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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: elaine on December 21, 2007, 03:38:28 pm

Title: BAH! Domains & ranges
Post by: elaine on December 21, 2007, 03:38:28 pm
This is driving me nuts. My tutor doesn't even know  >:(

What is the general rule for finding the domain and range of "composite functions" 

For example:

f: R R, f(x)= x^2 -4
g: R+ {0} R, g(x) =

a) Find domain and range of f g
Also, why doesn't g f exist??
________________________________________________________________________

Sorry about this, but I have no idea how to do this one as well:

f: {x:x 3} R, f(x)= 3 - x
g: R R, g(x)= x^2 - 1

Define a restirction of g* of g such that f g* is defined and find f g*
I don't even know what that means...what is the * for??

I know this is a big ask, so thank you so much in advance :)
Title: Re: BAH! Domains & ranges
Post by: cara.mel on December 21, 2007, 03:46:32 pm
 a) Domain = range of g(x) = (0,)
range = sub in domain as per usual. equation is x-4, so (-4, )

G of f doesnt exist as the range of f(x) is not a subset of the domain of g(x)

2nd question
the g* is just your book's way of indicating you have made a different function (might be a standard symbol, I dunno)
For f of g to be defined, again ran(g) must be subset of dom(f)
Domain of f = (-,3]
=> range of g* = (-,3]
=> domain of g* = [-2,2]

tell my if my working is right :(
Title: Re: BAH! Domains & ranges
Post by: Toothpaste on December 21, 2007, 03:51:41 pm
Existence of the composite function
A composite function will only exist if the range of the "second" function is equal to or is part of the domain of the first.

Remember:
for or to be defined the range of g domain of f
for or to be defined the range of f domain of g


i.e. "inner's" range belongs to the "outer's" domain.


Domains of composite functions
If the composite function exists:
New domain of the composite function = Domain of the inside function

Remember:
The domain of   = domain of
The domain of   = domain of

If you're asked to find the range of the composite function, you would use the new domain to find it. Since there isn't a set rule for finding the range like the domain.

Using your question as an example

{0}

= =

Domain
would have the same domain as   since that is the "inner" function.
So the domain of is {0}

Range
The range of can be found using the domain:  {0}
= , it's linear as you can see. We want the range between domain: . At 0, = -4.

So therefore the range is .

Proving existence 
For to exist, the range of f must be a subset "" of the domain of g.

It's easier at first to draw up a table similar to this (but prettier):
Code: [Select]
     domain        range
f(x)    R         [-4,infinite)
g(x)   R+U{0}     [0,infinite)


So for to exist, must be a subset of .

BUT it's not, since the [-4, -1] extra doesn't belong in :

. So it's undefined, i.e. doesn't exist.

I hope this is clear. If not, just tell me. (if anyone spots a mistake or something, tell me too)

Title: Re: BAH! Domains & ranges
Post by: /0 on December 21, 2007, 03:59:33 pm
This is a related question so I hope I can post it here? :p

For , does anyone have a proof that ? That's what the textbook says but I just can't get it into my thick head.

What I think is that it should be , which I think makes sense because the two 'checkpoints' goes through are and .

But every time I test it I get , against my intuition... it's really confusing!

Thanks
Title: Re: BAH! Domains & ranges
Post by: AppleXY on December 21, 2007, 04:14:38 pm
Think the functions of sets and Imagine the f and the g machine.

When a input value is given to the "f" set or the domain of f, it is transferred (think of it as "fed" over) to the "g" or the range of g which thus, brings its output.

