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giggles123 Group Title

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

  • 2 years ago
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  1. Hero Group Title
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    I hate these

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

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

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

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

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

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

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

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

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

    • 2 years ago
  11. giggles123 Group Title
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    what happened to the x?

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

    • 2 years ago
  13. myininaya Group Title
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    what do you mean what happen to x?

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

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

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

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

    • 2 years ago
  18. myininaya Group Title
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    this was given to us

    • 2 years ago
  19. myininaya Group Title
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    money and i applied these conditions to f'

    • 2 years ago
  20. giggles123 Group Title
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    I got it. Thank you so much!

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

    • 2 years ago
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