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§ 2.3 The First and Second Derivative Tests and Curve Sketching

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**Section Outline Curve Sketching Critical Values**

The First Derivative Test The Second Derivative Test Test for Inflection Points

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**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.

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**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 = Also notice that the function has a relative minimum and a relative maximum at these points, respectively.

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First Derivative Test

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**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:

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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.

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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.

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**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

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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.

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**Second Derivative Test**

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**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

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**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).

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**Second Derivative Test**

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

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**Test for Inflection Points**

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**Second Derivative Test**

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

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**Second Derivative Test**

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

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**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.

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**Second Derivative Test**

CONTINUED (0, 2)

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