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anonymous

  • 5 years ago

Find f. f ''(x) = 20x^3 + 12x^2 + 4 f(0) = 3 f(1) = 3 My Answer was x^5+x^4-2x+3 my answer was wrong, can someone tell me where I went wrong?

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  1. anonymous
    • 5 years ago
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    one sec

  2. myininaya
    • 5 years ago
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    f'(x)=20x^4/4+12x^3/3+4x+C=5x^4+4x^3+4x+C f(x)=5x^5/5+4x^4/4+4x^2/2+Cx+D=x^5+x^4+2x^2+Cx+D

  3. anonymous
    • 5 years ago
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    what math are you in Calc I or II

  4. anonymous
    • 5 years ago
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    1

  5. anonymous
    • 5 years ago
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    okay I used integration, I'm assuming that you know how?

  6. anonymous
    • 5 years ago
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    Yes

  7. myininaya
    • 5 years ago
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    f(0)=D=3 f(1)=1+1+2+C+3=3 so C=-4

  8. anonymous
    • 5 years ago
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    okay so take the integral of f double prime (c)

  9. anonymous
    • 5 years ago
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    i mean (x)

  10. anonymous
    • 5 years ago
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    so you get 20x^4/4 + 12x^3/3 + 4x

  11. anonymous
    • 5 years ago
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    then simplify and do it once more

  12. anonymous
    • 5 years ago
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    go it?

  13. anonymous
    • 5 years ago
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    Medal pleaser ^.^

  14. anonymous
    • 5 years ago
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    f(x)= x^5+x^4+C(x)+D

  15. anonymous
    • 5 years ago
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    Plug f(0) in for x and set it equal to 3 makes D= 3 Plug f(1) in for x and set it equal to 3 makes C= -2 plugging those in was wrong

  16. myininaya
    • 5 years ago
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    you are missing 2x^2 in f(x)

  17. anonymous
    • 5 years ago
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    x^5 + x^4 + 2x^2 + c

  18. anonymous
    • 5 years ago
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    myininaya solved it already

  19. anonymous
    • 5 years ago
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    yeah

  20. myininaya
    • 5 years ago
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    f(x)=x^5+x^4+2x^2+Cx+D

  21. anonymous
    • 5 years ago
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    Thanks all

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