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- 1. Section 4.3 Derivatives and the Shapes of Curves V63.0121, Calculus I March 25-26, 2009 . . Image credit: cobalt123 . . . . . .
- 2. Outline Monotonicity The Increasing/Decreasing Test Finding intervals of monotonicity The First Derivative Test Concavity Deﬁnitions Testing for Concavity The Second Derivative Test . . . . . .
- 3. 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). . . . . . .
- 4. 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. . . . . . .
- 5. Finding intervals of monotonicity I Example Find the intervals of monotonicity of f(x) = 2x − 5. . . . . . .
- 6. 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 (−∞, ∞). . . . . . .
- 7. 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). . . . . . .
- 8. 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 . . . . . .
- 9. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. . . . . . .
- 10. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solution f′ (x) = 2x, which is positive when x > 0 and negative when x is. . . . . . .
- 11. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solution f′ (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: .′ − f + . . 0 .. 0 . . . . . . .
- 12. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solution f′ (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: .′ − f + . . 0 .. ↘ ↗ 0 . f . . . . . . . . .
- 13. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solution f′ (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: .′ − f + . . 0 .. ↘ ↗ 0 . f . . . So f is decreasing on (−∞, 0) and increasing on (0, ∞). . . . . . .
- 14. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solution f′ (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: .′ − f + . . 0 .. ↘ ↗ 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, ∞) . . . . . .
- 15. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). . . . . . .
- 16. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − × + . .. . . −1/3 x 0 . − + . . 0 .. .x+4 5 − . 4/5 . . . . . .
- 17. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − × + . .. . . −1/3 x 0 . − + . . 0 .. .x+4 5 − . 4/5 . ′ (x) × f .. 0 .. − 0 . . 4/5 . (x) f . . . . . .
- 18. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − × + . .. . . −1/3 x 0 . − + . . 0 .. .x+4 5 − . 4/5 . ′ (x) 0−× f + + . .. . . . . ↗ − ↘. ↗ 0 . . 4/5 . . . (x) f . . . . . .
- 19. 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. . . . . . .
- 20. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solution f′ (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: .′ − f + . . 0 .. ↘ ↗ 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, ∞) . . . . . .
- 21. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solution f′ (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: .′ − f + . . 0 .. ↘ ↗ 0 . f . . . m . in So f is decreasing on (−∞, 0) and increasing on (0, ∞). In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞) . . . . . .
- 22. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − × + . .. . . −1/3 x 0 . − + . . 0 .. .x+4 5 − . 4/5 . ′ (x) 0−× f + + . .. . . . . ↗ − ↘. ↗ 0 . . 4/5 . . . (x) f . . . . . .
- 23. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − × + . .. . . −1/3 x 0 . − + . . 0 .. .x+4 5 − . 4/5 . ′ (x) 0−× f + + . .. . . . . ↗ − ↘. ↗ 0 . . 4/5 . . . (x) f m . ax m . in . . . . . .
- 24. Outline Monotonicity The Increasing/Decreasing Test Finding intervals of monotonicity The First Derivative Test Concavity Deﬁnitions Testing for Concavity The Second Derivative Test . . . . . .
- 25. Deﬁnition The graph of f is called concave up on and 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”. . . . . . .
- 26. Deﬁnition A point P on a curve y = f(x) is called an inﬂection point if f is continuous there and the curve changes from concave upward to concave downward at P (or vice versa). . concave up i .nﬂection point . . . concave down . . . . . .
- 27. Theorem (Concavity Test) If f′′ (x) > 0 for all x in 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 . . . . . .
- 28. Theorem (Concavity Test) If f′′ (x) > 0 for all x in 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) = f′ (b). So By MVT, there exists a b between a and x with x−a f(x) = f(a) + f′ (b)(x − a) ≥ f(a) + f′ (a)(x − a) = L(x) . . . . . .
- 29. Example Find the intervals of concavity for the graph of f(x) = x3 + x2 . . . . . . .
