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AravindG

  • 3 years ago

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

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  1. AravindG
    • 3 years ago
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    @hartnn

  2. hartnn
    • 3 years ago
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    what u got as first derivative ?

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

  4. AravindG
    • 3 years ago
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    should i take second derivative?

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

  6. hartnn
    • 3 years ago
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    mathematically, u should take 2nd derivative...

  7. AravindG
    • 3 years ago
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    max at x=-1 and minimum at x=1

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

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

  10. AravindG
    • 3 years ago
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    i see is my final answers ryt?

  11. AravindG
    • 3 years ago
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    ....

  12. AravindG
    • 3 years ago
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    u dr??

  13. hartnn
    • 3 years ago
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    sorry, got disconnected...yes, its correct

  14. AravindG
    • 3 years ago
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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

  15. AravindG
    • 3 years ago
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    ,,,

  16. hartnn
    • 3 years ago
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    yes, u need to..

  17. AravindG
    • 3 years ago
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    are answers 0 and 2?

  18. hartnn
    • 3 years ago
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    mim at x=2

  19. AravindG
    • 3 years ago
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    max at x=0 min at x=2 ryt?

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

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

  22. AravindG
    • 3 years ago
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    seeya gn8 sweeet drms

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