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Math Help - find the absolute extrema of the function

  1. #1
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    find the absolute extrema of the function

    f(x)=(x-3)^3 [0,4]



    f(x)=x^3-12X [1,4]
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  2. #2
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    1. f'(x)=[(x-3)^3]'=[x^3-9x^2+27x-27]'=3x^2-18x+27
    f'(x)=0 \Longrightarrow x=3
    Notice, that the derivative has a double root from what results, that the f function is a hyperbola with focus in x=3 \in [0,4],
    the extrema are then f(0) and f(4):
    f(0)=-27
    f(4)=1

    2. f'(x)=[x^3-12x]'=3x^2-12=3(x+2)(x-2), x=2 belongs to the interval,
    all we have to do now is to check the f function values for the variable of 1, 2 and 4 and pick the highest and the lowest one.
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  3. #3
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    was lost some saved me

    thank you was very lost
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