But right now I don't have time cause I gotta go :( Hope you get it.
Title: Re: BAH! Domains & ranges
Post by: Mao on December 21, 2007, 04:24:22 pm
dom g[f(x)] = dom f(x)
this is true only when g[f(x)] exist
in which case:
ran f(x) is a subset of dom g(x)
hence for all value of x in dom f(x), the composite function g[f(x)] exist

since g[f(x)] is composite function that still has the independent variable x (which is restrained by dom f(x))
therefore dom g[f(x)] = dom f(x)

EDIT: (end of work syndrome = lazy to put in any formal notation xD)
EDIT 2: HOLY GOSH TOOTHPICK U SUPERWOMAN!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Title: Re: BAH! Domains & ranges
Post by: cara.mel on December 21, 2007, 04:28:33 pm
oh, so I've forgotten stuff since I did this and gotten it wrong? :(
Ignore my post then 0=)

yeah I have. sorry :(
Title: Re: BAH! Domains & ranges
Post by: Mao on December 21, 2007, 04:40:37 pm
*whispers* i think we just overflowed elaine with too much information... =S

yeah I have. sorry :(
cheer up! no emo face! :D
Title: Re: BAH! Domains & ranges
Post by: elaine on December 21, 2007, 04:40:50 pm
Existence of the composite function
A composite function will only exist if the range of the "second" function is equal to or is part of the domain of the first.

for gof or to be defined the range of f domain of g



i.e. "inner's" range belongs to the "outer's" domain.


Domains of composite functions
If the composite function exists:
New domain of the composite function = Domain of the inside function

The domain of   = domain of
The domain of   = domain of

If you're asked to find the range of the composite function, you would use the new domain to find it. Since there isn't a set rule for finding the range like the domain.

Using your question as an example

{0}

would have the same domain as   since that is the "inner" function.
So the domain of is {0}

For to exist, the range of f must be a subset "" of the domain of g.

It's easier at first to draw up a table similar to this (but prettier):
Code: [Select]
     domain      |  range
f(x)    R         [-4,infinite)
g(x)   R+U{0}     [0,infinite)

  • The range of is .
    The domain of is . (same as {0} )

So for to exist, must be a subset of .

BUT it's not, since the [-4, -1] extra doesn't belong in :

. So it's undefined, i.e. doesn't exist.




Hang on finishing this off.....



omg toothpick you rock. In fact, all of you do! Thanks so much guys! :) I'm just going to go and print it off now and try to process it. I might need to ask for clarification later though lol.
Title: Re: BAH! Domains & ranges
Post by: Toothpaste on December 21, 2007, 04:42:46 pm
I haven't finished! Hold on!

EDIT: Okay, I'm done now. :)
Title: Re: BAH! Domains & ranges
Post by: elaine on December 21, 2007, 04:44:50 pm
*whispers* i think we just overflowed elaine with too much information... =S

no no I need all the help I can get! lol i just need process time lol. which might take a while haha.
Title: Re: BAH! Domains & ranges
Post by: Mao on December 21, 2007, 04:51:48 pm
toothpick looked after q1
...
f: {x:x 3} R, f(x)= 3 - x
g: R R, g(x)= x^2 - 1

Define a restirction of g* of g such that f g* is defined and find f g*
I don't even know what that means...what is the * for??

first: for f[g(x)] to exist, ran g must be a subset of dom f
hence we're restricting so that the maximum value is 3. doing this graphically (easiest method),
restriction:
by doing this, we have set the restriction to g, and we can refer to it as g* (g* is essentially the same as g, except it has a constraint)
to find the function f[g*(x)]: simply substitute g* into f

Title: Re: BAH! Domains & ranges
Post by: /0 on December 21, 2007, 05:18:01 pm
Awesome stuff, thanks  8)
Title: Re: BAH! Domains & ranges
Post by: Toothpaste on December 21, 2007, 05:20:19 pm
Awesome stuff, thanks  8)
Did we indirectly answer your question too?

:)
Title: Re: BAH! Domains & ranges
Post by: elaine on December 21, 2007, 05:27:51 pm
I haven't finished! Hold on!

EDIT: Okay, I'm done now. :)

rightio, printing off the latest version :) and thanks obsolete as well for anwering q2  ;D
loving all the pretty colours toothpick!
Title: Re: BAH! Domains & ranges
Post by: Toothpaste on December 21, 2007, 05:33:13 pm
Oh and here's something simple I found in my bound book that might be useful:

= Replace x in the equation with the number 2.

= Replace x in the equation with the expression for .

= Replace x in the equation with the value of

= Replace x in the equation with the number 2.
Title: Re: BAH! Domains & ranges
Post by: /0 on December 21, 2007, 08:03:57 pm
Awesome stuff, thanks  8)
Did we indirectly answer your question too?

