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calculus

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Given that f(x)=2/3x^3 + 3/2x^2 -2x + 5. Find if exist, the maximum and minimum point using Second Derivative Test.

  • calculus -

    f = 2/3 x^3 + 3/2 x^2 - 2x + 5
    f' = 2x^2 + 3x - 2
    f'' = 4x + 3

    f has min/max at f'=0, or x= -2 or 1/2
    f''(-2) < 0 so f has a maximum there
    f''(1/2) > 0 so f has a minimum there

    From what you know about the general shape of cubics, this should make sense.

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