![8
Concavity - Graphically
Notice in Figure 5 that the slopes of the tangent lines
increase from left to right on the interval (a, b), so f ' is
increasing and f is concave upward (abbreviated CU) on
(a, b). [It can be proved that this is equivalent to saying that
the graph of f lies above all of its tangent lines on (a, b).]
Figure 5](https://image.slidesharecdn.com/lecture16-graphing-section4-141111094621-conversion-gate01/85/Lecture-16-graphing-section-4-3-8-320.jpg)
Lecture 16 graphing - section 4.3
- 1. Derivatives and the 4.3 Shapes of Curves
- 2. 2 Increasing and Decreasing Functions
- 3. Increasing and Decreasing Functions 3
- 4. First Derivative Test for Local Max/Min 4
- 5. Increasing and Decreasing Functions It is easy to remember the First Derivative Test by visualizing diagrams such as those in Figure 4. 5 Figure 4
- 6. 6 Concavity
- 7. 7 Concavity - Definition
- 8. 8 Concavity - Graphically Notice in Figure 5 that the slopes of the tangent lines increase from left to right on the interval (a, b), so f ' is increasing and f is concave upward (abbreviated CU) on (a, b). [It can be proved that this is equivalent to saying that the graph of f lies above all of its tangent lines on (a, b).] Figure 5
- 9. 9 Concavity – Points of Inflection A point where a curve changes its direction of concavity is called an inflection point. The curve in Figure 5 changes from concave upward to concave downward at P and from concave downward to concave upward at Q, so both P and Q are inflection points.
- 10. Second Derivative Test for Concavity 10
- 11. Second Derivative Test for Max/Min 11
- 12. Concavity For instance, part (a) is true because f '' (x) > 0 near c and so f is concave upward near c. This means that the graph of f lies above its horizontal tangent at c and so f has a local minimum at c. (See Figure 6.) 12 Figure 6 f '' (c) > 0, f is concave upward
- 13. Example 4 – Analyzing a Curve using Derivatives Discuss the curve y = x4 – 4x3 with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve. 13 Solution: If f (x) = x4 – 4x3, then f ' (x) = 4x3 – 12x2 = 4x2(x – 3) f ' ' (x) = 12x2 – 24x = 12x(x – 2)
- 14. Example 4 – Solution To find the critical numbers we set f '(x) = 0 and obtain x = 0 and x = 3. To use the Second Derivative Test we evaluate f '' at these critical numbers: cont’d 14 f ''(0) = 0 f ''(3) = 36 > 0 Since f '(3) = 0 and f ''(3) > 0, f (3) = –27 is a local minimum. Since f ''(0) = 0, the Second Derivative Test gives no information about the critical number 0. But since f '(x) < 0 for x < 0 and also for 0 < x < 3, the First Derivative Test tells us that f does not have a local maximum or minimum at 0.
- 15. Example 4 – Solution Since f ''(x) = 0 when x = 0 or 2, we divide the real line into intervals with these numbers as endpoints and complete the following chart. cont’d The point (0, 0) is an inflection point since the curve changes from concave upward to concave downward there. Also (2, –16) is an inflection point since the curve changes from concave downward to concave upward there. 15
- 16. 16 Example 4 – Solution Using the local minimum, the intervals of concavity, and the inflection points, we sketch the curve in Figure 7. Figure 7 cont’d