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giggles123

  • 3 years ago

let f(x)=ax^3+6x^2+bx+4. Determine the constant a and b such that f has a relative min at x=-1 and a rel max at x=2@Algebra

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  1. Hero
    • 3 years ago
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    I hate these

  2. moneybird
    • 3 years ago
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    first derivative of f(x) first

  3. giggles123
    • 3 years ago
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    I did that then what? I got F'(x)=3ax^2+12x+b

  4. myininaya
    • 3 years ago
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    \[f'(x)=3a^2+12x+b\] \[f'(-1)=0 ;f'(2)=0\]

  5. myininaya
    • 3 years ago
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    \[f'(x)=3ax^2+12x+b*\]

  6. Hero
    • 3 years ago
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    I was just about to say....what happened to the x?

  7. myininaya
    • 3 years ago
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    \[f'(-1)=3a-12+b=0\]

  8. myininaya
    • 3 years ago
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    \[f'(2)=3a(4)+12(2)+b=0\]

  9. myininaya
    • 3 years ago
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    \[3a+b=12\] \[12a+b=-24\]

  10. moneybird
    • 3 years ago
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    3a + b =12 12a + 24 +b = 0 12a + b =-24 -(3a + b) = 12 9a = -36 a = -4 b =24

  11. giggles123
    • 3 years ago
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    what happened to the x?

  12. myininaya
    • 3 years ago
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    \[ 3a+b=12\] \[-(12a+b=-24)\] ------------------ -9a+0b=36 -9a=36 a=-4 -12+b=12 b=24 gj money

  13. myininaya
    • 3 years ago
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    what do you mean what happen to x?

  14. giggles123
    • 3 years ago
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    how did you go from f'(x)=3ax^2+12x+b* to f′(−1)=3a−12+b=0

  15. myininaya
    • 3 years ago
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    \[f(a)=-4x^3+6x^2+24x+4\]

  16. myininaya
    • 3 years ago
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    -1 and 2 are critical numbers this is a polynomial so our critical numbers only exist when f'=0

  17. myininaya
    • 3 years ago
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    f'(-1)=0 f'(2)=0

  18. myininaya
    • 3 years ago
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    this was given to us

  19. myininaya
    • 3 years ago
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    money and i applied these conditions to f'

  20. giggles123
    • 3 years ago
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    I got it. Thank you so much!

  21. luckey
    • 3 years ago
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    here b can take any value and a is rrestricted to any value in the interval (-2,2)

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