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Lesson 20: Derivatives and the Shape of Curves (Section 041 slides)

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The Mean Value Theorem gives us tests for determining the shape of curves between critical points.

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.. Section 4.2 Derivatives and the Shapes of Curves V63.0121.041, Calculus I New York University November 15, 2010 Announcements Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7 There is class on November 24 . . . . . .

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. . . . . . Announcements Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7 There is class on November 24 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 2 / 32

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. . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 3 / 32

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. . . . . . 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) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 4 / 32

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. . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 5 / 32

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. . . . . . 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 − a = f′ (c). ... a .. b V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32

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. . . . . . 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 − a = f′ (c). ... a .. b V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32

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. . . . . . 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 − a = f′ (c). ... a .. b .. c V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32

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. . . . . . 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 − a = f′ (c). ... a .. b .. c Another way to put this is that there exists a point c such that f(b) = f(a) + f′ (c)(b − a) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32

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. . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 7 / 32

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. . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 8 / 32

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. . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 9 / 32

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. . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 9 / 32

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. . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 10 / 32

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. . . . . . 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). V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 10 / 32

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. . . . . . Finding intervals of monotonicity I Example Find the intervals of monotonicity of f(x) = 2x − 5. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32

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. . . . . . 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 (−∞, ∞). V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32

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. . . . . . 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). V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32

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. . . . . . 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 Since f′ (x) = 1 1 + x2 is always positive, f(x) is always increasing. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32

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. . . . . . Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32

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. . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32

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. . . . . . 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. + V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32

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. . . . . . 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′ . f .− . ↘ .. 0 .0. +. ↗ V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32

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. . . . . . 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′ . f .− . ↘ .. 0 .0. +. ↗ So f is decreasing on (−∞, 0) and increasing on (0, ∞). V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32

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. . . . . . 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′ . f .− . ↘ .. 0 .0. +. ↗ So f is decreasing on (−∞, 0) and increasing on (0, ∞). In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32

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. . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32

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. . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32

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. . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32

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. . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32

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. . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + .. − V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32

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. . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + .. − .. + V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32

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. . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + . ↗ .. − .. + V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32

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. . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + . ↗ .. − . ↘ .. + V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32

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. . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + . ↗ .. − . ↘ .. + . ↗ V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32

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. . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 14 / 32

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. . . . . . 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′ . f .− . ↘ .. 0 .0. +. ↗ So f is decreasing on (−∞, 0) and increasing on (0, ∞). In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 15 / 32

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. . . . . . 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′ . f .− . ↘ .. 0 .0. +. ↗ . 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, ∞) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 15 / 32

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. . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + . ↗ .. − . ↘ .. + . ↗ V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 16 / 32

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. . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + . ↗ .. − . ↘ .. + . ↗ . max V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 16 / 32

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. . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + . ↗ .. − . ↘ .. + . ↗ . max . min V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 16 / 32

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. . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 17 / 32

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. . . . . . 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”. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 18 / 32

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. . . . . . Synonyms for concavity Remark “concave up” = “concave upwards” = “convex” “concave down” = “concave downwards” = “concave” V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 19 / 32

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. . . . . . 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 down . concave up .. inflection point V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 20 / 32

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. . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 21 / 32

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. . . . . . 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). V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 21 / 32

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. . . . . . Finding Intervals of Concavity I Example Find the intervals of concavity for the graph of f(x) = x3 + x2 . V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32

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. . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32

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. . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32

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. . . . . . 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) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32

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. . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32

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. . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32

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. . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32

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. . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × .. −− V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32

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. . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × .. −− .. −− V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32

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. . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × .. −− .. −− .. ++ V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32

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. . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × .. −− .. −− .. ++ . ⌢ V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32

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. . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × .. −− .. −− .. ++ . ⌢ . ⌢ V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32

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. . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × .. −− .. −− .. ++ . ⌢ . ⌢ . ⌣ V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32

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. . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × .. −− .. −− .. ++ . ⌢ . ⌢ . ⌣ . IP V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32

61

. . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 24 / 32

62

. . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 24 / 32

63

. . . . . . Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. .. f′′ = (f′ )′ . f′ ... c .+. f′ . f ... c . 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32

64

. . . . . . 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. .. f′′ = (f′ )′ . f′ ... c .+. f′ . f ... c . 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32

65

. . . . . . 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. .. f′′ = (f′ )′ . f′ ... c .+..+ . f′ . f ... c . 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32

66

. . . . . . 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. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. f′ . f ... c . 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32

67

. . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. f′ . f ... c . 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32

68

. . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . f′ . f ... c . 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32

69

. . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32

70

. . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32

71

. . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 .. − 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32

72

. . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 .. − .. + 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32

73

. . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 .. − .. + 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32

