if $f(x)=ln~x$ and $g(x)=9-x^2$, what is the domain of the $f(g(x))$?

Question

anonymous · Answer

the domain of $\ln(x)$ is $x>0$ so your job is to solve 
$$9-x^2>0$$

anonymous · Answer

$$x^2<9~~?$$

anonymous · Answer

or wait..

anonymous · Answer

the domain of $f(x)=ln~x$ is $x>0$
the domain of $g(x)=9-x^2$ is $-3<x<3$
so that would make the domain of $f(g(x))$ $0<x<3$

yes? :D

anonymous · Answer

@satellite73

anonymous · Answer

@Hero @ash2326

ash2326 · Answer

@yummydum Why do you say domain of g(x) is bounded?

anonymous · Answer

aand i'm confused all over again...

hero · Answer

@yummydum, ask yourself, "Would something like -18 work for the domain?".

hero · Answer

If so, then you need to expand your domain.

anonymous · Answer

okay the domain for g(x) is $x<\pm3$?

ash2326 · Answer

@yummydum There are no constraints for x in g(x), so the domain of g(x) should be?

anonymous · Answer

domain for f(x) 0<x
domain for g(x) x<3
domain for f(g(x)) 0<x<3 ....I'm right, yea?

anonymous · Answer

there is a constraint because 9-x^2 has to be greater than 0 ...so x has to be less than 3

ash2326 · Answer

Domain of g(x) is not correct, how do you arrive at this?

ash2326 · Answer

Is that mentioned in the question, g(x) >= 0 ?

anonymous · Answer

o.O ....the equation says $$9-x^2\large\color{red}{>0}$$

ash2326 · Answer

Nope, that's when you write f(g(x)
$$f(g(x)=\ln (9-x^2)$$
here, 9-x^2 >0
because logarithm can't have a negative no. or 0 inside it

anonymous · Answer

ok here we go
$$9-x^2>0$$ or $$(3-x)(3+x)>0$$

anonymous · Answer

the zeros are $-3$ and $3$ and since 
$$y=9-x^2$$ is a parabola that opens down, it is positive between the zeros and negative outsides of them

sirm3d · Answer

@yummydum  you found the answer, unfortunately you said it incorrectly with "the domain of g(x) is -3 < x < 3

anonymous · Answer

you want to know where it is positive, so you can take the log of the damn thing, so pick 
$$(-3,3)$$

anonymous · Answer

or if you prefer 
$$-3<x<3$$

anonymous · Answer

yea i know that part is wrong @sirm3d 
but the answer i said $0<x<3$ is correct...yes?

anonymous · Answer

no it is not correct

anonymous · Answer

x-x

sirm3d · Answer

as @satellite73 said, "your job is to solve $9-x^2>0$ "

anonymous · Answer

$x$ for example could be $-2$ because $9-(-2)^2=9-4=5$

anonymous · Answer

yes so x is less than 3

anonymous · Answer

yes?

ash2326 · Answer

@yummydum we have
$$9-x^2<0$$
$$(3-x)(3+x)<0$$
Let's draw number line|dw:1377057070109:dw|

anonymous · Answer

okay..

ash2326 · Answer

Let's choose a no. greater than 3, say 4
9-x^2
x=4
9-16=-7 obviously less than 0
So we know that x> 3 is not part of the solution
Let's choose a no. less than -3
say =-4
9-x^2
x=-4
9-16=-7
again this also not a part of the solution.
SO we have to check the region from -3 to 3.
Do you follow this?

anonymous · Answer

yes

anonymous · Answer

then what?

ash2326 · Answer

ok, check x =-2 and x=1
Check if they work?
9-x^2 >0

anonymous · Answer

yes they do

ash2326 · Answer

So that's your region or domain

anonymous · Answer

my choices are
$$x\le3$$$$|x|\le3$$$$|x|>3$$$$|x|<3$$$$0<x<3$$

ash2326 · Answer

ok if you have 
$$|x|>3$$
$$=> x> 3\ or -x >3$$
so
$$x>3 \ or\ x<-3$$
Is this match our answer?

anonymous · Answer

yes

ash2326 · Answer

oh really? try putting x>3 in 9-x^2. Is that greater then 0 ?

ash2326 · Answer

*than

anonymous · Answer

oh wait no...sorry got the signs mixed up

ash2326 · Answer

then choose the correct answer, and you should have a reason for doing so

anonymous · Answer

$$|x|<3$$ meaning x<3 and -x<3
-3<x<3
and thats the domain of 9-x^2

ash2326 · Answer

yes, correct. Did you understand ?

anonymous · Answer

yes but the ln part confuses me..i thought it meant that x cannot be 0

anonymous · Answer

or wait...nvm in this case it means 9-x^2 cannot be 0

ash2326 · Answer

yes, that's why only chose >0

hero · Answer

@ash, what happens if x = -18?

anonymous · Answer

i get it now, thanks @ash2326

ash2326 · Answer

9-x^2=> -315
obviously negative no. can't be inside log

hero · Answer

Yeah, I forgot you have to square afterwards, sorry.

hero · Answer

@yummydum, that's not nice. I was only trying to help.

ash2326 · Answer

@yummydum He was just trying to help, be polite and nice to other users.

anonymous · Answer

I mean stop as in thats confusing again..it took a while to register all that, wasn't trying to be rude

hero · Answer

I know I goofed up. Believe me, I'm cringing badly.

ash2326 · Answer

ha ha Anyways @yummydum It's clear now, isn't it?

anonymous · Answer

yea, thanks

anonymous · Answer

another question really quick:

$$\log_b\left({m\sqrt{n}\over p}\right)=\log_b m+{1\over2}\log_b n-\log_b p$$yes?

ash2326 · Answer

yes, right

anonymous · Answer

okay and with this question am i doing it right?$$2\sin^2\theta=\sin\theta$$$$\sin\theta(2\sin\theta-1)=0$$ $$sin\theta=0~~~~~~\sin\theta={1\over2}$$

ash2326 · Answer

yes

ash2326 · Answer

Anyways if you have other question, please post as a new one.

anonymous · Answer

thanks, will do