Math 3: Critical points and inflection points

1. Graph the function f(x) = x3 + 3x2 − 24x, labeling all critical points, roots, and inflection points.

Solution: The graph should look like this:

200 y

150

100

50 x −8 −6 −4 −2 2 4 6 8 −50

−100

−150

−200

First we factor f(x) = x(x2 + 3x − 24), and use the quadratic formula to get that x = 0, x = 3.62, and x = −6.62 (the last two being approximate) are the roots of f.

Next, we calculate that f 0(x) = 3x2 + 6x − 24, or f 0(x) = 3(x − 2)(x + 4). This means that we have critical points at x = 2 and x = −4. Plugging these values into f(x), we get that the critical points are (2, −28) and (−4, 80).

Finally, we calculate that f 00(x) = 6x + 6. This means that f 00(−1) = 0. We check that f 00 is negative to the left of x = −1 and positive to the right of x = −1, hence (−1, 26) is an inflection point of f. 1 2. Let f(x) = sin(x) − 2 x. Find all critical points of f and determine whether each critical point is a local maximum, a local minimum, or neither. Also find all points of inflection of f.

0 1 1 Solution: We have that f (x) = cos(x)− 2 , hence the critical points occur when cos(x) = 2 . π π This happens when x = 3 + 2πn or x = − 3 + 2πn, where n is any integer.

To determine the behavior at these points, we could use the first derivative test, but in this case the second derivative test is easier. We have that f 00(x) = − sin(x), therefore √ π  π  3 f 00 + 2πn = − sin + 2πn = − and 3 3 √2  π   π  3 f 00 − + 2πn = − sin − + 2πn = . 3 3 2 π This means that the function has local maxima at x = 3 + 2πn and local minima at π x = − 3 + 2πn.

Inflection points can occur when f 00(x) = 0; as sin(x) = 0 when x = πn for any integer n, these are the possible inflection points. By considering the graph of the sine function, we determine that f 00(x) does in fact switch signs at all of these points, hence they are all inflection points of f.

3. Show that a cubic function f(x) = ax3 + bx2 + cx + d always has exactly one inflection point. x1+x2+x3 If the 3 roots of f are x1, x2, and x3, show that the point of inflection is at x = 3 .

00 00 −b Solution: We have that f (x) = 6ax + 2b, hence f (x) = 0 when x = 3a . Furthermore, the third derivative of f is f (3)(x) = 6a, which is always either positive or negative (the −b leading coefficient a is not allowed to be 0 in a cubic). This means that x = 3a is in fact an inflection point, as f 00(x) switches from positive to negative or negative to positive at this point.

If f has 3 roots x1, x2, and x3, we can write f(x) = a(x − x1)(x − x2)(x − x3). The 2 coefficient of x when this is expanded is −a(x1 + x2 + x3), hence b = −a(x1 + x2 + x3). −b x1+x2+x3 Plugging this into x = 3a , we get x = 3 .

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