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anonymous

  • one year ago

someone please help medal award! Given f(x)=x^2-3 and g(x)=x+2, find the following (write in general form). a) f(g(x)) b) g(f(x)) c. f(f(x)) d. evaluate f(g(-5))

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  1. anonymous
    • one year ago
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    Okay so first things first. Put g(x) inside the f(x) equation.

  2. anonymous
    • one year ago
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    For a)

  3. anonymous
    • one year ago
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    would I have to multiply x^2-3 by x+2? for a?

  4. anonymous
    • one year ago
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    \[(x+2)^2 -3\]

  5. anonymous
    • one year ago
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    you replace the x from f(x) with the equation of g(x)

  6. anonymous
    • one year ago
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    oh i see I got x^2+4x+1

  7. anonymous
    • one year ago
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    Correct

  8. anonymous
    • one year ago
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    You do basically the opposite for B) You use the g(x) equation and replace the x with the f(x) equation

  9. anonymous
    • one year ago
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    but how would you set it up since theres a -3 and a 2?

  10. anonymous
    • one year ago
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    You replace X with (X^2 -3) +2

  11. anonymous
    • one year ago
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    +2 is from g(x) = x "+2"

  12. anonymous
    • one year ago
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    I got X^2-1

  13. anonymous
    • one year ago
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    and for f I believe it's set up as (x^2-3)-3

  14. anonymous
    • one year ago
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    for f(f(x))?

  15. anonymous
    • one year ago
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    the X^2-1 I got for problem b I simplified it, and I set up (x^2-3)+2 for problem c

  16. anonymous
    • one year ago
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    and for f I set it up as g(-5)=(-5+2)^2-3

  17. anonymous
    • one year ago
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    Yes for B that is correct, But for c) you use f(f(x)). You do not use g(x) in the question.

  18. anonymous
    • one year ago
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    so for c it would be (x^2-3)(x^2-3)?

  19. anonymous
    • one year ago
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    \[f(f(x)) = (x^2 - 3)^2-3\]

  20. anonymous
    • one year ago
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    and c factors out to x^4-6x^2+6 :)

  21. anonymous
    • one year ago
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    Correct! :)

  22. anonymous
    • one year ago
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    thanks so much :-) for d I got 6 I wasn't sure if I set it up correctly I set it up as (-5+2)^2-3

  23. anonymous
    • one year ago
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    That's what I got too, and np! :)

  24. anonymous
    • one year ago
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    thanks so much for all your help :D

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