anonymous
  • anonymous
find indicated sum, difference, product or quotient: f(x)=16-x^2; g(x)=4-x find (f+g)(x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
g(f(x)) = 4-(16-x^2) right so far?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
so then it's 4 -16 +x^2

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anonymous
  • anonymous
and -12 +x^2
anonymous
  • anonymous
yup! :)
anonymous
  • anonymous
cool, now do I plug in that g(x) for f(x)= 16-x^2
anonymous
  • anonymous
quite horrible, but just sub 4-x into the x.
anonymous
  • anonymous
like 16 -(4-x)
anonymous
  • anonymous
\(16 -(4-x)^2\)
anonymous
  • anonymous
oh shoot, I forgot the exponent
anonymous
  • anonymous
16-(4-x)^2 is 16 (-4-x)(-4-x) which is 32 +8x+x^2
anonymous
  • anonymous
and then add that to the -12+x^2 from g(x)?
anonymous
  • anonymous
hmm, very close, but something's off isn't it?
anonymous
  • anonymous
16-(4-x)^2 is not 16 (-4-x)(-4-x) it's 16 - (4-x)(4-x) though.
anonymous
  • anonymous
ohhh, that may fix it
anonymous
  • anonymous
yeah. something will cancel out.
anonymous
  • anonymous
yea, it left +8x -x^2
anonymous
  • 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
  • anonymous
note that (f+g)(x)=16-x^2 +4-x
anonymous
  • anonymous
(f-g)(x)=16-x^2 -(4-x)
anonymous
  • anonymous
hmm, so what do I need to do to solve it now?
anonymous
  • anonymous
well, first, lol which are you doing? addition or subtraction?
anonymous
  • anonymous
well, it says f+g, so I would assume addition
anonymous
  • anonymous
so, simplify this one and you're done. (f+g)(x)=16-x^2 +4-x
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
16-(12+x^2) +4-(12+x^2)? I think I screwed it up
anonymous
  • anonymous
or is it 16-(+8x -x^2) +4-(+8x -x^2) I'm not too sure of anything anymore
anonymous
  • anonymous
you don't need sub anything, just simplify as it's (f+g)(x) =f(x)+g(x)
anonymous
  • anonymous
oh, cuz I've already done all that. duh.
anonymous
  • anonymous
but when I add them together, I only get 20, the x^2's cancel out
anonymous
  • anonymous
I have an option for the answer to be -x^2-x+20
anonymous
  • anonymous
add 16-x^2 and 4-x only.
anonymous
  • anonymous
ok, well, thats it, obviously. but what happened to the 8x and +/-2x^2's? were those just wrong?
anonymous
  • anonymous
they weren't needed? :)
anonymous
  • anonymous
jesus christ. haha. well ok. sorry for being such a pain in the retrice thanks for the help though
anonymous
  • anonymous
hmm, well thats an odd substitution for @$$.... whatever lol
anonymous
  • anonymous
lol you're welcome

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