find indicated sum, difference, product or quotient: f(x)=16-x^2; g(x)=4-x
find (f+g)(x)

- anonymous

find indicated sum, difference, product or quotient: f(x)=16-x^2; g(x)=4-x
find (f+g)(x)

- katieb

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- anonymous

g(f(x)) = 4-(16-x^2) right so far?

- anonymous

yes.

- anonymous

so then it's 4 -16 +x^2

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- anonymous

and -12 +x^2

- anonymous

yup! :)

- anonymous

cool, now do I plug in that g(x) for f(x)= 16-x^2

- anonymous

quite horrible, but just sub 4-x into the x.

- anonymous

like 16 -(4-x)

- anonymous

\(16 -(4-x)^2\)

- anonymous

oh shoot, I forgot the exponent

- anonymous

16-(4-x)^2 is 16 (-4-x)(-4-x) which is 32 +8x+x^2

- anonymous

and then add that to the -12+x^2 from g(x)?

- anonymous

hmm, very close, but something's off isn't it?

- anonymous

16-(4-x)^2 is not 16 (-4-x)(-4-x)
it's 16 - (4-x)(4-x) though.

- anonymous

ohhh, that may fix it

- anonymous

yeah. something will cancel out.

- anonymous

yea, it left +8x -x^2

- anonymous

but I still don't have it exactly right. I should be getting -x^2 +x +12 I think, but I've got a -12

- anonymous

note that (f+g)(x)=16-x^2 +4-x

- anonymous

(f-g)(x)=16-x^2 -(4-x)

- anonymous

hmm, so what do I need to do to solve it now?

- anonymous

well, first, lol which are you doing? addition or subtraction?

- anonymous

well, it says f+g, so I would assume addition

- anonymous

so, simplify this one and you're done. (f+g)(x)=16-x^2 +4-x

- anonymous

oh ok

- anonymous

16-(12+x^2) +4-(12+x^2)? I think I screwed it up

- anonymous

or is it 16-(+8x -x^2) +4-(+8x -x^2) I'm not too sure of anything anymore

- anonymous

you don't need sub anything, just simplify as it's (f+g)(x) =f(x)+g(x)

- anonymous

oh, cuz I've already done all that. duh.

- anonymous

but when I add them together, I only get 20, the x^2's cancel out

- anonymous

I have an option for the answer to be -x^2-x+20

- anonymous

add 16-x^2 and 4-x only.

- anonymous

ok, well, thats it, obviously. but what happened to the 8x and +/-2x^2's? were those just wrong?

- anonymous

they weren't needed? :)

- anonymous

jesus christ. haha. well ok. sorry for being such a pain in the retrice thanks for the help though

- anonymous

hmm, well thats an odd substitution for @$$.... whatever lol

- anonymous

lol you're welcome

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