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mathcalculus Group Title

H-E-L-P so frustrating!! Determine the extrema of below on the given interval f(x)=5x^3-61x^2+16x+3 (a) on [0,4] The minimum is ?? and the maximum is ?? (b) on [-9,9] The minimum is ?? and the maximum is ?? please solve so i can know which answer i got wrong.

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    can someone please tell me the correct answer to this. it seems like one of my answers is incorrect.

    • one year ago
  2. satellite73 Group Title
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    check \(f(0), f(4)\) and also find the critical point and find \(f\) of that number

    • one year ago
  3. mathcalculus Group Title
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    a) on [0,4] The minimum is -589 and the maximum is 4.06074 (b) on [-9,9] The minimum is -8727 and the maximum is -1149

    • one year ago
  4. satellite73 Group Title
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    the derivative is \( 15 x^2-122 x+16\)

    • one year ago
  5. satellite73 Group Title
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    which, by some miracle factors as \((x-8) (15 x-2)\)

    • one year ago
  6. satellite73 Group Title
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    zeros of the derivative are \(\frac{2}{15}\) and \(8\)

    • one year ago
  7. satellite73 Group Title
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    so \(\frac{2}{15}\) is the \(x\) coordinate of the local max and \(8\) is the \(x\) coordinate of the local min

    • one year ago
  8. mathcalculus Group Title
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    so for the intervals [0,4] max is: 2/15 and 8 is the min? what about [-9,9]

    • one year ago
  9. mathcalculus Group Title
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    i just want to make sure because it submitted the answer and it keeps saying its wrong.

    • one year ago
  10. mathcalculus Group Title
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    @satellite73

    • one year ago
  11. mathcalculus Group Title
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    any one! this is important!

    • one year ago
  12. mathcalculus Group Title
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    never mind! i found it. thanks to those who actually helped.

    • one year ago
  13. Peter14 Group Title
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    f'(x)=10x^2 - 122x +16 0=10x^2 - 122x + 16 (just working here because I don't have paper handy)

    • one year ago
  14. satellite73 Group Title
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    careful here the max is the \(y\) value not the \(x\) value

    • one year ago
  15. Peter14 Group Title
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    just finding critical points

    • one year ago
  16. Peter14 Group Title
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    oh, sorry you already found them

    • one year ago
  17. satellite73 Group Title
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    8 is not in the interval \([0,4]\) for that one, you need to check \(f(0), f(4), f(\frac{2}{15})\) the largest is the max and the smallest is the min

    • one year ago
  18. satellite73 Group Title
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    for \([-9,9]\) you need to check \[f(-9),f(\frac{2}{15}), f(8), f(9)\]

    • one year ago
  19. mathcalculus Group Title
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    got it. thank you!

    • one year ago
  20. satellite73 Group Title
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    yw

    • one year ago
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