- 30. Example Find the intervals of concavity for the graph of f(x) = x3 + x2 . Solution We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2. . . . . . .
- 31. Example Find the intervals of concavity for the graph of f(x) = x3 + x2 . Solution We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2. This is negative when x < −1/3, positive when x > −1/3, and 0 when x = −1/3 . . . . . .
- 32. Example Find the intervals of concavity for the graph of f(x) = x3 + x2 . Solution We have f′ (x) = 3x2 + 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) . . . . . .
- 33. Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). . . . . . .
- 34. Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution 10 −1/3 4 −4/3 2 −4/3 f′′ (x) = −x (5x − 2) =x x 9 9 9 . . . . . .
- 35. Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution 10 −1/3 4 −4/3 2 −4/3 f′′ (x) = −x (5x − 2) =x x 9 9 9 x−4/3 is always positive, so the concavity is determined by the 5x − 2 factor . . . . . .
- 36. Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution 10 −1/3 4 −4/3 2 −4/3 f′′ (x) = −x (5x − 2) =x x 9 9 9 x−4/3 is always positive, so the concavity is determined by the 5x − 2 factor So f is concave down on (−∞, 2/5), concave up on (2/5, ∞), and has an inﬂection point when x = 2/5 . . . . . .
- 37. 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 f(c) is a local maximum. If f′′ (c) > 0, then f(c) is a local minimum. If f′′ (c) = 0, the second derivative test is inconclusive (this does not mean c is neither; we just don’t know yet). . . . . . .
- 38. Example Find the local extrema of f(x) = x3 + x2 . . . . . . .
- 39. Example Find the local extrema of f(x) = x3 + x2 . Solution f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3. . . . . . .
- 40. Example Find the local extrema of f(x) = x3 + x2 . Solution f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3. Remember f′′ (x) = 6x + 2 . . . . . .
- 41. Example Find the local extrema of f(x) = x3 + x2 . Solution f′ (x) = 3x2 + 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. . . . . . .
- 42. Example Find the local extrema of f(x) = x3 + x2 . Solution f′ (x) = 3x2 + 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. . . . . . .
- 43. Example Find the local extrema of f(x) = x2/3 (x + 2) . . . . . .
- 44. Example Find the local extrema of f(x) = x2/3 (x + 2) Solution 1 −1/3 Remember f′ (x) = (5x + 4) which is zero when x = −4/5 x 3 . . . . . .
- 45. Example Find the local extrema of f(x) = x2/3 (x + 2) Solution 1 −1/3 Remember f′ (x) = (5x + 4) which is zero when x = −4/5 x 3 10 −4/3 Remember f′′ (x) = (5x − 2), which is negative when x 9 x=− 4/5 . . . . . .
- 46. Example Find the local extrema of f(x) = x2/3 (x + 2) Solution 1 −1/3 Remember f′ (x) = (5x + 4) which is zero when x = −4/5 x 3 10 −4/3 Remember f′′ (x) = (5x − 2), which is negative when x 9 x=− 4/5 So x = −4/5 is a local maximum. . . . . . .
- 47. Example Find the local extrema of f(x) = x2/3 (x + 2) Solution 1 −1/3 Remember f′ (x) = (5x + 4) which is zero when x = −4/5 x 3 10 −4/3 Remember f′′ (x) = (5x − 2), which is negative when x 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. . . . . . .
- 48. Graph Graph of f(x) = x2/3 (x + 2): y . . −4/5, 1.03413) ( . . ( . 2/5, 1.30292) . . x . . −2, 0) ( ( . 0, 0) . . . . . .
- 49. 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? . . . . . .
- 50. 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: g(x) = −x4 f(x) = x4 h(x) = x3 . . . . . .
- 51. 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: g(x) = −x4 f(x) = x4 h(x) = x3 All of them have f′′ (0) = 0. 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. . . . . . .
- 52. Summary 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 . . . . . .
- 53. Summary 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 week: Graphing functions . . . . . .

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