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

Question

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

anonymous · Answer

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

anonymous · Answer

yes.

anonymous · Answer

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

anonymous · Answer

and -12 +x^2

anonymous · Answer

yup! :)

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

like 16 -(4-x)

anonymous · Answer

$16 -(4-x)^2$

anonymous · Answer

oh shoot, I forgot the exponent

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

ohhh, that may fix it

anonymous · Answer

yeah. something will cancel out.

anonymous · Answer

yea, it left +8x -x^2

anonymous · Answer

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 · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

oh ok

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

they weren't needed? :)

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

lol you're welcome