Presentation on theme: "§ 2.3 The First and Second Derivative Tests and Curve Sketching."— Presentation transcript:
1 § 2.3The First and Second Derivative Tests and Curve Sketching
2 Section Outline Curve Sketching Critical Values The First Derivative TestThe Second Derivative TestTest for Inflection Points
3 Curve Sketching A General Approach to Curve Sketching 1) Starting with f (x), we compute2) 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.
4 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.
6 First Derivative Test Find the local maximum and minimum points of EXAMPLEFind the local maximum and minimum points ofSOLUTIONFirst 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 arex = 1/3 and x = -1/2.Substituting the critical values into the expression of f:
7 First Derivative TestCONTINUEDThus 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.
8 First Derivative TestCONTINUEDDivide 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 usingA 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.
9 First Derivative Test + + + + + CONTINUED -1/2 1/3 Critical Points, Intervalsx < -1/2-1/2 < x < 1/3x > 1/3____+9x - 3__++2x + 1+__+Increasing onDecreasing onIncreasing onLocal maximumLocal minimum
10 First Derivative TestCONTINUEDYou 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.
12 Second Derivative Test EXAMPLELocate all possible relative extreme points on the graph of the functionCheck the concavity at these points and use this informationto sketch the graph of f (x).SOLUTIONWe haveThe easiest way to find the critical values is to factor the expression for
13 Second Derivative Test CONTINUEDFrom 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 computeTherefore, 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).
14 Second Derivative Test CONTINUEDThe following is a sketch of the function.(-3, 0)(-1, -4)
16 Second Derivative Test EXAMPLESketch the graph ofSOLUTIONWe haveWe set and solve for x.(critical values)
17 Second Derivative Test CONTINUEDSubstituting these values of x back into f (x), we find thatWe now compute(local minimum)(local maximum).
18 Second Derivative Test CONTINUEDSince the concavity reverses somewhere between , theremust be at least one inflection point. If we set , we find thatSo the inflection point must occur at x = 0. In order to plot the inflection point, we computeThe final sketch of the graph is given below.