:)

Yeah, I think so
Title: Re: BAH! Domains & ranges
Post by: AppleXY on December 21, 2007, 08:15:56 pm
I might also add to the existance of a composite function with high-res images illustrating it :)
Title: Re: BAH! Domains & ranges
Post by: lanvins on December 23, 2007, 11:36:59 pm
How do u find the domain of the following?: A rectangular piece of cardboard has dimensions 20cm by 36cm. Four squares each x cm by x cm are cut from the corners. An open box is formed by folding up the flaps
Title: Re: BAH! Domains & ranges
Post by: phagist_ on December 23, 2007, 11:55:28 pm
well I'm not going to straight out tell you the answer =P

but... what are the rules (in terms of x) governing the length of each side (i.e from there look at the possible values x can take)
Title: Re: BAH! Domains & ranges
Post by: Mao on December 23, 2007, 11:57:57 pm
lol hehe i am :P

this is an "implied domain" question
as the equation is a model based on real life:
we cannot have 0 or negative width remaining, nor can we cut 0 or negative width

that constrains x to be (0,10)
Title: Re: BAH! Domains & ranges
Post by: lanvins on December 24, 2007, 12:00:06 am
i don't get why it can't be (0,18)
Title: Re: BAH! Domains & ranges
Post by: phagist_ on December 24, 2007, 12:08:55 am
ok, how'd you arrive at those set of values?

I got 20-2x as the equation to one side and 36-2x as the equation to the other.

From 20-2x it has to be greater than zero.. as you cannot have a negative length...

Title: Re: BAH! Domains & ranges
Post by: lanvins on December 24, 2007, 12:14:14 am
i got: 20-2x> 0,          36-2x>,                   x>0

which became:

-2x>-20,     -2x>-36,                   x>0

which became:
x<10,             x<18,                     x>0     
Title: Re: BAH! Domains & ranges
Post by: Collin Li on December 24, 2007, 12:18:56 am


You have to take the intersection (overlap) of all your inequalities. The result is: (0,10)

(The reason why you take the intersection of them all, is because you need to satisfy all those conditions for the model to work)
Title: Re: BAH! Domains & ranges
Post by: phagist_ on December 24, 2007, 12:21:11 am
ahh I see... well what happens when you sub values from (0,18) into 20-2x>0 ?

It works fine from (0,10) but once you exceed that, you hit negative territory - hence it cannot work.

Try drawing a diagram and visualize what happens has x increases... once it approaches 10 one side becomes very very small, and should it exceed it, the side would have a negative length - which is impossible.

Title: Re: BAH! Domains & ranges
Post by: lanvins on December 24, 2007, 12:28:09 am
This might be a dumb question but to phagist,um, how did you know to choose 20-2x>0 to sub into and not the others? thanks
Title: Re: BAH! Domains & ranges
Post by: Collin Li on December 24, 2007, 12:35:29 am
If you're going to use maths, do what I said. Find all the inequalities, and find the intersection of them all.



You have to take the intersection (overlap) of all your inequalities. The result is: (0,10)

(The reason why you take the intersection of them all, is because you need to satisfy all those conditions for the model to work)

However, this is the "commonsense" way to think of it:

Code: [Select]
_________
|_|     |_|
|         |
|_       _|
|_|_____|_|

If you have a rectangle: 20x36, then if the squares have a side length larger than 10, you'll run into troubles, because the squares will start to overlap each other. On the shorter side (the length 20 side), you'll have two squares of equal size that simply cannot be more than 10 cm each, or the squares will begin to overlap (10+10 = 20). What happens on the length 36 side does not matter, because that will allow a bigger square size, but we have already reasoned that any square size larger than 10 is incompatible with a rectangle that has a side length of 20.
Title: Re: BAH! Domains & ranges
Post by: phagist_ on December 24, 2007, 12:35:59 am
ok from your 2 equations 20-2x>0 and 36-2x>0 you solve both.

from 20-2x>0 you get x = (0,10) and from 36-2x>0 you get x = (0,18)
remember you have to satisfy both equations

so you know x = (0,10) satisfies both, what happens if we choose say, 13?

well it satisfies 36-2x>0, but hold on a sec it doesn't work with 20-2x>0 - it yields a negative answer, which we know it is impossible to have a negative length.

