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Lesson 18: Derivatives and the Shapes of Curves

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  • 1. Section 4.2 Derivatives and the Shapes of Curves V63.0121.006/016, Calculus I New York University March 30, 2010 Announcements Quiz 3 on Friday (Sections 2.6–3.5) Midterm Exam scores have been updated If your Midterm Letter Grade on Blackboard differs from your midterm grade on Albert, trust Blackboard
  • 2. Announcements Quiz 3 on Friday (Sections 2.6–3.5) Midterm Exam scores have been updated If your Midterm Letter Grade on Blackboard differs from your midterm grade on Albert, trust Blackboard V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 2 / 28
  • 3. Outline Recall: The Mean Value Theorem Monotonicity The Increasing/Decreasing Test Finding intervals of monotonicity The First Derivative Test Concavity Definitions Testing for Concavity The Second Derivative Test V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 3 / 28
  • 4. Recall: The Mean Value Theorem Theorem (The Mean Value Theorem) Let f be continuous on [a, b] and differentiable on (a, b). Then there exists a point c in (a, b) such that f (b) − f (a) b = f (c). b−a a V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 4 / 28
  • 5. Recall: The Mean Value Theorem Theorem (The Mean Value Theorem) Let f be continuous on [a, b] and differentiable on (a, b). Then there exists a point c in (a, b) such that f (b) − f (a) b = f (c). b−a a V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 4 / 28
  • 6. Recall: The Mean Value Theorem Theorem (The Mean Value Theorem) c Let f be continuous on [a, b] and differentiable on (a, b). Then there exists a point c in (a, b) such that f (b) − f (a) b = f (c). b−a a V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 4 / 28
  • 7. Why the MVT is the MITC Most Important Theorem In Calculus! Theorem Let f = 0 on an interval (a, b). Then f is constant on (a, b). Proof. Pick any points x and y in (a, b) with x < y . Then f is continuous on [x, y ] and differentiable on (x, y ). By MVT there exists a point z in (x, y ) such that f (y ) − f (x) = f (z) = 0. y −x So f (y ) = f (x). Since this is true for all x and y in (a, b), then f is constant. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 5 / 28
  • 8. Outline Recall: The Mean Value Theorem Monotonicity The Increasing/Decreasing Test Finding intervals of monotonicity The First Derivative Test Concavity Definitions Testing for Concavity The Second Derivative Test V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 6 / 28
  • 9. What does it mean for a function to be increasing? Definition A function f is increasing on (a, b) if f (x) < f (y ) whenever x and y are two points in (a, b) with x < y . V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 7 / 28
  • 10. What does it mean for a function to be increasing? Definition A function f is increasing on (a, b) if f (x) < f (y ) whenever x and y are two points in (a, b) with x < y . An increasing function “preserves order.” Write your own definition (mutatis mutandis) of decreasing, nonincreasing, nondecreasing V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 7 / 28
  • 11. The Increasing/Decreasing Test Theorem (The Increasing/Decreasing Test) If f > 0 on (a, b), then f is increasing on (a, b). If f < 0 on (a, b), then f is decreasing on (a, b). V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 8 / 28
  • 12. The Increasing/Decreasing Test Theorem (The Increasing/Decreasing Test) If f > 0 on (a, b), then f is increasing on (a, b). If f < 0 on (a, b), then f is decreasing on (a, b). Proof. It works the same as the last theorem. Pick two points x and y in (a, b) with x < y . We must show f (x) < f (y ). By MVT there exists a point c in (x, y ) such that f (y ) − f (x) = f (c) > 0. y −x So f (y ) − f (x) = f (c)(y − x) > 0. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 8 / 28
  • 13. Finding intervals of monotonicity I Example Find the intervals of monotonicity of f (x) = 2x − 5. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 9 / 28
  • 14. Finding intervals of monotonicity I Example Find the intervals of monotonicity of f (x) = 2x − 5. Solution f (x) = 2 is always positive, so f is increasing on (−∞, ∞). V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 9 / 28
  • 15. Finding intervals of monotonicity I Example Find the intervals of monotonicity of f (x) = 2x − 5. Solution f (x) = 2 is always positive, so f is increasing on (−∞, ∞). Example Describe the monotonicity of f (x) = arctan(x). V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 9 / 28
  • 16. Finding intervals of monotonicity I Example Find the intervals of monotonicity of f (x) = 2x − 5. Solution f (x) = 2 is always positive, so f is increasing on (−∞, ∞). Example Describe the monotonicity of f (x) = arctan(x). Solution 1 Since f (x) = is always positive, f (x) is always increasing. 1 + x2 V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 9 / 28
  • 17. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f (x) = x 2 − 1. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
  • 18. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f (x) = x 2 − 1. Solution f (x) = 2x, which is positive when x > 0 and negative when x is. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
