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Have you searched the web? I havn't studied this yet /:

I have.. I can't find anything similar :/

@qpHalcy0n can you help me?

@jim_thompson5910 can you please help? :)

f(x) + g(x) literally means "add the functions f(x) and g(x)"

but since f(x) = sqrt(2+x) and g(x) = sqrt(2-x), we know that
f(x) + g(x) = sqrt(2+x) + sqrt(2-x)

yeah it seems too simple that it looks like a trick question, but it's not

the domain is the set of allowable inputs, or x values

remember that you can't take the square root of a negative number, so
2+x >= 0
and
2-x >= 0

those two inequalities become
x >= -2
and
x <= 2
which combine to
-2 <= x <= 2

so, in interval notation.. it would be (-inf,-2)U(2,inf) ?

no, [-2, 2]

that represents \(\Large -2 \le x \le 2\)

oh okay... and would this be correct? f(x) - g(x) = sqrt(2+x) - sqrt(2-x)

yes that's 100% correct

the domain is the same because the radicands (the stuff in the square roots) are the same

f(x) * g(x) = sqrt(2+x) * sqrt(2-x) ?

yes

domains are the same (for the same reason explained above)

domain is*

f(x)/g(x) = sqrt(2+x)/sqrt(2-x) and same domain?

the first part is correct, but the domain will be slightly different

so now what is the difference between f+g and fog(x)?

f+g means "add f and g"
fog(x) means "start with f(x), and plug in g(x) as the input"

fog(x) is the same as f(g(x))

well it's with different numbers, I just wanted to know the difference.

i gotcha, do you see how I'm getting that?

f(g(x))=2+sqrt(2-x)?

f(x) = sqrt(2+x) and g(x) = sqrt(2-x) correct?

yes.

Then
\[\Large f(g(x)) = \sqrt{2+\sqrt{2-x}}\]
or
\[\Large f\circ g(x) = \sqrt{2+\sqrt{2-x}}\]

can you draw that? cause its not showing up correct.,. and im a lil confused

|dw:1346027228896:dw|

oh my bad, there should be a closing parenthesis after g(x)

but everything else is correct

okay so if f(x)=1/x and g(x)= 9x+1 fog(x) would be... 1/(9x+1)?

yes perfect

and its domain would be all real numbers except ...?

what makes 9x+1 equal to zero?

anything less than zero?

9x +1 = 0
x = ???

-1/9

good, so x can be any number but -1/9

[-1/9,inf] ?

(-inf, -1/9) U (-1/9, inf)

for the domain it it says.. its domain is all real numbers except __ so would I just put in -1/9?

yes

that's in a more understandable way (in my opinion)

interval notation can be a little tricky to grasp

haha ya! gof(x) would be 9(1/x)+1?

yes, which becomes 9/x + 1

the 1 isn't part of the fraction

then to find the domain would I do x+1=0 x=-1?

no, the denominator is just x

\[\Large f \circ g(x) = \frac{9}{x}+1\]

oh its (9/x)+1?

yes

so how would you do that domain then? would it be.. x cannot be zero?

yes, x can be any number but 0

would fof(x) be 1/(1/x).. so 1?

1/(1/x) = (1/1)/(1/x)
1/(1/x) = (1/1)*(x/1)
1/(1/x) = x/1
1/(1/x) = x

So 1/(1/x) simplifies to x

fof(x)=x and its domain is all real numbers except 0?

yes

you nailed it

gog(x)=9(9x+1)+1=81x+9+1=81x+10?

perfect

domain: all real numbers

since the result is a polynomial

it says... its domain is (__,__)

-inf, inf?

yes it is

yyayyy! thank you!

you're welcome

one more thing :/ ?

what's that

How do I find the corresponding function values.
1. f(g(-2))
2. f(g(0))

1)
what is g(-2)??

that's the point at (2,-2)?

look at g(x) and not f(x)

(-2,2)?

you are correct, the point (-2,2) is on g(x)

so
f(g(-2)) = f(2)

so what would be the corresponding function value? how do I put that together?

cause I tried f(2) and f(1) for the next one and they're both wrong.

notice how (2,-2) is on f(x)

yes

so f(2) = -2

which means overall
f(g(-2)) = -2

this example may be a bit confusing with all the 2's...

so for the second one.. f(g(0)) (0,1)?

close
g(0) = 1
f(g(0)) = f(1)
f(g(0)) = -1

all right thank you!

sure thing