As I said draw it out and visualize what happens when x increases...

say x=9 (which satisfies both equations) you will now have a 2x18
you can see one side (the one governed by 20-2x>0) will reach zero before the other... anything after that (x=10,11,12,etc...) the side will have negative length (which not possible!) thus your rectangle is ruined.

as coblin said, you need to find the intersection of solutions so that there is a maximal domain which satisfies both equations.
Title: Re: BAH! Domains & ranges
Post by: Mao on December 24, 2007, 09:51:59 am
ok from your 2 equations 20-2x>0 and 36-2x>0 you solve both.

from 20-2x>0 you get x = (0,10) and from 36-2x>0 you get x = (0,18)
remember you have to satisfy both equations

so you know x = (0,10) satisfies both, what happens if we choose say, 13?

well it satisfies 36-2x>0, but hold on a sec it doesn't work with 20-2x>0 - it yields a negative answer, which we know it is impossible to have a negative length.

As I said draw it out and visualize what happens when x increases...

say x=9 (which satisfies both equations) you will now have a 2x18
you can see one side (the one governed by 20-2x>0) will reach zero before the other... anything after that (x=10,11,12,etc...) the side will have negative length (which not possible!) thus your rectangle is ruined.

as coblin said, you need to find the intersection of solutions so that there is a maximal domain which satisfies both equations.
Lol
Moral of the day: when asked for implied domain, always use the shorter side :D
Title: Re: BAH! Domains & ranges
Post by: starbuckscoffee on December 24, 2007, 12:22:21 pm
HELP! (from the essential maths methods book Ex 1H)

1. The dimension of an enclosure are shown (the diagram at the bottom). The perimeter of the enclosure is 160m.
a. Find a rule for the area, A m^2, of the enclosure in terms of x (i got -x^2 + 76x -528 but that's wrong)

b. State a suitable domain of the function A(x)

2. A cuboid tank is open at the top and the internal dimension of its base are x m and 2x m. The height is h m. The volume of the tank is V cubic metres and the volume is fixed. Let S m^2 denotes the internal surface area of the tank.
a. i. Find S in terms of x and h,  ii. V and x

Thanks, (ps i've spent the last 3 hrs trying 2 figure it out with no luck)   

Title: Re: BAH! Domains & ranges
Post by: /0 on December 24, 2007, 01:25:56 pm
EDIT: In the book the side length on the bottom is 20m. I am using that for my answer.

1.

a)

Area can be given by:

Also, the perimeter gives
                       

Solving for y gives:

And substituting this into gives:



b) From the diagram, we must have . We also must have , by so the domain of the Area function is

2.

a)
i)

ii) The volume of the tank is

Since we only need it to be in terms of V and x, we don't need h, so we solve for h.





Title: Re: BAH! Domains & ranges
Post by: phagist_ on December 24, 2007, 01:33:38 pm
Quesion 1
Well you know the perimeter is 160 and from the diagram you can find that in terms of x and y.

I got as 160=2x+2y.

Then find the Area of the said shape in terms of x and y:



Now re-arrange first equation in terms of y;


Sub that into the Area formula you came up with to get;
which after expanding comes to;



Question 1b
You know
Looking at the diagram and thus the implied tomain is

I'll do the other one when I got time, but gotta run now.

Title: Re: BAH! Domains & ranges
Post by: Mao on December 24, 2007, 02:31:15 pm
b) From the diagram, we must have . We also must have , by so the domain of the Area function is

It should be therefore :P
Title: Re: BAH! Domains & ranges
Post by: /0 on December 24, 2007, 08:36:57 pm
b) From the diagram, we must have . We also must have , by so the domain of the Area function is

It should be therefore :P

I decided to answer the question he referred to in the book.
Title: Re: BAH! Domains & ranges
Post by: droodles on March 08, 2008, 08:58:58 pm
im doing that enclosure question right now, god i really hate it