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# Lesson20 -derivatives_and_the_shape_of_curves_021_slides

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## Lesson20 -derivatives_and_the_shape_of_curves_021_slidesPresentation Transcript

• Section 4.2 Derivatives and the Shapes of Curves V63.0121.021, Calculus I New York University . November 16, 2010 Announcements Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7 There is class on November 23. The homework is due on November 24. Turn in homework to my mailbox or bring to class on November 23. . . . . . .
• Announcements Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7 There is class on November 23. The homework is due on November 24. Turn in homework to my mailbox or bring to class on November 23. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 2 / 32
• Objectives Use the derivative of a function to determine the intervals along which the function is increasing or decreasing (The Increasing/Decreasing Test) Use the First Derivative Test to classify critical points of a function as local maxima, local minima, or neither. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 3 / 32
• Objectives Use the second derivative of a function to determine the intervals along which the graph of the function is concave up or concave down (The Concavity Test) Use the first and second derivative of a function to classify critical points as local maxima or local minima, when applicable (The Second Derivative Test) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 4 / 32
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 5 / 32
• 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 b f(b) − f(a) = f′ (c). . b−a a . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 6 / 32
• 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 b f(b) − f(a) = f′ (c). . b−a a . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 6 / 32
• 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 b f(b) − f(a) = f′ (c). . b−a a . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 6 / 32
• 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 b f(b) − f(a) = f′ (c). . b−a a Another way to put this is that there exists a point c such that f(b) = f(a) + f′ (c)(b − a) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 6 / 32
• 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)(y − x) So f(y) = f(x). Since this is true for all x and y in (a, b), then f is constant. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 7 / 32
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 8 / 32
• What does it mean for a function to be increasing? Definition A function f is increasing on the interval I if f(x) < f(y) whenever x and y are two points in I with x < y. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 9 / 32
• What does it mean for a function to be increasing? Definition A function f is increasing on the interval I if f(x) < f(y) whenever x and y are two points in I with x < y. An increasing function “preserves order.” I could be bounded or infinite, open, closed, or half-open/half-closed. Write your own definition (mutatis mutandis) of decreasing, nonincreasing, nondecreasing . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 9 / 32
• The Increasing/Decreasing Test Theorem (The Increasing/Decreasing Test) If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0 on an interval, then f is decreasing on that interval. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 10 / 32
• The Increasing/Decreasing Test Theorem (The Increasing/Decreasing Test) If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0 on an interval, then f is decreasing on that interval. Proof. It works the same as the last theorem. Assume f′ (x) > 0 on an interval I. Pick two points x and y in I 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)(y − x) > 0. So f(y) > f(x). . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 10 / 32
• Finding intervals of monotonicity I Example Find the intervals of monotonicity of f(x) = 2x − 5. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 11 / 32
• 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 (−∞, ∞). . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 11 / 32
• 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). . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 11 / 32
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 11 / 32
• Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 12 / 32
• 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 12 / 32
• 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: − 0. + f′ 0 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 12 / 32
• 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: − 0. + f′ ↘ 0 ↗ f . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 12 / 32
• 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: − 0. + f′ ↘ 0 ↗ f So f is decreasing on (−∞, 0) and increasing on (0, ∞). . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 12 / 32
• 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: − 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, ∞) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 12 / 32
• Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 13 / 32
• 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 = 3 x−1/3 (5x + 4) 3 1 The critical points are 0 and and −4/5. − ×. + x−1/3 0 − 0 + 5x + 4 −4/5 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 13 / 32
• 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 = 3 x−1/3 (5x + 4) 3 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) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 13 / 32
• 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 = 3 x−1/3 (5x + 4) 3 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) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 13 / 32
• 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 = 3 x−1/3 (5x + 4) 3 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) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 13 / 32
• 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 = 3 x−1/3 (5x + 4) 3 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) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 13 / 32
• 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 = 3 x−1/3 (5x + 4) 3 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) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 13 / 32
