Cracking The Ap Calculus ab Exam 2018

(Marvins-Underground-K-12) #1

function has either a vertical tangent or a cusp at x = 0. We’ll be able to determine which after


we take the second derivative. Notice also that the derivative is negative for x < 0 and positive


for x > 0. Therefore, the curve is decreasing for x < 0 and increasing for x > 0. Next, we take


the second derivative: . If we set this equal to zero, there is no solution. The


second derivative is always negative, which tells us that the curve is always concave down


and that the curve has a cusp at x = 0. Note that if it had switched concavity there, then x = 0


would be a vertical tangent. Now, we can draw the curve. It looks like the following:

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