^2-2x+2]/[x^3]

The first thing that I notice is that we're dividing by x^3, which means if x^3 is equal to zero, we're going to have a problem.  So I ask myself, when is x^3 equal to zero; the answer (only answer actually, it's a triple-repeated root for the polynomial x^3) is x = zero.  So we automatically know that x = 0 is not in the domain.  Other than that, there are no problems, so all real numbers (or complex, depending on what your class is working with) would be the domain.

EDIT:  I would assume all real numbers.  That will also be the domain of every other function since there's no division problem that would preclude the use of a particular number.  Polynomials are nice like that; they can take in every real number, unless when they're composed in a way that makes them a rational function in which case the dividing polynomial will determine the domain restriction).
« Last Edit: October 11, 2010, 10:15:19 PM by Jude »

Offline WolfyTopic starter

Re: I suppose this would fall here...I need help. o-o
« Reply #6 on: October 11, 2010, 10:20:47 PM »
So...your saying that the domain would be (-infinity,Infinity)?

Or would it be (-infinity,1)U(1,Infinity)?

Since it can't be zero, I'm guessing the latter.

Also, I'm assuming for the rest, that you should just plug them in, thus making the first one, for example, 5x^2-2x+2+x^3?

I'm also guessing that if that's the case, then the domain would be (-infinity,infinity)

Offline Trieste

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Re: I suppose this would fall here...I need help. o-o
« Reply #7 on: October 11, 2010, 10:24:08 PM »
For compositions, you just plug them in. They've given you the easy format, so that it's easy to tell where f(x) and g(x) goes.

The proper notation for "all real numbers except for 0" is x =/= 0.

Offline Jude

Re: I suppose this would fall here...I need help. o-o
« Reply #8 on: October 11, 2010, 10:26:52 PM »
So...your saying that the domain would be (-infinity,Infinity)?

Or would it be (-infinity,1)U(1,Infinity)?

Since it can't be zero, I'm guessing the latter.

Also, I'm assuming for the rest, that you should just plug them in, thus making the first one, for example, 5x^2-2x+2+x^3?

I'm also guessing that if that's the case, then the domain would be (-infinity,infinity)
If the domain is all reals except x = 0, then we need to construct a set that includes all reals except for x = 0.  The problem with choosing (-infinity,1)U(1,Infinity) is that while it covers almost everything, we lose all of the values from 1 to 0; we only want to toss out zero.

Another wrong answer would be:  (-infinity,0]U[0,Infinity), which is actually just negative infinity to positive infinity.  It's wrong because the brackets mean it includes 0.  If we make it (-infinity,0)U(0,Infinity) with no brackets included, then we include all values except the endpoints.  Thus, we're in the clear and x = 0 isn't included.

A sidenote, in mathematics infinity has the following definition:  infinity is > all numbers.  Infinity itself isn't actually a number, but a concept, which is why you don't include it in the interval either, so you use a parenthesis on it too.

EDIT:  I suggest you stick with the set notation that you used; it seems to be the proper notation for the level of math you're taking (yes, the way we describe domain varies wildly depending on the subject matter and course level unfortunately).
« Last Edit: October 11, 2010, 10:28:40 PM by Jude »

Offline WolfyTopic starter

Re: I suppose this would fall here...I need help. o-o
« Reply #9 on: October 11, 2010, 10:29:16 PM »
For compositions, you just plug them in. They've given you the easy format, so that it's easy to tell where f(x) and g(x) goes.

The proper notation for "all real numbers except for 0" is x =/= 0.

I know that's the proper notation...but...I've given the option that it gives me to pick out the answer. O-o

Buuuut....The multiplication one has me slighltly confused...though I'd take a stab and say the answer is 5x^5-2x^4+2x^3?

And...would that one be all real numbers as Well? Hmm....

Offline Jude

Re: I suppose this would fall here...I need help. o-o
« Reply #10 on: October 11, 2010, 10:31:51 PM »
f(x)=5x^2-2x+2 and g(x)=x^3

f*g = (5x^2-2x+2)(x^3) = 5x^5-2x^4+2x^3, and it's a polynomial so there are no problems with domain, so all reals works just fine.

When determining the domain of the function you need to glance at it and ask yourself "is there any number that, when plugged in, will result in a value which cannot be determined."  Whenever putting in x as a particular value makes you divide by zero, take the square root of a negative number in a real-number only set, or something like that, then that particular value isn't in the domain.
« Last Edit: October 11, 2010, 10:34:23 PM by Jude »

Offline WolfyTopic starter

Re: I suppose this would fall here...I need help. o-o
« Reply #11 on: October 11, 2010, 10:33:18 PM »
f(x)=5x^2-2x+2 and g(x)=x^3

f*g = (5x^2-2x+2)(x^3) = 5x^5-2x^4+2x^3

So I was right then. :D

So then, wouldn't every domain but f/g be (-infinity,infinity) ?
« Last Edit: October 11, 2010, 10:34:27 PM by Wolfy »

Offline Jude

Re: I suppose this would fall here...I need help. o-o
« Reply #12 on: October 11, 2010, 10:35:24 PM »
Composition of functions -- is this precalc/college algebra, algebra II, or early in a calc class?  Given the time of year, I'd guess College Alg?

