Cracking The Ap Calculus ab Exam 2018

(Marvins-Underground-K-12) #1

(0, 0). Note also that the function is positive for x < 0 and negative for x > 0, so the curve is


increasing for x < 0 and decreasing for x > 0. Next, we take the second derivative: =


    =    .  If  we  set this    equal   to  zero,   we  get

no solution, which means that there is no point of inflection at the origin. Notice that the second


derivative is positive for x < −2, negative for −2 < x < 2, and positive for x > 2. Therefore, the


curve is concave up for x < −2 and x > 2, and concave down for −2 < x < 2. 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, so (0, 0) is a maximum. Now, we can draw the


curve. It looks like the following:

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