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dan18

  • 2 years ago

The maximum value of the function

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  1. hartnn
    • 2 years ago
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    can be found by taking the derivative of that funtion and equating it to 0.

  2. dan18
    • 2 years ago
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    y=\[\sqrt{2-x}\]

  3. hartnn
    • 2 years ago
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    so did u find y' ?

  4. dan18
    • 2 years ago
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    y=-4[\sqrt{2-x}\]

  5. hartnn
    • 2 years ago
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    could u differentiate y ?

  6. miteshchvm
    • 2 years ago
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    differentiate the function, then put the differentiated equation = 0 you will have either one answer or 2, if you get two answers, the one which has negative value will be the maximum value of the function ;)

  7. dan18
    • 2 years ago
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    is it the y value?

  8. hartnn
    • 2 years ago
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    have u learned to differentiate x^n ?

  9. dan18
    • 2 years ago
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    y'= nx^n-1

  10. hartnn
    • 2 years ago
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    so couldn't u differentiate y=-4 sqrt(2-x) ??

  11. dan18
    • 2 years ago
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    does the max. value asked here is the x or the y value?

  12. hartnn
    • 2 years ago
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    y value

  13. dan18
    • 2 years ago
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    y'= 2/(2-x)^-(1/2)

  14. akash_809
    • 2 years ago
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    @dan18 , the correct way is to differentiate and equate to zero , but just by looking at this question once can tell it's maximum value is infinity for x=-infinity....pretty weird question i guess...are u sure u hv typed the question correctly

  15. hartnn
    • 2 years ago
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    wouldn't it be y'= 2/((2-x)^(1/2))

  16. dan18
    • 2 years ago
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    I get x=2 @ y', then subts. it to the eq. does the max. value = 0?

  17. akash_809
    • 2 years ago
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    @dan18 is the question correct ?

  18. hartnn
    • 2 years ago
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    but actual answer is infact Max. value = 0 when x=2 http://www.wolframalpha.com/input/?i=y%3D-4+sqrt%282-x%29

  19. dan18
    • 2 years ago
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    \[y=-4\sqrt{2-x}\]

  20. dan18
    • 2 years ago
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    thak you guys!

  21. akash_809
    • 2 years ago
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    now u hv got the correct question, yes the max would be zero

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