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Mathematics 21 Online
OpenStudy (anonymous):

Maximal domain of g(x)... g(x) = (3) - (4/(x+2)) Is it x + 2 = 0.. x = - 2... Is that how we get the maximal domain??

OpenStudy (anonymous):

The maximal domain is the set of all x that allows your function to exist.

OpenStudy (anonymous):

So here, your function is not defined at the point x=-2 since there, the denominator is 0.

OpenStudy (anonymous):

so it can't equal to zero because it's over something

OpenStudy (anonymous):

oh okay thanks

OpenStudy (anonymous):

So your domain is the set of all x real except for x=0.

OpenStudy (anonymous):

i think real, as long as X not equal to 0

OpenStudy (anonymous):

x+2 not equal to zero

OpenStudy (anonymous):

so is the answer...? R / -2 ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

\[\mathbb{R} - \left\{ -2 \right\}\]

OpenStudy (anonymous):

Thanks :)

OpenStudy (anonymous):

np

OpenStudy (anonymous):

Also what is it's range? + it wants the inverse...

OpenStudy (anonymous):

is the range infinity to 3 ?

OpenStudy (anonymous):

negative infinity?

OpenStudy (anonymous):

u want this (-infinity,2),( 2, infinity)? but i never see a form like that, u just write R-{-2}

OpenStudy (anonymous):

Okay, do you know how to get the inverse?

OpenStudy (anonymous):

You can use the definition for the inverse. g is the inverse of f if \[f(g(x))=x=g(f(x))\]

OpenStudy (anonymous):

The inverse of your function can be found then as,\[f(g(x))=3-\frac{4}{g(x)+2}=x\]and solve for g(x).

OpenStudy (anonymous):

do we put the w hole g(x) under that 4 ?

OpenStudy (anonymous):

?

OpenStudy (anonymous):

oh sorry.. do we re arrange the g(x) to the other side or something?

OpenStudy (anonymous):

yeah, you solve for g(x)

OpenStudy (anonymous):

4 = x (g(x)+2)

OpenStudy (anonymous):

where'd you 3 go?

OpenStudy (anonymous):

oh lol, -1 = x (g(x) + 2)

OpenStudy (anonymous):

\[3-\frac{4}{g(x)+2}=x \rightarrow g(x)=\frac{4}{3-x}-2\]

OpenStudy (anonymous):

check, i'm being distracted by different people

OpenStudy (anonymous):

g(x) = 3- 4/x+2.... and inverse is 3 - 4/g(x) +2... can you just leave it like that?

OpenStudy (anonymous):

y = 3- (4/x+2) x = 3-(4/y+2) and then simplify

OpenStudy (anonymous):

Take your original function. Put g(x) in place of x and replace f(x) with x Solve (rearrange) so you have g(x) = blah blah

OpenStudy (anonymous):

how to simplify after... -1 = x(g(x) + 2) ?

OpenStudy (anonymous):

expand

OpenStudy (anonymous):

x(g(x)+2) = xg(x) + 2x

OpenStudy (anonymous):

subtract 2x from both sides

OpenStudy (anonymous):

divide by x

OpenStudy (anonymous):

but i think you've made a mistake

OpenStudy (anonymous):

I'll write it out

OpenStudy (anonymous):

it wants the inverse g^-1... of g(x) = (3) - (4/(x+2))

OpenStudy (anonymous):

y = 3- (4/x+2) x = 3-(4/y+2) x(y+2) = 3-4 x(y+2)=-1 y+2 = -1/x y = (-1/x) -2 can this make sense?

OpenStudy (anonymous):

Yeah, same thing, different notation.

OpenStudy (anonymous):

yeah it makes sense except the G^-1.... the inverse... it looks different.

OpenStudy (anonymous):

y = g^-1(x)

OpenStudy (anonymous):

I stuffed the g(x) calc. because I'm writing it out on this damn thing, but the procedure is correct

OpenStudy (anonymous):

=) Okay so should I just replace the y's with g(x).. as the answer?

OpenStudy (anonymous):

You should end up with\[g^{-1}(x)=2\frac{x-1}{3-x}\]

OpenStudy (anonymous):

no no ... don't just replace the y's with g(x)... Do what I said... Take your original function. Put g^-1(x) in place of x and replace f(x) with x Solve (rearrange) so you have g(x) = blah blah

OpenStudy (anonymous):

it looks difficult to do but okay :)

OpenStudy (anonymous):

You know what, there's a lot of confusion here. I'm going to write it out and scan.

OpenStudy (anonymous):

Thanks :D :D :D I'll wait

OpenStudy (anonymous):

Lokisan are you still there?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

scanning

OpenStudy (anonymous):

Ok alright

OpenStudy (anonymous):

OpenStudy (anonymous):

Ive reformatted my pc so ill down pdf reader now, thanks for the link.

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Lokisan, did you use the retro version of this site?

OpenStudy (anonymous):

?

OpenStudy (anonymous):

retro

OpenStudy (anonymous):

Yes, there was another version to this site

OpenStudy (anonymous):

That means you joined it after the changes have taken place

OpenStudy (anonymous):

Well, I think the retro was better, as you could type there realtime

OpenStudy (anonymous):

I mean, it was more like a black board, and we didn't have to wait for the whole typing to be over

OpenStudy (anonymous):

oh

OpenStudy (anonymous):

Further more we could erase typed things and we could also type inbetween others' typings

OpenStudy (anonymous):

yeah, this is a pain, and i can't use this site on my phone - not practical

OpenStudy (anonymous):

oh that's heaps better

OpenStudy (anonymous):

Would you mind visiting this one https://docs.google.com/document/d/1-_5IUkf1O4EFkBw-kDEZV1T8THk4lzrpmQEzVKtfHHo/edit?hl=en_GB#

OpenStudy (anonymous):

I get this: Google Docs has encountered a server error. If reloading the page doesn't help, please contact us. To discuss this or other issues, visit the Google Docs Help forum. To see the list of known problems, go to the Google Docs Known Issues page

OpenStudy (anonymous):

Just try it once again or try refreshing it

OpenStudy (anonymous):

That makes sense now! :D Thanks a lot and nice hand writing :)

OpenStudy (anonymous):

No probs

OpenStudy (anonymous):

Fan me!

OpenStudy (anonymous):

Okay, I'm a fan already :)

OpenStudy (anonymous):

Good :)

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