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

1. brandonloves

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

yes.

3. brandonloves

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

4. brandonloves

and -12 +x^2

yup! :)

6. brandonloves

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

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

8. brandonloves

like 16 -(4-x)

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

10. brandonloves

oh shoot, I forgot the exponent

11. brandonloves

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

12. brandonloves

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

13. brandonloves

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

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

15. brandonloves

ohhh, that may fix it

yeah. something will cancel out.

17. brandonloves

yea, it left +8x -x^2

18. brandonloves

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

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

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

21. brandonloves

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

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

23. brandonloves

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

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

25. brandonloves

oh ok

26. brandonloves

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

27. brandonloves

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

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

29. brandonloves

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

30. brandonloves

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

31. brandonloves

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

33. brandonloves

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

they weren't needed? :)

35. brandonloves

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

36. brandonloves

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