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anonymous

  • 5 years ago

how do you find the abs max of func x^3-12x on the interval 0<x<4

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  1. amistre64
    • 5 years ago
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    take the first and second derivatives....

  2. amistre64
    • 5 years ago
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    y = x^3 - 12x y' = 3x^2 - 12 y' = 0 will give critical points; y'' = 6x ; plug those points into y'' to see if they are (+) or (-) to find max and min

  3. amistre64
    • 5 years ago
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    y' = 3x^2 - 12x = 0 x(3x - 12) = 0 when x = 0 or x = 4 are the critical points of your equation... but I have to wonder, does your interval INCLUDE 0 and 4? because the way you have it written it means everything between 0 and 4 EXCEPT for 0 and 4.

  4. anonymous
    • 5 years ago
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    yes it does so i just plug in 0 and 4 into the second deritave to find the max?

  5. anonymous
    • 5 years ago
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    because the answer is 16 and i have no idea how that it

  6. amistre64
    • 5 years ago
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    you plug it into the second derivative see how it "acts"; and according to its behaviour, there are a few possibilities of what it could be... But first, plug in 0 and 4 into your original equation to see what value they give..

  7. amistre64
    • 5 years ago
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    y(0) = 0 y(4) = 16. 16 is obviously higher than 0

  8. amistre64
    • 5 years ago
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    so that would be the maximium point within the interval....

  9. anonymous
    • 5 years ago
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    thanks haha i feel dumb now i knew that

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