Cracking The Ap Calculus ab Exam 2018

(Marvins-Underground-K-12) #1
When    x   =   0,  y   =   0

When    x   =    ,  y   =   1

When    x   =   − , y   =   1

Thus, there are critical points at (0, 0), , and .


Take the second derivative to find any points of inflection.


    =   16  −   192x^2

This equals zero at x = and x = −.


Next, plug x = and x = − into the original equation to find the y-coordinates of the points of


inflection, which are at and . Now determine whether the points are maxima or


minima.


At x = 0, we have a minimum; the curve is concave up there.


At x = , it’s a maximum, and the curve is concave down.


At x = − , it’s also a maximum (still concave down).


Now plot.

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