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calculuss review for exam

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use the second derivate test to locate the maxima and minima of y = x^2 + 2x - 3

  • calculuss review for exam - ,

    since y'' = 2 any max/min will be a min.

  • calculuss review for exam - ,

    what do you mean by that? :/

  • calculuss review for exam - ,

    Take a look at the explanation of the 2nd derivative test. If y'' < 0 you have a max, and if y'' > 0 you have a min where y'=0.

  • calculuss review for exam - ,

    if y' = 0 doesn't that mean there is no max or min? im sorry but i really don't understand this.. please help me

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