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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 107 § 2.4 Curve Sketching (Conclusion)

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Presentation on theme: "Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 107 § 2.4 Curve Sketching (Conclusion)"— Presentation transcript:

1 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 107 § 2.4 Curve Sketching (Conclusion)

2 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 2 of 107 The Second Derivative Test Graphs With Asymptotes Section Outline

3 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 3 of 107 Second Derivative TestEXAMPLE SOLUTION Sketch the graph of We have We set and solve for x. (critical value)

4 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 4 of 107 Second Derivative Test Since, we know nothing about the graph at x = 2. However, the test for inflection points suggests that we have an inflection point at x = 2. First, lets verify that we indeed have an inflection point at x = 2. If this proves to be not the case, we would use a similar method (using the first derivative) to see if we have a relative extremum at x = 2. We will look at the left- and right-hand limits of and see if the values of the second derivative are oppositely signed as x approaches 2. CONTINUED x 6x

5 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 5 of 107 Second Derivative Test From the values of in the table, we find that, as x approaches 2, the second derivative not only has a limit of 0, but the values of the function to the left of x = 2, remain negative. Also, the values of the function to the right of x = 2 remain positive. Therefore, by definition, f (x) has an inflection point at x = 2 (since the graph of f (x) is concave down to the left of x = 2 and concave up to the right of x = 2). CONTINUED Notice, x = 2 was the only candidate for generating a relative extremum. Therefore, there are no relative extrema. We will now find the y-coordinate for the inflection point. So, the only inflection point is at (2, 3).

6 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 6 of 107 Second Derivative Test Now we will look for intercepts. Lets first look for a y-intercept by evaluating f (0). CONTINUED So, we have a y-intercept at (0, -5). To find any x-intercepts, we replace f (x) with 0. Since this equation does not factor, and the quadratic formula cannot help us either, we attempt to use the Rational Roots Theorem from algebra. In doing so we find that there are no rational roots (x-intercepts). So, if there is an x- intercept, it will be an irrational number. Below, we show some of the work employed in estimating the x-intercept.

7 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 7 of 107 Second Derivative TestCONTINUED x f (x) Notice that the y-values corresponding to x = 0.54 and x = 0.55 are below the x- axis and the y-values corresponding to x = 0.56 and x = 0.57 are above the x- axis. Therefore, in between x = 0.55 and x = 0.56, there must be an x-intercept. For the sake of brevity, well just take x = 0.56 for our x-intercept since, out of the four x-values above, it has the y-value closest to zero. Therefore, the point of our x-intercept is (0.56, 0). Now we will sketch a graph of the function.

8 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 8 of 107 Second Derivative TestCONTINUED (0, 5) (0.56, 0)(2, 3)

9 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 9 of 107 f (x) yields information about where things are on a graph. yields information about slope on a graph. yields information about concavity on a graph.

10 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 10 of 107 Graphs With AsymptotesEXAMPLE SOLUTION Sketch the graph of We have We set and solve for x. (critical values)

11 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 11 of 107 Graphs With Asymptotes (We exclude the case x = -2 since this is outside our specified domain.) The graph has a horizontal tangent at (2, f (2)) = (2, 13). Since, the graph is concave up at x = 2 and, by the second derivative test, (2, 13) is a relative minimum. In fact, for all positive x, and therefore the graph is concave up at all points. CONTINUED Before sketching the graph, notice that as x approaches zero the term 12/x in the formula for f (x) is dominant. That is, this term becomes arbitrarily large, whereas the terms 3x + 1 contribute a diminishing proportion to the function value as x approaches 0. Thus f (x) has the y-axis as an asymptote. For large values of x, the term 3x is dominant. The value of f (x) is only slightly larger than 3x since the terms 12/x + 1 has decreasing significance as x becomes arbitrarily large; that is, the graph of f (x) is slightly above the graph of y = 3x + 1. As x increases, the graph of f (x) has the line y = 3x + 1 as an asymptote.

12 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 12 of 107 Graphs With AsymptotesCONTINUED (2, 13) y = 3x + 1

13 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 13 of 107 Curve Sketching Techniques

14 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 14 of 107 Curve Sketching Techniques


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