  • 19. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f (x) = x 2 − 1. Solution f (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: − 0 + f 0 V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
  • 20. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f (x) = x 2 − 1. Solution f (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: − 0 + f 0 f V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
  • 21. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f (x) = x 2 − 1. Solution f (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: − 0 + f 0 f So f is decreasing on (−∞, 0) and increasing on (0, ∞). V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
  • 22. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f (x) = x 2 − 1. Solution f (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: − 0 + f 0 f So f is decreasing on (−∞, 0) and increasing on (0, ∞). In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞) V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
  • 23. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f (x) = x 2/3 (x + 2). V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
  • 24. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f (x) = x 2/3 (x + 2). Solution f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4) 2 1 The critical points are 0 and and −4/5. − × + x −1/3 0 − 0 + 5x + 4 −4/5 V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
  • 25. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f (x) = x 2/3 (x + 2). Solution f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4) 2 1 The critical points are 0 and and −4/5. − × + x −1/3 0 − 0 + 5x + 4 −4/5 0 × f (x) −4/5 0 f (x) V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
  • 26. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f (x) = x 2/3 (x + 2). Solution f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4) 2 1 The critical points are 0 and and −4/5. − × + x −1/3 0 − 0 + 5x + 4 −4/5 + 0 × f (x) −4/5 0 f (x) V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
  • 27. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f (x) = x 2/3 (x + 2). Solution f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4) 2 1 The critical points are 0 and and −4/5. − × + x −1/3 0 − 0 + 5x + 4 −4/5 + 0 − × f (x) −4/5 0 f (x) V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
  • 28. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f (x) = x 2/3 (x + 2). Solution f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4) 2 1 The critical points are 0 and and −4/5. − × + x −1/3 0 − 0 + 5x + 4 −4/5 + 0 − × + f (x) −4/5 0 f (x) V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
  • 29. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f (x) = x 2/3 (x + 2). Solution f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4) 2 1 The critical points are 0 and and −4/5. − × + x −1/3 0 − 0 + 5x + 4 −4/5 + 0 − × + f (x) −4/5 0 f (x) V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
  • 30. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f (x) = x 2/3 (x + 2). Solution f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4) 2 1 The critical points are 0 and and −4/5. − × + x −1/3 0 − 0 + 5x + 4 −4/5 + 0 − × + f (x) −4/5 0 f (x) V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
  • 31. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f (x) = x 2/3 (x + 2). Solution f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4) 2 1 The critical points are 0 and and −4/5. − × + x −1/3 0 − 0 + 5x + 4 −4/5 + 0 − × + f (x) −4/5 0 f (x) V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
  • 32. The First Derivative Test Theorem (The First Derivative Test) Let f be continuous on [a, b] and c a critical point of f in (a, b). If f (x) > 0 on (a, c) and f (x) < 0 on (c, b), then c is a local maximum. If f (x) < 0 on (a, c) and f (x) > 0 on (c, b), then c is a local minimum. If f (x) has the same sign on (a, c) and (c, b), then c is not a local extremum. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 12 / 28
  • 33. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f (x) = x 2 − 1. Solution f (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: − 0 + f 0 f So f is decreasing on (−∞, 0) and increasing on (0, ∞). In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞) V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 13 / 28
  • 34. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f (x) = x 2 − 1. Solution f (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: − 0 + f 0 f min So f is decreasing on (−∞, 0) and increasing on (0, ∞). In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞) V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 13 / 28
  • 35. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f (x) = x 2/3 (x + 2). Solution f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4) 2 1 The critical points are 0 and and −4/5. − × + x −1/3 0 − 0 + 5x + 4 −4/5 + 0 − × + f (x) −4/5 0 f (x) V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 14 / 28
  • 36. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f (x) = x 2/3 (x + 2). Solution f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4) 2 1 The critical points are 0 and and −4/5. − × + x −1/3 0 − 0 + 5x + 4 −4/5 + 0 − × + f (x) −4/5 0 f (x) max V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 14 / 28
  • 37. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f (x) = x 2/3 (x + 2). Solution f (x) = 3 x −1/3 (x + 2) + x 2/3 = 3 x −1/3 (5x + 4) 2 1 The critical points are 0 and and −4/5. − × + x −1/3 0 − 0 + 5x + 4 −4/5 + 0 − × + f (x) −4/5 0 f (x) max min V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 14 / 28
