monroe17
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
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AndreaYoung1
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Have you searched the web? I havn't studied this yet /:
monroe17
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I have.. I can't find anything similar :/
monroe17
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@qpHalcy0n can you help me?
monroe17
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@jim_thompson5910 can you please help? :)
jim_thompson5910
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f(x) + g(x) literally means "add the functions f(x) and g(x)"
jim_thompson5910
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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)
monroe17
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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
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yeah it seems too simple that it looks like a trick question, but it's not
jim_thompson5910
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the domain is the set of allowable inputs, or x values
jim_thompson5910
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remember that you can't take the square root of a negative number, so
2+x >= 0
and
2-x >= 0
jim_thompson5910
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those two inequalities become
x >= -2
and
x <= 2
which combine to
-2 <= x <= 2
monroe17
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so, in interval notation.. it would be (-inf,-2)U(2,inf) ?
jim_thompson5910
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no, [-2, 2]
jim_thompson5910
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that represents \(\Large -2 \le x \le 2\)
monroe17
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oh okay... and would this be correct? f(x) - g(x) = sqrt(2+x) - sqrt(2-x)
jim_thompson5910
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yes that's 100% correct
jim_thompson5910
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the domain is the same because the radicands (the stuff in the square roots) are the same
monroe17
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f(x) * g(x) = sqrt(2+x) * sqrt(2-x) ?
jim_thompson5910
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domains are the same (for the same reason explained above)
jim_thompson5910
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domain is*
monroe17
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f(x)/g(x) = sqrt(2+x)/sqrt(2-x) and same domain?
jim_thompson5910
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the first part is correct, but the domain will be slightly different
jim_thompson5910
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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)
monroe17
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so now what is the difference between f+g and fog(x)?
jim_thompson5910
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f+g means "add f and g"
fog(x) means "start with f(x), and plug in g(x) as the input"
jim_thompson5910
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fog(x) is the same as f(g(x))
jim_thompson5910
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\[\Large f(x) = \sqrt{2+x}\]
\[\Large f(g(x)) = \sqrt{2+g(x)}\]
\[\Large f(g(x)) = \sqrt{2+\sqrt{2-x}}\]
monroe17
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well it's with different numbers, I just wanted to know the difference.
jim_thompson5910
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i gotcha, do you see how I'm getting that?
monroe17
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f(g(x))=2+sqrt(2-x)?
jim_thompson5910
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f(x) = sqrt(2+x) and g(x) = sqrt(2-x) correct?
monroe17
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yes.
jim_thompson5910
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Then
\[\Large f(g(x)) = \sqrt{2+\sqrt{2-x}}\]
or
\[\Large f\circ g(x) = \sqrt{2+\sqrt{2-x}}\]
monroe17
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can you draw that? cause its not showing up correct.,. and im a lil confused
jim_thompson5910
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|dw:1346027228896:dw|
jim_thompson5910
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oh my bad, there should be a closing parenthesis after g(x)
jim_thompson5910
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but everything else is correct
monroe17
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okay so if f(x)=1/x and g(x)= 9x+1 fog(x) would be... 1/(9x+1)?
jim_thompson5910
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yes perfect
monroe17
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and its domain would be all real numbers except ...?
jim_thompson5910
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what makes 9x+1 equal to zero?
monroe17
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anything less than zero?
jim_thompson5910
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9x +1 = 0
x = ???
monroe17
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-1/9
jim_thompson5910
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good, so x can be any number but -1/9
monroe17
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[-1/9,inf] ?
jim_thompson5910
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(-inf, -1/9) U (-1/9, inf)
jim_thompson5910
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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)
monroe17
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for the domain it it says.. its domain is all real numbers except __ so would I just put in -1/9?
jim_thompson5910
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that's in a more understandable way (in my opinion)
jim_thompson5910
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interval notation can be a little tricky to grasp
monroe17
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haha ya! gof(x) would be 9(1/x)+1?
jim_thompson5910
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yes, which becomes 9/x + 1
jim_thompson5910
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the 1 isn't part of the fraction
monroe17
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then to find the domain would I do x+1=0 x=-1?
jim_thompson5910
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no, the denominator is just x
jim_thompson5910
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\[\Large f \circ g(x) = \frac{9}{x}+1\]
monroe17
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oh its (9/x)+1?
monroe17
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so how would you do that domain then? would it be.. x cannot be zero?
jim_thompson5910
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yes, x can be any number but 0
monroe17
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would fof(x) be 1/(1/x).. so 1?
jim_thompson5910
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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
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So 1/(1/x) simplifies to x
jim_thompson5910
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We have to add the restriction that x can't be zero to make sure that the two expressions are completely equivalent
monroe17
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fof(x)=x and its domain is all real numbers except 0?
jim_thompson5910
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you nailed it
monroe17
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gog(x)=9(9x+1)+1=81x+9+1=81x+10?
jim_thompson5910
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perfect
jim_thompson5910
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domain: all real numbers
jim_thompson5910
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since the result is a polynomial
monroe17
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it says... its domain is (__,__)
monroe17
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-inf, inf?
jim_thompson5910
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yes it is
monroe17
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yyayyy! thank you!
jim_thompson5910
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you're welcome
monroe17
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one more thing :/ ?
jim_thompson5910
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what's that
monroe17
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How do I find the corresponding function values.
1. f(g(-2))
2. f(g(0))
jim_thompson5910
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1)
what is g(-2)??
monroe17
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that's the point at (2,-2)?
jim_thompson5910
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look at g(x) and not f(x)
monroe17
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(-2,2)?
jim_thompson5910
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you are correct, the point (-2,2) is on g(x)
jim_thompson5910
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so
f(g(-2)) = f(2)
monroe17
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so what would be the corresponding function value? how do I put that together?
monroe17
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cause I tried f(2) and f(1) for the next one and they're both wrong.
jim_thompson5910
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notice how (2,-2) is on f(x)
monroe17
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yes
jim_thompson5910
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so f(2) = -2
jim_thompson5910
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which means overall
f(g(-2)) = -2
jim_thompson5910
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this example may be a bit confusing with all the 2's...
jim_thompson5910
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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
monroe17
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so for the second one.. f(g(0)) (0,1)?
jim_thompson5910
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close
g(0) = 1
f(g(0)) = f(1)
f(g(0)) = -1
monroe17
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all right thank you!
jim_thompson5910
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sure thing