- anonymous

Given that f(x)=sqrt{2 + x} and g(x)=sqrt{2 - x}, find formulas for the following functions, and their domains. In each case, enter the domain using interval notation.
(a) f+g= and its domain is

- schrodinger

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- anonymous

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

- anonymous

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

- anonymous

@qpHalcy0n can you help me?

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## More answers

- anonymous

@jim_thompson5910 can you please help? :)

- jim_thompson5910

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

- jim_thompson5910

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)

- anonymous

that's what I got.. But, I wasn't sure if f(x) + g(x) = sqrt(2+x) + sqrt(2-x) was correct or not.. how would I find the domain?

- jim_thompson5910

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

- jim_thompson5910

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

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

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

- jim_thompson5910

no, [-2, 2]

- jim_thompson5910

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

- anonymous

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

- jim_thompson5910

yes that's 100% correct

- jim_thompson5910

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

- anonymous

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

- jim_thompson5910

yes

- jim_thompson5910

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

- jim_thompson5910

domain is*

- anonymous

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

- jim_thompson5910

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

- jim_thompson5910

now we have to avoid dividing by zero, so 2-x can't be zero
2-x = 0
x = 2
so if x = 2, then 2-x is zero. So we toss out x = 2 from the domain
So the domain is now [-2, 2)

- anonymous

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

- jim_thompson5910

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

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

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

- jim_thompson5910

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

- anonymous

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

- jim_thompson5910

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

- anonymous

yes.

- jim_thompson5910

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

- anonymous

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

- jim_thompson5910

|dw:1346027228896:dw|

- jim_thompson5910

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

- jim_thompson5910

but everything else is correct

- anonymous

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

- jim_thompson5910

yes perfect

- anonymous

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

- jim_thompson5910

what makes 9x+1 equal to zero?

- anonymous

anything less than zero?

- jim_thompson5910

9x +1 = 0
x = ???

- anonymous

-1/9

- jim_thompson5910

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

- anonymous

[-1/9,inf] ?

- jim_thompson5910

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

- jim_thompson5910

start with (-inf, inf), which is the entire number line
then poke a hole at -1/9 to get (-inf, -1/9) U (-1/9, inf)

- anonymous

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

- jim_thompson5910

yes

- jim_thompson5910

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

- jim_thompson5910

interval notation can be a little tricky to grasp

- anonymous

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

- jim_thompson5910

yes, which becomes 9/x + 1

- jim_thompson5910

the 1 isn't part of the fraction

- anonymous

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

- jim_thompson5910

no, the denominator is just x

- jim_thompson5910

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

- anonymous

oh its (9/x)+1?

- jim_thompson5910

yes

- anonymous

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

- jim_thompson5910

yes, x can be any number but 0

- anonymous

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

- jim_thompson5910

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

- jim_thompson5910

So 1/(1/x) simplifies to x

- jim_thompson5910

We have to add the restriction that x can't be zero to make sure that the two expressions are completely equivalent

- anonymous

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

- jim_thompson5910

yes

- jim_thompson5910

you nailed it

- anonymous

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

- jim_thompson5910

perfect

- jim_thompson5910

domain: all real numbers

- jim_thompson5910

since the result is a polynomial

- anonymous

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

- anonymous

-inf, inf?

- jim_thompson5910

yes it is

- anonymous

yyayyy! thank you!

- jim_thompson5910

you're welcome

- anonymous

one more thing :/ ?

- jim_thompson5910

what's that

- anonymous

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

##### 1 Attachment

- jim_thompson5910

1)
what is g(-2)??

- anonymous

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

- jim_thompson5910

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

- anonymous

(-2,2)?

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

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

- anonymous

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

- jim_thompson5910

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

- anonymous

yes

- jim_thompson5910

so f(2) = -2

- jim_thompson5910

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

- jim_thompson5910

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

- jim_thompson5910

but the idea is you start with g(x), plug in x = -2
then you get some result, which you plug into f(x) to get your answer

- anonymous

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

- jim_thompson5910

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

- anonymous

all right thank you!

- jim_thompson5910

sure thing

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