anonymous
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Have you searched the web? I havn't studied this yet /:
anonymous
  • anonymous
I have.. I can't find anything similar :/
anonymous
  • anonymous
@qpHalcy0n can you help me?

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anonymous
  • anonymous
@jim_thompson5910 can you please help? :)
jim_thompson5910
  • jim_thompson5910
f(x) + g(x) literally means "add the functions f(x) and g(x)"
jim_thompson5910
  • 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
  • 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
  • jim_thompson5910
yeah it seems too simple that it looks like a trick question, but it's not
jim_thompson5910
  • jim_thompson5910
the domain is the set of allowable inputs, or x values
jim_thompson5910
  • 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
  • jim_thompson5910
those two inequalities become x >= -2 and x <= 2 which combine to -2 <= x <= 2
anonymous
  • anonymous
so, in interval notation.. it would be (-inf,-2)U(2,inf) ?
jim_thompson5910
  • jim_thompson5910
no, [-2, 2]
jim_thompson5910
  • jim_thompson5910
that represents \(\Large -2 \le x \le 2\)
anonymous
  • anonymous
oh okay... and would this be correct? f(x) - g(x) = sqrt(2+x) - sqrt(2-x)
jim_thompson5910
  • jim_thompson5910
yes that's 100% correct
jim_thompson5910
  • jim_thompson5910
the domain is the same because the radicands (the stuff in the square roots) are the same
anonymous
  • anonymous
f(x) * g(x) = sqrt(2+x) * sqrt(2-x) ?
jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • jim_thompson5910
domains are the same (for the same reason explained above)
jim_thompson5910
  • jim_thompson5910
domain is*
anonymous
  • anonymous
f(x)/g(x) = sqrt(2+x)/sqrt(2-x) and same domain?
jim_thompson5910
  • jim_thompson5910
the first part is correct, but the domain will be slightly different
jim_thompson5910
  • 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
  • anonymous
so now what is the difference between f+g and fog(x)?
jim_thompson5910
  • 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
  • jim_thompson5910
fog(x) is the same as f(g(x))
jim_thompson5910
  • 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
  • anonymous
well it's with different numbers, I just wanted to know the difference.
jim_thompson5910
  • jim_thompson5910
i gotcha, do you see how I'm getting that?
anonymous
  • anonymous
f(g(x))=2+sqrt(2-x)?
jim_thompson5910
  • jim_thompson5910
f(x) = sqrt(2+x) and g(x) = sqrt(2-x) correct?
anonymous
  • anonymous
yes.
jim_thompson5910
  • jim_thompson5910
Then \[\Large f(g(x)) = \sqrt{2+\sqrt{2-x}}\] or \[\Large f\circ g(x) = \sqrt{2+\sqrt{2-x}}\]
anonymous
  • anonymous
can you draw that? cause its not showing up correct.,. and im a lil confused
jim_thompson5910
  • jim_thompson5910
|dw:1346027228896:dw|
jim_thompson5910
  • jim_thompson5910
oh my bad, there should be a closing parenthesis after g(x)
jim_thompson5910
  • jim_thompson5910
but everything else is correct
anonymous
  • anonymous
okay so if f(x)=1/x and g(x)= 9x+1 fog(x) would be... 1/(9x+1)?
jim_thompson5910
  • jim_thompson5910
yes perfect
anonymous
  • anonymous
and its domain would be all real numbers except ...?
jim_thompson5910
  • jim_thompson5910
what makes 9x+1 equal to zero?
anonymous
  • anonymous
anything less than zero?
jim_thompson5910
  • jim_thompson5910
9x +1 = 0 x = ???
anonymous
  • anonymous
-1/9
jim_thompson5910
  • jim_thompson5910
good, so x can be any number but -1/9
anonymous
  • anonymous
[-1/9,inf] ?
jim_thompson5910
  • jim_thompson5910
(-inf, -1/9) U (-1/9, inf)
jim_thompson5910
  • 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
  • anonymous
for the domain it it says.. its domain is all real numbers except __ so would I just put in -1/9?
jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • jim_thompson5910
that's in a more understandable way (in my opinion)
jim_thompson5910
  • jim_thompson5910
interval notation can be a little tricky to grasp
anonymous
  • anonymous
haha ya! gof(x) would be 9(1/x)+1?
jim_thompson5910
  • jim_thompson5910
yes, which becomes 9/x + 1
jim_thompson5910
  • jim_thompson5910
the 1 isn't part of the fraction
anonymous
  • anonymous
then to find the domain would I do x+1=0 x=-1?
jim_thompson5910
  • jim_thompson5910
no, the denominator is just x
jim_thompson5910
  • jim_thompson5910
\[\Large f \circ g(x) = \frac{9}{x}+1\]
anonymous
  • anonymous
oh its (9/x)+1?
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
so how would you do that domain then? would it be.. x cannot be zero?
jim_thompson5910
  • jim_thompson5910
yes, x can be any number but 0
anonymous
  • anonymous
would fof(x) be 1/(1/x).. so 1?
jim_thompson5910
  • 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
  • jim_thompson5910
So 1/(1/x) simplifies to x
jim_thompson5910
  • 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
  • anonymous
fof(x)=x and its domain is all real numbers except 0?
jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • jim_thompson5910
you nailed it
anonymous
  • anonymous
gog(x)=9(9x+1)+1=81x+9+1=81x+10?
jim_thompson5910
  • jim_thompson5910
perfect
jim_thompson5910
  • jim_thompson5910
domain: all real numbers
jim_thompson5910
  • jim_thompson5910
since the result is a polynomial
anonymous
  • anonymous
it says... its domain is (__,__)
anonymous
  • anonymous
-inf, inf?
jim_thompson5910
  • jim_thompson5910
yes it is
anonymous
  • anonymous
yyayyy! thank you!
jim_thompson5910
  • jim_thompson5910
you're welcome
anonymous
  • anonymous
one more thing :/ ?
jim_thompson5910
  • jim_thompson5910
what's that
anonymous
  • anonymous
How do I find the corresponding function values. 1. f(g(-2)) 2. f(g(0))
jim_thompson5910
  • jim_thompson5910
1) what is g(-2)??
anonymous
  • anonymous
that's the point at (2,-2)?
jim_thompson5910
  • jim_thompson5910
look at g(x) and not f(x)
anonymous
  • anonymous
(-2,2)?
jim_thompson5910
  • jim_thompson5910
you are correct, the point (-2,2) is on g(x)
jim_thompson5910
  • jim_thompson5910
so f(g(-2)) = f(2)
anonymous
  • anonymous
so what would be the corresponding function value? how do I put that together?
anonymous
  • anonymous
cause I tried f(2) and f(1) for the next one and they're both wrong.
jim_thompson5910
  • jim_thompson5910
notice how (2,-2) is on f(x)
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
so f(2) = -2
jim_thompson5910
  • jim_thompson5910
which means overall f(g(-2)) = -2
jim_thompson5910
  • jim_thompson5910
this example may be a bit confusing with all the 2's...
jim_thompson5910
  • 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
  • anonymous
so for the second one.. f(g(0)) (0,1)?
jim_thompson5910
  • jim_thompson5910
close g(0) = 1 f(g(0)) = f(1) f(g(0)) = -1
anonymous
  • anonymous
all right thank you!
jim_thompson5910
  • jim_thompson5910
sure thing

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