§ 2.3 The First and Second Derivative Tests and Curve Sketching

Section Outline Curve Sketching Critical Values The First Derivative Test The Second Derivative Test Test for Inflection Points

Curve Sketching A General Approach to Curve Sketching 1) Starting with f (x), we compute 2) Next, we locate all relative maximum and relative minimum points and make a partial sketch. 3) We study the concavity of f (x) and locate all inflection points. 4) We consider other properties of the graph, such as the intercepts, and complete the sketch.

Critical Values Definition Example Critical Values: Given a function f (x), a number a in the domain such that either or is undefined. For the function below, notice that the slope of the function is 0 at x = -2 and the slope is undefined at x = -0.4. Also notice that the function has a relative minimum and a relative maximum at these points, respectively.

First Derivative Test

First Derivative Test Find the local maximum and minimum points of EXAMPLE Find the local maximum and minimum points of SOLUTION First we find the critical values and critical points of f: The first derivative if 9x – 3 = 0 or 2x + 1 = 0. Thus the critical values are x = 1/3 and x = -1/2. Substituting the critical values into the expression of f:

First Derivative Test CONTINUED Thus the critical points are (1/3, 43/18) and (-1/2, 33/8). To tell whether we have a relative maximum, minimum, or neither at a critical point we shall apply the first derivative test. This requires a careful study of the sign of , which can be facilitated with the aid of a chart. Here is how we can set up the chart.

First Derivative Test CONTINUED Divide the real line into intervals with the critical values as endpoints. Since the sign of depends on the signs of its two factors 9x – 3 and 2x + 1, determine the signs of the factors of over each interval. Usually this is done by testing the sign of a factor at points selected from each interval. In each interval, use a plus sign if the factor is positive and a minus sign if the factor is negative. Then determine the sign of over each interval by multiplying the signs of the factors and using A plus sign of corresponds to an increasing portion of the graph f and a minus sign to a decreasing portion. Denote an increasing portion with an upward arrow and a decreasing portion with a downward arrow. The sequence of arrows should convey the general shape of the graph and, in particular, tell you whether or not your critical values correspond to extreme points.

First Derivative Test + + + + + CONTINUED -1/2 1/3 Critical Points, Intervals x < -1/2 -1/2 < x < 1/3 x > 1/3 __ __ + 9x - 3 __ + + 2x + 1 + __ + Increasing on Decreasing on Increasing on Local maximum Local minimum

First Derivative Test CONTINUED You can see from the chart that the sign of varies from positive to negative at x = -1/2. Thus, according to the first derivative test, f has a local maximum at x = -1/2. Also, the sign of varies from negative to positive at x = 1/3; and so f has a local minimum at x = 1/3. In conclusion, f has a local maximum at (-1/2, 33/8) and a local minimum at (1/3, 43/18). NOTE: Upon the analyzing the various intervals, had any two consecutive intervals not alternated between “increasing” and “decreasing”, there would not have been a relative maximum or minimum at the value for x separating those two intervals.

Second Derivative Test

Second Derivative Test EXAMPLE Locate all possible relative extreme points on the graph of the function Check the concavity at these points and use this information to sketch the graph of f (x). SOLUTION We have The easiest way to find the critical values is to factor the expression for

Second Derivative Test CONTINUED From this factorization it is clear that will be zero if and only if x = -3 or x = -1. In other words, the graph will have horizontal tangent lines when x = -3 and x = -1, and no where else. To plot the points on the graph where x = -3 and x = -1, we substitute these values back into the original expression for f (x). That is, we compute Therefore, the slope of f (x) is 0 at the points (-3, 0) and (-1, -4). Next, we check the sign of at x = -3 and at x = -1 and apply the second derivative test: (local maximum) (local minimum).

Second Derivative Test CONTINUED The following is a sketch of the function. (-3, 0) (-1, -4)

Test for Inflection Points

Second Derivative Test EXAMPLE Sketch the graph of SOLUTION We have We set and solve for x. (critical values)

Second Derivative Test CONTINUED Substituting these values of x back into f (x), we find that We now compute (local minimum) (local maximum).

Second Derivative Test CONTINUED Since the concavity reverses somewhere between , there must be at least one inflection point. If we set , we find that So the inflection point must occur at x = 0. In order to plot the inflection point, we compute The final sketch of the graph is given below.

Second Derivative Test CONTINUED (0, 2)