− − 2. Next, we set the derivative equal to zero to find the critical points. There is one
solution: x = 1. We plug this value into the original equation to find the y-coordinate of the
critical point: When x = 1, y = 3(1) − 2(1) = 1. Thus, we have a critical point at (1, 1). But,
notice that the derivative is not defined at x = 0. This means that the 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 for x > 1 and positive for 0
< x < 1. Therefore, the curve is decreasing for x < 0 and for x > 1, and increasing for 0 < x < 1.
Next, we take the second derivative: . If we set this equal to zero, there is no
solution. Therefore, there is no point of inflection. 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. Next, we
need to determine if each critical point is maximum, minimum, or something else. If we plug x
= 1 into the second derivative, the value is negative, so (1, 1) is a maximum. Now, we can
draw the curve. It looks like the following: