calculuss review for exam
posted by Amarjeet .
use the second derivate test to locate the maxima and minima of y = x^2 + 2x - 3
since y'' = 2 any max/min will be a min.
what do you mean by that? :/
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.
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