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coolaidd Group Title

..

  • one year ago
  • one year ago

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  1. ParthKohli Group Title
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    Just subtract for finding (f - g) and add for finding (f + g).

    • one year ago
  2. tjones89 Group Title
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    (f-g)(x) = f(x) - g(x)

    • one year ago
  3. ParthKohli Group Title
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    \[\rm (f - g)(x) = (2x - 2)-(x + 4)\]\[\rm (f + g)(x) = (2x + 2) + (x + 4)\]

    • one year ago
  4. ParthKohli Group Title
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    Oh, then it gets a little difficult... you have to plug g(x) into f(x).

    • one year ago
  5. ParthKohli Group Title
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    For example, if you got \(\rm f(x) = x^2\) and \(\rm g(x) = x+1\), then\[\rm f(g(x)) = f(x + 1) = (x + 1)^2\]

    • one year ago
  6. coolaidd Group Title
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    oh..ok :/

    • one year ago
  7. coolaidd Group Title
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    soo what would this be?

    • one year ago
  8. bluebrandon Group Title
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    for f(g(x)) just replace all the x in 2x-2 with (x+4) and for g(f(x)) just replace all the x in x+4 with (2x-2)

    • one year ago
  9. jiteshmeghwal9 Group Title
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    \(x(f-g)\)=\(f(x)-g(x)\) now put the values of \(f(x)\) & \(g(x)\) & solve them.

    • one year ago
  10. nubeer Group Title
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    yup.. thats how its done.

    • one year ago
  11. coolaidd Group Title
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    thanks!

    • one year ago
  12. nubeer Group Title
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    welcome.

    • one year ago
  13. coolaidd Group Title
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    so would it then be 2x+8-2 8x-8?

    • one year ago
  14. nubeer Group Title
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    how u got 8x-8?

    • one year ago
  15. coolaidd Group Title
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    2x times +4 and-2 times +4

    • one year ago
  16. nubeer Group Title
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    noooo... (2x-2) +4 u have to just open the bracket.. if there is a positive or negative sign outside the bracket that means its addint or subtracting not multiplying. your expression will be like 2x-2 +4

    • one year ago
  17. coolaidd Group Title
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    so it would just be 2x-2 +4? i dont have to solve it?

    • one year ago
  18. nubeer Group Title
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    (2x-2)4 if it was written liek this then u had to multiplied.. and yes that would be your answer just simplify it 2x+2 would be your answer for gf(x)

    • one year ago
  19. coolaidd Group Title
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    thanks!

    • one year ago
  20. nubeer Group Title
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    welcome ..

    • one year ago
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