  • 38. Outline Recall: The Mean Value Theorem Monotonicity The Increasing/Decreasing Test Finding intervals of monotonicity The First Derivative Test Concavity Definitions Testing for Concavity The Second Derivative Test V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 15 / 28
  • 39. Concavity Definition The graph of f is called concave up on an interval I if it lies above all its tangents on I . The graph of f is called concave down on I if it lies below all its tangents on I . concave up concave down We sometimes say a concave up graph “holds water” and a concave down graph “spills water”. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 16 / 28
  • 40. Inflection points indicate a change in concavity Definition A point P on a curve y = f (x) is called an inflection point if f is continuous at P and the curve changes from concave upward to concave downward at P (or vice versa). concave up inflection point concave down V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 17 / 28
  • 41. Theorem (Concavity Test) If f (x) > 0 for all x in an interval I , then the graph of f is concave upward on I . If f (x) < 0 for all x in I , then the graph of f is concave downward on I . V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 18 / 28
  • 42. Theorem (Concavity Test) If f (x) > 0 for all x in an interval I , then the graph of f is concave upward on I . If f (x) < 0 for all x in I , then the graph of f is concave downward on I . Proof. Suppose f (x) > 0 on I . This means f is increasing on I . Let a and x be in I . The tangent line through (a, f (a)) is the graph of L(x) = f (a) + f (a)(x − a) f (x) − f (a) By MVT, there exists a c between a and x with = f (c). So x −a f (x) = f (a) + f (c)(x − a) ≥ f (a) + f (a)(x − a) = L(x) . V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 18 / 28
  • 43. Finding Intervals of Concavity I Example Find the intervals of concavity for the graph of f (x) = x 3 + x 2 . V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 19 / 28
  • 44. Finding Intervals of Concavity I Example Find the intervals of concavity for the graph of f (x) = x 3 + x 2 . Solution We have f (x) = 3x 2 + 2x, so f (x) = 6x + 2. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 19 / 28
  • 45. Finding Intervals of Concavity I Example Find the intervals of concavity for the graph of f (x) = x 3 + x 2 . Solution We have f (x) = 3x 2 + 2x, so f (x) = 6x + 2. This is negative when x < −1/3, positive when x > −1/3, and 0 when x = −1/3 V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 19 / 28
  • 46. Finding Intervals of Concavity I Example Find the intervals of concavity for the graph of f (x) = x 3 + x 2 . Solution We have f (x) = 3x 2 + 2x, so f (x) = 6x + 2. This is negative when x < −1/3, positive when x > −1/3, and 0 when x = −1/3 So f is concave down on (−∞, −1/3), concave up on (−1/3, ∞), and has an inflection point at (−1/3, 2/27) V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 19 / 28
  • 47. Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f (x) = x 2/3 (x + 2). V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
  • 48. Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f (x) = x 2/3 (x + 2). Solution 10 −1/3 4 −4/3 2 −4/3 f (x) = x − x = x (5x − 2) 9 9 9 V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
  • 49. Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f (x) = x 2/3 (x + 2). Solution 10 −1/3 4 −4/3 2 −4/3 f (x) = x − x = x (5x − 2) 9 9 9 The second derivative f (x) is not defined at 0 V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
  • 50. Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f (x) = x 2/3 (x + 2). Solution 10 −1/3 4 −4/3 2 −4/3 f (x) = x − x = x (5x − 2) 9 9 9 The second derivative f (x) is not defined at 0 Otherwise, x −4/3 is always positive, so the concavity is determined by the 5x − 2 factor V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
  • 51. Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f (x) = x 2/3 (x + 2). Solution 10 −1/3 4 −4/3 2 −4/3 f (x) = x − x = x (5x − 2) 9 9 9 The second derivative f (x) is not defined at 0 Otherwise, x −4/3 is always positive, so the concavity is determined by the 5x − 2 factor So f is concave down on (−∞, 0], concave down on [0, 2/5), concave up on (2/5, ∞), and has an inflection point when x = 2/5 V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
  • 52. The Second Derivative Test Theorem (The Second Derivative Test) Let f , f , and f be continuous on [a, b]. Let c be be a point in (a, b) with f (c) = 0. If f (c) < 0, then c is a local maximum. If f (c) > 0, then c is a local minimum. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 21 / 28
  • 53. The Second Derivative Test Theorem (The Second Derivative Test) Let f , f , and f be continuous on [a, b]. Let c be be a point in (a, b) with f (c) = 0. If f (c) < 0, then c is a local maximum. If f (c) > 0, then c is a local minimum. Remarks If f (c) = 0, the second derivative test is inconclusive (this does not mean c is neither; we just don’t know yet). We look for zeroes of f and plug them into f to determine if their f values are local extreme values. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 21 / 28
