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d i E Concavity and Curve Sketching a l l u d Dr. Abdulla - PowerPoint PPT Presentation

Section 13.3 d i E Concavity and Curve Sketching a l l u d Dr. Abdulla Eid b A College of Science . r D MATHS 104: Mathematics for Business II Dr. Abdulla Eid (University of Bahrain) Concavity 1 / 18 Concavity Increasing


  1. Section 13.3 d i E Concavity and Curve Sketching a l l u d Dr. Abdulla Eid b A College of Science . r D MATHS 104: Mathematics for Business II Dr. Abdulla Eid (University of Bahrain) Concavity 1 / 18

  2. Concavity Increasing Function has three cases d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Concavity 2 / 18

  3. Question: How to distinguish between these three types of behavior? Answer: Recall: If g ( x ) is increasing, then g ′ ( x ) > 0. d i E a l l u d b 1 If f ′′ ( x ) > 0, then the curve is concave upward (CU). A 2 If f ′′ ( x ) < 0, then the curve is concave downward (CD). . r D 3 If f ′′ ( x ) = 0 (for all x ), then f ( x ) has no curvature (line). Dr. Abdulla Eid (University of Bahrain) Concavity 3 / 18

  4. Inflection Points d i E Definition a A number c is called an inflection point of f ( x ) if at these point, the l l u function changes from concave upward to downward and vice verse. The d candidates are the points c , where b A f ′′ ( c ) = 0 or f ′′ ( c ) does not exist . r D Dr. Abdulla Eid (University of Bahrain) Concavity 4 / 18

  5. Example Discuss the following curve with respect to concavity and inflection points. f ( x ) = x 3 Solution: d i We find the derivatives first which are E a f ′ ( x ) = 3 x 2 l l u f ′′ ( x ) = 6 x d b To find the inflection points, we find where the second derivative equal to A zero or does not exist. . r D f ′ ( x ) does not exist f ′′ ( x ) = 0 denominator = 0 numerator = 0 1 = 0 6 x = 0 Always False x = 0 No Solution Dr. Abdulla Eid (University of Bahrain) Concavity 5 / 18

  6. Number Line d i E a l l u d b A . r 1 f is CD in ( − ∞ , 0 ) . D 2 f is CU in ( 0, ∞ ) . 3 f has inflection point at x = 0 with value f ( 0 ) = 0. Dr. Abdulla Eid (University of Bahrain) Concavity 6 / 18

  7. Example Discuss the following curve with respect to concavity and inflection points. f ( x ) = 2 + ln x Solution: d i We find the derivatives first which are E f ′ ( x ) = 1 a l l x u f ′′ ( x ) = − 1 d b x 2 A To find the inflection points, we find where the second derivative equal to . r zero or does not exist. D f ′′ ( x ) = 0 f ′ ( x ) does not exist numerator = 0 denominator = 0 − 1 = 0 x 2 = 0 Always False x = 0 No Solution Dr. Abdulla Eid (University of Bahrain) Concavity 7 / 18

  8. Number Line d i E a l l u d b A . r D 1 f is CD in ( 0, ∞ ) . 2 f has no inflection point. Dr. Abdulla Eid (University of Bahrain) Concavity 8 / 18

  9. Example (Old Exam Question) Discuss the following curve with respect to concavity and inflection points. f ( x ) = x 4 − 3 x 3 + 3 x 2 − 5 d i E Solution: We find the derivatives first which are a l l f ′ ( x ) = 4 x 3 − 9 x 2 + 6 x u d f ′′ ( x ) = 12 x 2 − 18 x + 6 b A To find the inflection points, we find where the second derivative equal to . r zero or does not exist. D f ′ ( x ) does not exist f ′′ ( x ) = 0 denominator = 0 numerator = 0 1 = 0 12 x 2 − 18 x + 6 = 0 Always False x = 1 or x = 1 No Solution Dr. Abdulla Eid (University of Bahrain) Concavity 9 / 18 2

  10. Number Line d i E a l l u d b A . 1 f is CD in ( 1 2 , 1 ) . r D 2 f is CU in ( − ∞ , 1 2 ) ∪ ( 1, ∞ ) . 3 f has inflection point at x = 1 2 with value f ( 1 2 ) = and at x = 1 with value f ( 1 ) = . Dr. Abdulla Eid (University of Bahrain) Concavity 10 / 18