Offline WolfyTopic starter

Re: I suppose this would fall here...I need help. o-o
« Reply #13 on: October 11, 2010, 10:37:08 PM »
Yes, College Alg. o3o

Offline Trieste

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Re: I suppose this would fall here...I need help. o-o
« Reply #14 on: October 11, 2010, 10:38:15 PM »
I know that's the proper notation...but...I've given the option that it gives me to pick out the answer. O-o

Whoops, sorry Wolfy. The only time I've ever run into set notation is on nitpicky online homework; most professors thus far have been fine with x=/=0 and leaving it at that.

:)

Offline WolfyTopic starter

Re: I suppose this would fall here...I need help. o-o
« Reply #15 on: October 11, 2010, 10:40:35 PM »
Whoops, sorry Wolfy. The only time I've ever run into set notation is on nitpicky online homework; most professors thus far have been fine with x=/=0 and leaving it at that.

:)

Indeed..Online homework. XP

Also, if anyone knows a good site that can be used as a Graphing Calculator, I'd much appreciate it...part of my problem with this class is that I don't have the $100 for the fricken required calculator. >_<

Offline Jude

Re: I suppose this would fall here...I need help. o-o
« Reply #16 on: October 11, 2010, 10:40:50 PM »
Now in exchange for the assistance I gave you, you will watch this:  Antoine Dodson (RAW footage)

http://www.coolmath.com/graphit/ - for your graphing calculator needs.

Offline WolfyTopic starter

Re: I suppose this would fall here...I need help. o-o
« Reply #17 on: October 11, 2010, 10:43:27 PM »
I've tried that site..and as far as I know, it can't do circles. D:...though then again, perhaps I just didn't type it in right.

Offline Jude

Re: I suppose this would fall here...I need help. o-o
« Reply #18 on: October 11, 2010, 10:45:20 PM »
You could try writing the circle as two functional halves instead?  A lot of graphical calculators can't do relations, so, yeaaah.

Offline WolfyTopic starter

Re: I suppose this would fall here...I need help. o-o
« Reply #19 on: October 11, 2010, 10:46:01 PM »
Problem:

f(x)=2x+3
g(x)=x^3
f times g(-4)

So then...

first:

-4^3, so -64

then 2(-64)+3


so f times g(-4)=-125?

Offline Jude

Re: I suppose this would fall here...I need help. o-o
« Reply #20 on: October 11, 2010, 10:49:25 PM »
f(x)=2x+3
g(x)=x^3
f times g(-4)
Man, that's a strange question.  I've never seen composition questions where they have a number plugged in... Lets assume that they want x = -4 for both functions instead of f(x)*g(-4) (which is equal to something else entirely).

f(-4) = -5, g(-4) = -64.  So, f*g(-4) = -5*64, which is like... off the top of my head, 320?

If it was the other way around it would be f(x)*g(-4) = (2x+3)*-64 = -128x - 192

EDIT:  To be more specific, the error you made is that you did f o g(-4) [that is, f composed with g, which is substituting g(-4) in for the x in f(x)] not f times g(-4).

EDIT2:  I missed a negative sign.
« Last Edit: October 11, 2010, 10:54:30 PM by Jude »

Offline WolfyTopic starter

Re: I suppose this would fall here...I need help. o-o
« Reply #21 on: October 11, 2010, 10:52:39 PM »
lol...I can't check which way they want it done, so I'm not sure myself. xD

But...I remember our professor writing down f*g can be translated into [f(g(x))]

or something similar to that. O-o

Offline Jude

Re: I suppose this would fall here...I need help. o-o
« Reply #22 on: October 11, 2010, 10:57:50 PM »
Your professor may be using different notation than I was expecting.

Because if according to him f*g = [f(g(x))], that means f*g = f o g.  (the o is called simply called circle when we're speaking outloud).

In that case, you would've done it right, and it explains why the way I did it is... simply mindnumbing and obtuse.  I'd ask him for clarification on the notation, because that's weiiiird.

Offline WolfyTopic starter

Re: I suppose this would fall here...I need help. o-o
« Reply #23 on: October 11, 2010, 10:59:46 PM »
Well...O-o...I suppose I should have mentioned that it does use an open circle.

I assumed that was a multiplication symbol, though. Lol.

Offline Jude

Re: I suppose this would fall here...I need help. o-o
« Reply #24 on: October 11, 2010, 11:03:41 PM »
Nah, f times g, and f circle g are different things.  The circle means composed, the times means you simply multiply the functions together.

f o g means that you take g and you put it inside of f, thus f o g = f[g(x)]

For example, if g = 2x, and f = 4x, then f o g = 4(2(x)) = 8x.  If you have a specific value of x to calculate for, then you can either find the general f o g then plug in the x after, or do it the way you did it in calculating g(that value) first.  Either is fine; I suggest doing the former, that way you'll get more algebraic practice for when you have to do it symbolically in Calculus and higher (which is quite often).

Composition of functions is absolutely essential to understand symbolically (with letters, not concrete numbers) for the first topic in Calculus, differentiation.