Cracking The Ap Calculus ab Exam 2018

3 x^2   −   18x +   24  =   0

3(x ²   −   6x  +   8)  =   0

3(x −   4)(x    −   2)  =   0

x   =   2,  4

Plug x = 2 and x = 4 into the original equation to find the y-coordinates of the critical points.

When    x   =   2,  y   =   10

When    x   =   4,  y   =   6

Thus, we have critical points at (2, 10) and (4, 6).

In our third step, the second derivative indicates any points of inflection.

    =   6x  −   18

This equals zero at x = 3.

Next, plug x = 3 into the original equation to find the y-coordinates of the point of inflection, which is at
(3, 8). Plug the critical values into the second derivative to determine whether each is a maximum or a
minimum.

6(2)    −   18  =   −6

This is negative, so the curve has a maximum at (2, 10), and the curve is concave down there.

6(4)    −   18  =   6

This is positive, so the curve has a minimum at (4, 6), and the curve is concave up there.

It’s graph-plotting time.