Cracking The Ap Calculus ab Exam 2018

(Marvins-Underground-K-12) #1

, then y = . So there are points of inflection at


and  .   Next,   we  need    to  determine   if  each    critical    point   is

maximum, minimum, or something else. If we plug x = 0 into the second derivative, the value is


negative, sois a maximum. If we plug x = 0 into the second derivative, the value is positive, so


(2, −4) is a minimum. If we plug x = 2 into the second derivative, the value is negative, so (−2,


−4) is a minimum. Now, we can draw the curve. It looks like the following:

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