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October 04, 2025, 07:52:21 am

Author Topic: BAH! Domains & ranges  (Read 13777 times)  Share 

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elaine

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BAH! Domains & ranges
« on: December 21, 2007, 03:38:28 pm »
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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 :)
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cara.mel

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Re: BAH! Domains & ranges
« Reply #1 on: December 21, 2007, 03:46:32 pm »
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 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 :(
« Last Edit: December 21, 2007, 04:31:40 pm by cara.mel »

Toothpaste

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Re: BAH! Domains & ranges
« Reply #2 on: December 21, 2007, 03:51:41 pm »
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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)

  • 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.

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

« Last Edit: March 14, 2010, 02:01:42 pm by Toothpaste »

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Re: BAH! Domains & ranges
« Reply #3 on: December 21, 2007, 03:59:33 pm »
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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

AppleXY

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Re: BAH! Domains & ranges
« Reply #4 on: December 21, 2007, 04:14:38 pm »
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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.
« Last Edit: December 21, 2007, 04:21:25 pm by AppleXY »

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Mao

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Re: BAH! Domains & ranges
« Reply #5 on: December 21, 2007, 04:24:22 pm »
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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!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
« Last Edit: December 21, 2007, 04:32:08 pm by Obsolete Chaos »
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cara.mel

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Re: BAH! Domains & ranges
« Reply #6 on: December 21, 2007, 04:28:33 pm »
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oh, so I've forgotten stuff since I did this and gotten it wrong? :(
Ignore my post then 0=)

yeah I have. sorry :(
« Last Edit: December 21, 2007, 04:32:09 pm by cara.mel »

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Re: BAH! Domains & ranges
« Reply #7 on: December 21, 2007, 04:40:37 pm »
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*whispers* i think we just overflowed elaine with too much information... =S

yeah I have. sorry :(
cheer up! no emo face! :D
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elaine

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Re: BAH! Domains & ranges
« Reply #8 on: December 21, 2007, 04:40:50 pm »
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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.
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Toothpaste

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Re: BAH! Domains & ranges
« Reply #9 on: December 21, 2007, 04:42:46 pm »
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I haven't finished! Hold on!

EDIT: Okay, I'm done now. :)
« Last Edit: December 21, 2007, 04:59:48 pm by Toothpick »

elaine

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Re: BAH! Domains & ranges
« Reply #10 on: December 21, 2007, 04:44:50 pm »
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*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.
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Mao

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Re: BAH! Domains & ranges
« Reply #11 on: December 21, 2007, 04:51:48 pm »
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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

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/0

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Re: BAH! Domains & ranges
« Reply #12 on: December 21, 2007, 05:18:01 pm »
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Awesome stuff, thanks  8)

Toothpaste

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Re: BAH! Domains & ranges
« Reply #13 on: December 21, 2007, 05:20:19 pm »
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Awesome stuff, thanks  8)
Did we indirectly answer your question too?

:)

elaine

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Re: BAH! Domains & ranges
« Reply #14 on: December 21, 2007, 05:27:51 pm »
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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!
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