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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 107 § 2.3 The First and Second Derivative Tests and Curve Sketching

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 2 of 107 Curve Sketching Critical Values The First Derivative Test The Second Derivative Test Test for Inflection Points Section Outline

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 3 of 107 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. A General Approach to Curve Sketching

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 4 of 107 Critical Values DefinitionExample 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 = Also notice that the function has a relative minimum and a relative maximum at these points, respectively.

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 5 of 107 First Derivative Test

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 6 of 107 First Derivative TestEXAMPLE SOLUTION Find the local maximum and minimum points of 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:

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 7 of 107 First Derivative TestCONTINUED 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.

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 8 of 107 First Derivative TestCONTINUED 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.

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 9 of 107 First Derivative TestCONTINUED Critical Points, Intervals 9x - 3 2x + 1 x < -1/2-1/2 < x < 1/3x > 1/3 __ Increasing on Decreasing on -1/2 1/3 Local maximum Local minimum

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 10 of 107 First Derivative TestCONTINUED 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.

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 11 of 107 Second Derivative Test

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 12 of 107 Second Derivative TestEXAMPLE SOLUTION Locate all possible relative extreme points on the graph of the function We have Check the concavity at these points and use this information to sketch the graph of f (x). The easiest way to find the critical values is to factor the expression for

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 13 of 107 Second Derivative Test 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). CONTINUED Next, we check the sign of at x = -3 and at x = -1 and apply the second derivative test: (local maximum) (local minimum).

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 14 of 107 Second Derivative Test The following is a sketch of the function. CONTINUED (-3, 0) (-1, -4)

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 15 of 107 Test for Inflection Points

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 16 of 107 Second Derivative TestEXAMPLE SOLUTION Sketch the graph of We have We set and solve for x. (critical values)

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 17 of 107 Second Derivative Test Substituting these values of x back into f (x), we find that We now compute CONTINUED (local maximum). (local minimum)

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 18 of 107 Second Derivative Test 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 CONTINUED The final sketch of the graph is given below.

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 19 of 107 Second Derivative TestCONTINUED (0, 2)

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