  11. Exercise (All in All) Find the intervals where the function is increasing/decreasing, concave upward, concave downward, find all local max/min, find inflection points and sketch the graph of the function. f ( x ) = x 5 − 4 x 4 d i E Solution: We find the derivative first which is a l l f ′ ( x ) = 5 x 4 − 20 x 3 u d f ′′ ( x ) = 20 x 3 − 60 x 2 b A To find the critical points, we find where the derivative equal to zero or . r does not exist. D f ′ ( x ) does not exist f ′ ( x ) = 0 denominator = 0 numerator = 0 1 = 0 5 x 4 − 20 x 3 = 0 Always False x = 0 or x = 4 No Solution Dr. Abdulla Eid (University of Bahrain) Concavity 11 / 18

  12. Number Line d i E a l l u d b A . 1 f is increasing in ( − ∞ , 0 ) ∪ ( 4, ∞ ) . r D 2 f is decreasing in ( 0, 4 ) . 3 f has a local maximum at x = 0 with value f ( 0 ) = 0. 4 f has a local minimum at x = 4 with value f ( 4 ) = . Dr. Abdulla Eid (University of Bahrain) Concavity 12 / 18

  13. Recall that the derivatives are f ′ ( x ) = 5 x 4 − 20 x 3 f ′′ ( x ) = 20 x 3 − 60 x 2 d i E To find the inflection points, we find where the second derivative equal to a zero or does not exist. l l u d f ′ ( x ) does not exist b A f ′′ ( x ) = 0 denominator = 0 . numerator = 0 1 = 0 r D 20 x 3 − 60 x 2 = 0 Always False x = 1 or x = 3 No Solution Dr. Abdulla Eid (University of Bahrain) Concavity 13 / 18

  14. Number Line d i E a l l u d b A . r 1 f is CD in ( − ∞ , 3 ) . D 2 f is CU in ( 3, ∞ ) . 3 f has inflection point at x = 3 with value f ( 3 ) = . Dr. Abdulla Eid (University of Bahrain) Concavity 14 / 18

  15. Exercise (All in All) Find the intervals where the function is increasing/decreasing, concave upward, concave downward, find all local max/min, find inflection points and sketch the graph of the function. f ( x ) = x 3 − 6 x 2 + 9 x + 1 d i E Solution: We find the derivative first which is a l l f ′ ( x ) = 3 x 2 − 12 x + 9 u d f ′′ ( x ) = 6 x − 12 b A To find the critical points, we find where the derivative equal to zero or . does not exist. r D f ′ ( x ) does not exist f ′ ( x ) = 0 denominator = 0 numerator = 0 1 = 0 3 x 2 − 12 x + 9 = 0 Always False x = 1 or x = 3 No Solution Dr. Abdulla Eid (University of Bahrain) Concavity 15 / 18

  16. Number Line d i E a l l u d b A . 1 f is increasing in ( − ∞ , 1 ) ∪ ( 3, ∞ ) . r D 2 f is decreasing in ( 1, 3 ) . 3 f has a local maximum at x = 1 with value f ( 1 ) = 5. 4 f has a local minimum at x = 3 with value f ( 3 ) = 1. Dr. Abdulla Eid (University of Bahrain) Concavity 16 / 18

  17. Recall that the derivatives are f ′ ( x ) = 3 x 2 − 12 x + 9 f ′′ ( x ) = 6 x − 12 d i E To find the inflection points, we find where the second derivative equal to a zero or does not exist. l l u d f ′ ( x ) does not exist b A f ′′ ( x ) = 0 denominator = 0 . numerator = 0 1 = 0 r D 6 x − 12 = 0 Always False x = 2 No Solution Dr. Abdulla Eid (University of Bahrain) Concavity 17 / 18

  18. Number Line d i E a l l u d b A . r 1 f is CD in ( − ∞ , 2 ) . D 2 f is CU in ( 2, ∞ ) . 3 f has inflection point at x = 2 with value f ( 2 ) = 3. Dr. Abdulla Eid (University of Bahrain) Concavity 18 / 18

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