• 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 = 3 x−1/3 (5x + 4) 3 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) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 13 / 32
• 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 = 3 x−1/3 (5x + 4) 3 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) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 13 / 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′ changes from positive to negative at c, then c is a local maximum. If f′ changes from negative to positive at c, then c is a local minimum. If f′ (x) has the same sign on either side of c, then c is not a local extremum. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 14 / 32
• 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: − 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, ∞) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 15 / 32
• 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: − 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, ∞) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 15 / 32
• 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 = 3 x−1/3 (5x + 4) 3 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) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 16 / 32
• 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 = 3 x−1/3 (5x + 4) 3 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 16 / 32
• 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 = 3 x−1/3 (5x + 4) 3 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 16 / 32
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 17 / 32
• Concavity Definition The graph of f is called concave upwards on an interval if it lies above all its tangents on that interval. The graph of f is called concave downwards on an interval if it lies below all its tangents on that interval. . . concave up concave down We sometimes say a concave up graph “holds water” and a concave down graph “spills water”. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 18 / 32
• Synonyms for concavity Remark “concave up” = “concave upwards” = “convex” “concave down” = “concave downwards” = “concave” . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 19 / 32
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 20 / 32
• Theorem (Concavity Test) If f′′ (x) > 0 for all x in an interval, then the graph of f is concave upward on that interval. If f′′ (x) < 0 for all x in an interval, then the graph of f is concave downward on that interval. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 21 / 32
• Theorem (Concavity Test) If f′′ (x) > 0 for all x in an interval, then the graph of f is concave upward on that interval. If f′′ (x) < 0 for all x in an interval, then the graph of f is concave downward on that interval. Proof. Suppose f′′ (x) > 0 on the interval I (which could be infinite). 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) By MVT, there exists a c between a and x with f(x) = f(a) + f′ (c)(x − a) Since f′ is increasing, f(x) > L(x). . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 21 / 32
• Finding Intervals of Concavity I Example Find the intervals of concavity for the graph of f(x) = x3 + x2 . . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 22 / 32
• Finding Intervals of Concavity I 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 22 / 32
• Finding Intervals of Concavity I 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 22 / 32
• Finding Intervals of Concavity I 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 the open interval (−∞, −1/3), concave up on the open interval (−1/3, ∞), and has an inflection point at the point (−1/3, 2/27) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 22 / 32
• Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 23 / 32
• Finding Intervals of Concavity II 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 We have f′′ (x) = x − x = x (5x − 2). 9 9 9 + × . + x−4/3 0 − 0 + 5x − 2 2/5 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 23 / 32
• Finding Intervals of Concavity II 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 We have f′′ (x) = x − x = x (5x − 2). 9 9 9 + × . + x−4/3 0 − 0 + 5x − 2 2/5 × 0 f′′ (x) 0 2/5 f(x) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 23 / 32
• Finding Intervals of Concavity II 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 We have f′′ (x) = x − x = x (5x − 2). 9 9 9 + × . + x−4/3 0 − 0 + 5x − 2 2/5 −− × 0 f′′ (x) 0 2/5 f(x) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 23 / 32
• Finding Intervals of Concavity II 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 We have f′′ (x) = x − x = x (5x − 2). 9 9 9 + ×. + x−4/3 0 − 0 + 5x − 2 2/5 −− ×−− 0 f′′ (x) 0 2/5 f(x) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 23 / 32
• Finding Intervals of Concavity II 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 We have f′′ (x) = x − x = x (5x − 2). 9 9 9 + ×. + x−4/3 0 − 0 + 5x − 2 2/5 −− ×−− 0 ++ f′′ (x) 0 2/5 f(x) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 23 / 32
• Finding Intervals of Concavity II 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 We have f′′ (x) = x − x = x (5x − 2). 9 9 9 + ×. + x−4/3 0 − 0 + 5x − 2 2/5 −− ×−− 0 ++ f′′ (x) ⌢ 0 2/5 f(x) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 23 / 32
• Finding Intervals of Concavity II 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 We have f′′ (x) = x − x = x (5x − 2). 9 9 9 + ×. + x−4/3 0 − 0 + 5x − 2 2/5 −− ×−− 0 ++ f′′ (x) ⌢ 0 ⌢ 2/5 f(x) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 23 / 32
• Finding Intervals of Concavity II 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 We have f′′ (x) = x − x = x (5x − 2). 9 9 9 + ×. + x−4/3 0 − 0 + 5x − 2 2/5 −− ×−− 0 ++ f′′ (x) ⌢ 0 ⌢ 2/5 ⌣ f(x) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 23 / 32
• Finding Intervals of Concavity II 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 We have f′′ (x) = x − x = x (5x − 2). 9 9 9 + ×. + x−4/3 0 − 0 + 5x − 2 2/5 −− ×−− 0 ++ f′′ (x) ⌢ 0 ⌢ 2/5 ⌣ f(x) IP . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 23 / 32
• 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 24 / 32
• 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 24 / 32
• Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. + f′′ = (f′ )′ . c f′ 0 f′ c f . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 25 / 32
• Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is continuous, f′′ = (f′ )′ + f′′ (x) > 0 for all x . sufficiently close to c. c f′ 0 f′ c f . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 25 / 32
• Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is continuous, f′′ = (f′ )′ + + f′′ (x) > 0 for all x . sufficiently close to c. c f′ 0 f′ c f . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 25 / 32
• Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is continuous, f′′ = (f′ )′ + + + f′′ (x) > 0 for all x . sufficiently close to c. c f′ 0 f′ c f . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 25 / 32
• Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is continuous, f′′ = (f′ )′ + + + f′′ (x) > 0 for all x . sufficiently close to c. c f′ 0 f′ Since f′′ = (f′ )′ , we know f′ c f is increasing near c. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 25 / 32
• Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is continuous, f′′ = (f′ )′ + + + f′′ (x) > 0 for all x . sufficiently close to c. ↗ c f′ 0 f′ Since f′′ = (f′ )′ , we know f′ c f is increasing near c. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 25 / 32
• Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is continuous, f′′ = (f′ )′ + + + f′′ (x) > 0 for all x . sufficiently close to c. ↗ c ↗ f′ 0 f′ Since f′′ = (f′ )′ , we know f′ c f is increasing near c. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 25 / 32
• Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is continuous, f′′ = (f′ )′ + + + f′′ (x) > 0 for all x . sufficiently close to c. ↗ c ↗ f′ 0 f′ Since f′′ = (f′ )′ , we know f′ c f is increasing near 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 25 / 32
• Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is continuous, f′′ = (f′ )′ + + + f′′ (x) > 0 for all x . sufficiently close to c. ↗ c ↗ f′ − 0 f′ Since f′′ = (f′ )′ , we know f′ c f is increasing near 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 25 / 32
• Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is continuous, f′′ = (f′ )′ + + + f′′ (x) > 0 for all x . sufficiently close to c. ↗ c ↗ f′ − 0 + f′ Since f′′ = (f′ )′ , we know f′ c f is increasing near 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 25 / 32
• Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is continuous, f′′ = (f′ )′ + + + f′′ (x) > 0 for all x . sufficiently close to c. ↗ c ↗ f′ − 0 + f′ Since f′′ = (f′ )′ , we know f′ c f is increasing near 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 25 / 32
• Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is continuous, f′′ = (f′ )′ + + + f′′ (x) > 0 for all x . sufficiently close to c. ↗ c ↗ f′ − 0 + f′ Since f′′ = (f′ )′ , we know f′ ↘ c f is increasing near 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 25 / 32
• Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is continuous, f′′ = (f′ )′ + + + f′′ (x) > 0 for all x . sufficiently close to c. ↗ c ↗ f′ − 0 + f′ Since f′′ = (f′ )′ , we know f′ ↘ c ↗ f is increasing near 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 25 / 32
• Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is continuous, f′′ = (f′ )′ + + + f′′ (x) > 0 for all x . sufficiently close to c. ↗ c ↗ f′ − 0 + f′ Since f′′ = (f′ )′ , we know f′ ↘ c ↗ f is increasing near c. min 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 25 / 32
• Using the Second Derivative Test I Example Find the local extrema of f(x) = x3 + x2 . . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 26 / 32
• Using the Second Derivative Test I 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 26 / 32
• Using the Second Derivative Test I 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 26 / 32
• Using the Second Derivative Test I 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 26 / 32
• Using the Second Derivative Test I 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 26 / 32
• Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 27 / 32
• Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solution 1 −1/3 Remember f′ (x) = x (5x + 4) which is zero when x = −4/5 3 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 27 / 32
• Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solution 1 −1/3 Remember f′ (x) = x (5x + 4) which is zero when x = −4/5 3 10 −4/3 Remember f′′ (x) = x (5x − 2), which is negative when 9 x = −4/5 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 27 / 32
• Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solution 1 −1/3 Remember f′ (x) = x (5x + 4) which is zero when x = −4/5 3 10 −4/3 Remember f′′ (x) = x (5x − 2), which is negative when 9 x = −4/5 So x = −4/5 is a local maximum. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 27 / 32
• Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solution 1 −1/3 Remember f′ (x) = x (5x + 4) which is zero when x = −4/5 3 10 −4/3 Remember f′′ (x) = x (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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 27 / 32
• Using the Second Derivative Test II: Graph Graph of f(x) = x2/3 (x + 2): y (−4/5, 1.03413) (2/5, 1.30292) . x (−2, 0) (0, 0) . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 28 / 32
• When the second derivative is zero Remark At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0 If f′′ (c) = 0, must f have an inflection point at c? . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 29 / 32
• When the second derivative is zero Remark At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0 If f′′ (c) = 0, must f have an inflection point at c? Consider these examples: f(x) = x4 g(x) = −x4 h(x) = x3 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 29 / 32
• When first and second derivative are zero function derivatives graph type f′ (x) = 4x3 , f′ (0) = 0 f(x) = x4 min f′′ (x) = 12x2 , f′′ (0) = 0 . . g′ (x) = −4x3 , g′ (0) = 0 g(x) = −x4 max g′′ (x) = −12x2 , g′′ (0) = 0 h′ (x) = 3x2 , h′ (0) = 0 h(x) = x3 infl. h′′ (x) = 6x, h′′ (0) = 0 . . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 30 / 32
• When the second derivative is zero Remark At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0 If f′′ (c) = 0, must f have an inflection point at c? Consider these examples: f(x) = x4 g(x) = −x4 h(x) = x3 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 31 / 32
• 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 2010 32 / 32