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swin2013

  • 2 years ago

Second derivative test to find the local extrema for the function?

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  1. swin2013
    • 2 years ago
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    y = x^5-80x+100

  2. timo86m
    • 2 years ago
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    i think it has to do with finding the f''(x)=0

  3. swin2013
    • 2 years ago
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    I know that

  4. swin2013
    • 2 years ago
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    y'(x) = 5x^4 -80

  5. abb0t
    • 2 years ago
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    \[y' = 5x^4 -8\] \[y'' = 20x^3\]

  6. swin2013
    • 2 years ago
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    y''(x) = 20x^3

  7. abb0t
    • 2 years ago
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    set y'' = 0 and solve.

  8. swin2013
    • 2 years ago
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    i got 0... so i wasn't sure if that's right

  9. swin2013
    • 2 years ago
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    0 = 20x^3

  10. abb0t
    • 2 years ago
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    Yes.

  11. swin2013
    • 2 years ago
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    Okay thanks!

  12. swin2013
    • 2 years ago
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    Oh i forgot, is it the local minimum (x=0)?

  13. swin2013
    • 2 years ago
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    @abb0t

  14. abb0t
    • 2 years ago
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    If the second derivative is zero then the critical point can be anything. Look @ the graph of ur function.

  15. swin2013
    • 2 years ago
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    so it's just the local extrema is located at x=0

  16. swin2013
    • 2 years ago
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    and if it's like... 3 and -3 as the local extrema. I don't know if it's a local max or min correct?

  17. abb0t
    • 2 years ago
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    Only if it's zero. if < 0 = relative min if > 0 = relative max.

  18. swin2013
    • 2 years ago
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    ohhhh ok! Thank you so much!

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