  • 54. Proof of the Second Derivative Test Proof. Suppose f (c) = 0 and f (c) > 0. Since f is continuous, f (x) > 0 for all x sufficiently close to c. So f is increasing on an interval containing c. Since f (c) = 0 and f is increasing, f (x) < 0 for x close to c and less than c, and f (x) > 0 for x close to c and more than c. This means f changes sign from negative to positive at c, which means (by the First Derivative Test) that f has a local minimum at c. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 22 / 28
  • 55. Using the Second Derivative Test I Example Find the local extrema of f (x) = x 3 + x 2 . V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
  • 56. Using the Second Derivative Test I Example Find the local extrema of f (x) = x 3 + x 2 . Solution f (x) = 3x 2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
  • 57. Using the Second Derivative Test I Example Find the local extrema of f (x) = x 3 + x 2 . Solution f (x) = 3x 2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3. Remember f (x) = 6x + 2 V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
  • 58. Using the Second Derivative Test I Example Find the local extrema of f (x) = x 3 + x 2 . Solution f (x) = 3x 2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3. Remember f (x) = 6x + 2 Since f (−2/3) = −2 < 0, −2/3 is a local maximum. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
  • 59. Using the Second Derivative Test I Example Find the local extrema of f (x) = x 3 + x 2 . Solution f (x) = 3x 2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3. Remember f (x) = 6x + 2 Since f (−2/3) = −2 < 0, −2/3 is a local maximum. Since f (0) = 2 > 0, 0 is a local minimum. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
  • 60. Using the Second Derivative Test II Example Find the local extrema of f (x) = x 2/3 (x + 2) V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
  • 61. Using the Second Derivative Test II Example Find the local extrema of f (x) = x 2/3 (x + 2) Solution 1 Remember f (x) = x −1/3 (5x + 4) which is zero when x = −4/5 3 V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
  • 62. Using the Second Derivative Test II Example Find the local extrema of f (x) = x 2/3 (x + 2) Solution 1 Remember f (x) = x −1/3 (5x + 4) which is zero when x = −4/5 3 10 Remember f (x) = x −4/3 (5x − 2), which is negative when 9 x = −4/5 V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
  • 63. Using the Second Derivative Test II Example Find the local extrema of f (x) = x 2/3 (x + 2) Solution 1 Remember f (x) = x −1/3 (5x + 4) which is zero when x = −4/5 3 10 Remember f (x) = x −4/3 (5x − 2), which is negative when 9 x = −4/5 So x = −4/5 is a local maximum. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
  • 64. Using the Second Derivative Test II Example Find the local extrema of f (x) = x 2/3 (x + 2) Solution 1 Remember f (x) = x −1/3 (5x + 4) which is zero when x = −4/5 3 10 Remember f (x) = x −4/3 (5x − 2), which is negative when 9 x = −4/5 So x = −4/5 is a local maximum. Notice the Second Derivative Test doesn’t catch the local minimum x = 0 since f is not differentiable there. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
  • 65. Using the Second Derivative Test II: Graph Graph of f (x) = x 2/3 (x + 2): y (−4/5, 1.03413) (2/5, 1.30292) x (−2, 0) (0, 0) V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 25 / 28
  • 66. When the second derivative is zero At inflection points c, if f is differentiable at c, then f (c) = 0 Is it necessarily true, though? V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 26 / 28
  • 67. When the second derivative is zero At inflection points c, if f is differentiable at c, then f (c) = 0 Is it necessarily true, though? Consider these examples: f (x) = x 4 g (x) = −x 4 h(x) = x 3 V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 26 / 28
  • 68. When first and second derivative are zero function derivatives graph type f (x) = 4x 3 , f (0) = 0 f (x) = x 4 min f (x) = 12x 2 , f (0) = 0 g (x) = −4x 3 , g (0) = 0 g (x) = −x 4 max g (x) = −12x 2 , g (0) = 0 h (x) = 3x 2 , h (0) = 0 h(x) = x 3 infl. h (x) = 6x, h (0) = 0 V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 27 / 28
  • 69. When the second derivative is zero At inflection points c, if f is differentiable at c, then f (c) = 0 Is it necessarily true, though? Consider these examples: f (x) = x 4 g (x) = −x 4 h(x) = x 3 All of them have critical points at zero with a second derivative of zero. But the first has a local min at 0, the second has a local max at 0, and the third has an inflection point at 0. This is why we say 2DT has nothing to say when f (c) = 0. V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 28 / 28
  • 70. What have we learned today? Concepts: Mean Value Theorem, monotonicity, concavity Facts: derivatives can detect monotonicity and concavity Techniques for drawing curves: the Increasing/Decreasing Test and the Concavity Test Techniques for finding extrema: the First Derivative Test and the Second Derivative Test V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 29 / 28
  • 71. What have we learned today? Concepts: Mean Value Theorem, monotonicity, concavity Facts: derivatives can detect monotonicity and concavity Techniques for drawing curves: the Increasing/Decreasing Test and the Concavity Test Techniques for finding extrema: the First Derivative Test and the Second Derivative Test Next time: Graphing functions! V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 29 / 28