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anonymous

  • 5 years ago

find: a) (f+g)(x) b) (f-g)(x) c) (fg)(x) d) (f/g)(x) what is the domain of f/g? f(x)=x/x+1 g(x)=x^3

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  1. anonymous
    • 5 years ago
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    Right, we'll try this out. Slowly.

  2. anonymous
    • 5 years ago
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    can you substitute (f+g) with the actual functions?

  3. anonymous
    • 5 years ago
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    a)x^4/(x+1) +1

  4. anonymous
    • 5 years ago
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    a: ((x/x+1)+(x^3))(x) Distribute the outside x. x^2/x+1 + x^4

  5. anonymous
    • 5 years ago
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    b) We're going to replace f and g again. (f-g)(x) ((x/x+1)-(x^3))(x) Distribute again. x^2/x+1 - x^4

  6. anonymous
    • 5 years ago
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    Can you handle part C now?

  7. anonymous
    • 5 years ago
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    Give it a try.

  8. anonymous
    • 5 years ago
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    excuse me mate is it like that (f+g)(x) should be f(x)+g(x)

  9. anonymous
    • 5 years ago
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    so it should be just x/(x+1)+x^3 for the question a

  10. anonymous
    • 5 years ago
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    Ohh haha. Yes. That does make sense. I didn't even think to consider she was saying f plus g OF x. I was just doing multiplication.

  11. anonymous
    • 5 years ago
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    No no, ignore me. That is right. What was I thinking.

  12. anonymous
    • 5 years ago
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    b) f(x) - g(x) Therefore x/(x+1) -x^3 (x-(x^3(x+1)))/(x+1) (x-x^4-x^3)/(x+1)

  13. anonymous
    • 5 years ago
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    c)f(g(x)) x^3/(x^3+1) because we are substituting x^3 in the function f(x)

  14. anonymous
    • 5 years ago
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    did u understand c jane

  15. anonymous
    • 5 years ago
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    d) f(x) / g(x) (x/(x+1))/x^3) 1/(x^2(x+1))

  16. anonymous
    • 5 years ago
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    domain for the question is all real values of x excluding 0 and -1, because if we put 0 or -1 for x then f(x)/g(x) would become infinity since we get a 0 in the dinominator

  17. anonymous
    • 5 years ago
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    that's all,its easy

  18. anonymous
    • 5 years ago
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    Yes easy

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