Calculus

Consider the function f(x)=xsqrt(36−x^2), −1≤x≤6.

This function has an absolute minimum value equal to what?

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  1. f(x) = √(36-x^2) = (36-x^2)^(1/2)
    f ' (x) = (1/2)(36-x^2)^(-1/2) (-2x)
    = 0 for a max/min
    thus x = 0 , f(0) = 6
    but f(1) = √(36-1)
    = √35 which is < 6

    f(6) = √0 = 0

    So the absolute minimum is 0

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    posted by Reiny

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