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Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)

f(x)=x^3+3x^2-1 on [-3,1]

  • math -

    f(-3) = -1
    f(1) = 3

    f'(x) = 3x^2 + 6x = 3x(x+2)
    so, there are relative max/min at x= -2,0

    f(-2) = 3
    f(0) = -1

    So, on [-3,1] we have
    min = -1
    max = 3

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