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# Calc 3.4

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### Calc 3.4

1. 1. The slope of the slope is how fast the slope is changing. 3.4 Concavity and the 2 nd Derivative Test
2. 2. We learned how to locate intervals where a function is increasing or decreasing with the first derivative test. Locating intervals where f ’ increases or decreases determines where graph is curving upward or downward.
3. 3. To find open intervals where the function is concave upward or downward, we need to find intervals on which f ‘ is increasing or decreasing.
4. 4. So to use this test for concavity, we need to locate x-values where f “(x) = 0 or is undefined. These values separate the intervals. Finally, test the sign of the test values in these intervals.
5. 5. Ex 1 p. 191 Determining Concavity Determine the open intervals on which the graph of is concave upward or downward. Solution: The function is continuous on the real numbers (no domain problems!) Next, find the second derivative. Because f “ (x) = 0 at now we know interval borders
6. 6. Interval Test Value x = -3 x = 0 x =3 Sign of f ”(x) f ”(-3) > 0 f ”(0) < 0 f ”(3) > 0 Conclusion Concave up Concave down Concave up
7. 7. Ex 2 p 192 Determining Concavity Determine intervals on which the graph of is concave upward or downward. Solution. Differentiate twice, find zeros or undefined values Use x = -2 and 2 to split intervals
8. 8. Intervals (- ∞, -2) (-2, 2) (2, ∞) Test value x = -3 x = 0 x = 3 Sign of f “ f “(-3) > 0 f “(0) < 0 f “(3) > 0 Conclusion Concave up Concave down Concave up
9. 10. Our last problem had two points of inflection, which can be thought of as points where the graph changes from concave upward to concave downward or vice versa.
10. 11. Ex 3 p. 193 Points of Inflection Determine points of inflection and discuss concavity of when x=0, 2, so possible points of inflection are when x = 0, or 2 Points of inflection happen when x = 0, and 2 Interval (- ∞, 0) (0, 2) (2, ∞) Test value x =-1 x = 1 x = 3 Sign of f “ f “(-1) > 0 f “(1) < 0 f “(3) > 0 Conclusion Concave up Concave down Concave up
11. 13. Ex 4. p195 Using the Second Derivative Test to find relative max or mins Find the relative extrema of f(x) = -3x 5 + 5x 3 Find critical numbers first (what makes first derivative = 0) So critical numbers are Using f “(x) = -60x 3 + 30x, apply 2 nd Derivative test. Point on f(x) (-1, -2) (1, 2) (0, 0) Sign of f”(x) f ” (-1) > 0 f “(1) < 0 f “(0) = 0 Conclusion Concave up so relative min Concave down so relative max Test fails
12. 14. Looking on either side of x = 0, the first derivative is positive, so at x = 0 is neither a max or min.
13. 15. <ul><li>To summarize: </li></ul><ul><li>Concavity </li></ul><ul><li>look for zeros of f” and see where it is zero or function is not continuous. </li></ul><ul><li>Set up intervals with those values </li></ul><ul><li>Test intervals – if f”>0 it is concave up. If f” < 0 it is concave down in interval </li></ul><ul><li>2 nd Derivative test </li></ul><ul><li>Look for critical numbers of FIRST derivative </li></ul><ul><li>Evaluate SECOND derivative at those values </li></ul><ul><li>If f” > 0 then that critical number is relative min </li></ul><ul><li>If f” < 0 then that critical number is relative max </li></ul><ul><li>If f” = 0, then test fails and you’ll have to look at 1 st Derivative test to determine max, min or neither. </li></ul>
14. 16. You’ll need to get in your head the 1 st der. Test and 2 nd der. Test and how they are used! 3.4 Assign. P. 195/ 1-69 every other odd, 79-82