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AravindG

find maximum and minimum values for the function g(x)=x^3-3x

  • one year ago
  • one year ago

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  1. AravindG
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    @hartnn

    • one year ago
  2. hartnn
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    what u got as first derivative ?

    • one year ago
  3. AravindG
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    3x^2-3=0 x^2=1 x=1,-1

    • one year ago
  4. AravindG
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    should i take second derivative?

    • one year ago
  5. hartnn
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    correct. here u can just put the values in g(x) and by inspection u can tell which is max

    • one year ago
  6. hartnn
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    mathematically, u should take 2nd derivative...

    • one year ago
  7. AravindG
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    max at x=-1 and minimum at x=1

    • one year ago
  8. AravindG
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    bt then why we took second derivative in last qn?

    • one year ago
  9. hartnn
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    because we can't find the max or min just by putting the points and inspection, can we ?

    • one year ago
  10. AravindG
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    i see is my final answers ryt?

    • one year ago
  11. AravindG
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    ....

    • one year ago
  12. AravindG
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    u dr??

    • one year ago
  13. hartnn
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    sorry, got disconnected...yes, its correct

    • one year ago
  14. AravindG
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    one last qn and i am done ,,well for this qn i think i gt crct anser bt i want t o knw if i need to take second derivative "find maximum and minimum value of 3x^4-8x^3+12x^2-48x+25

    • one year ago
  15. AravindG
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    ,,,

    • one year ago
  16. hartnn
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    yes, u need to..

    • one year ago
  17. AravindG
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    are answers 0 and 2?

    • one year ago
  18. hartnn
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    mim at x=2

    • one year ago
  19. AravindG
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    max at x=0 min at x=2 ryt?

    • one year ago
  20. hartnn
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    yes, both are correct..i had doubt with max at x=0

    • one year ago
  21. AravindG
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    thanx a lot!!!! it really helped cant tell u hw thankful i am !!

    • one year ago
  22. AravindG
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    seeya gn8 sweeet drms

    • one year ago
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