74

. . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 .. − .. + . ↘ 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32

75

. . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 .. − .. + . ↘ . ↗ 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32

76

. . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 .. − .. + . ↘ . ↗ . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32

77

. . . . . . Using the Second Derivative Test I Example Find the local extrema of f(x) = x3 + x2 . V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32

78

. . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32

79

. . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32

80

. . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32

81

. . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32

82

. . . . . . Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32

83

. . . . . . Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solution Remember f′ (x) = 1 3 x−1/3 (5x + 4) which is zero when x = −4/5 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32

84

. . . . . . Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solution Remember f′ (x) = 1 3 x−1/3 (5x + 4) which is zero when x = −4/5 Remember f′′ (x) = 10 9 x−4/3 (5x − 2), which is negative when x = −4/5 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32

85

. . . . . . Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solution Remember f′ (x) = 1 3 x−1/3 (5x + 4) which is zero when x = −4/5 Remember f′′ (x) = 10 9 x−4/3 (5x − 2), which is negative when x = −4/5 So x = −4/5 is a local maximum. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32

86

. . . . . . Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solution Remember f′ (x) = 1 3 x−1/3 (5x + 4) which is zero when x = −4/5 Remember f′′ (x) = 10 9 x−4/3 (5x − 2), which is negative when 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32

87

. . . . . . Using the Second Derivative Test II: Graph Graph of f(x) = x2/3 (x + 2): .. x. y .. (−4/5, 1.03413) .. (0, 0) .. (2/5, 1.30292) .. (−2, 0) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 28 / 32

88

. . . . . . 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? V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 29 / 32

89

. . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 29 / 32

90

. . . . . . When first and second derivative are zero function derivatives graph type f(x) = x4 f′ (x) = 4x3, f′ (0) = 0 . min f′′ (x) = 12x2, f′′ (0) = 0 g(x) = −x4 g′(x) = −4x3, g′(0) = 0 . max g′′(x) = −12x2, g′′(0) = 0 h(x) = x3 h′ (x) = 3x2, h′ (0) = 0 . infl. h′′ (x) = 6x, h′′ (0) = 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 30 / 32

91

. . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 31 / 32

92

. . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 32 / 32

More Related Content

Lesson 20: Derivatives and the Shape of Curves (Section 041 slides)

  • 1. .. Section 4.2 Derivatives and the Shapes of Curves V63.0121.041, Calculus I New York University November 15, 2010 Announcements Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7 There is class on November 24 . . . . . .
  • 2. . . . . . . Announcements Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7 There is class on November 24 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 2 / 32
  • 3. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 3 / 32
  • 4. . . . . . . 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) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 4 / 32
  • 5. . . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 5 / 32
  • 6. . . . . . . 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 − a = f′ (c). ... a .. b V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32
  • 7. . . . . . . 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 − a = f′ (c). ... a .. b V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32
  • 8. . . . . . . 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 − a = f′ (c). ... a .. b .. c V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32
  • 9. . . . . . . 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 − a = f′ (c). ... a .. b .. c Another way to put this is that there exists a point c such that f(b) = f(a) + f′ (c)(b − a) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32
  • 10. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 7 / 32
  • 11. . . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 8 / 32
  • 12. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 9 / 32
  • 13. . . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 9 / 32
  • 14. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 10 / 32
  • 15. . . . . . . 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). V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 10 / 32
  • 16. . . . . . . Finding intervals of monotonicity I Example Find the intervals of monotonicity of f(x) = 2x − 5. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32
  • 17. . . . . . . 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 (−∞, ∞). V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32
  • 18. . . . . . . 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). V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32
  • 19. . . . . . . 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 Since f′ (x) = 1 1 + x2 is always positive, f(x) is always increasing. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32
  • 20. . . . . . . Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
  • 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
  • 22. . . . . . . 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. + V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
  • 23. . . . . . . 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′ . f .− . ↘ .. 0 .0. +. ↗ V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
  • 24. . . . . . . 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′ . f .− . ↘ .. 0 .0. +. ↗ So f is decreasing on (−∞, 0) and increasing on (0, ∞). V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
  • 25. . . . . . . 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′ . f .− . ↘ .. 0 .0. +. ↗ So f is decreasing on (−∞, 0) and increasing on (0, ∞). In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
  • 26. . . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
  • 27. . . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
  • 28. . . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
  • 29. . . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
  • 30. . . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + .. − V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
  • 31. . . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + .. − .. + V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
  • 32. . . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + . ↗ .. − .. + V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
  • 33. . . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + . ↗ .. − . ↘ .. + V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
  • 34. . . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + . ↗ .. − . ↘ .. + . ↗ V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
  • 35. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 14 / 32
  • 36. . . . . . . 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′ . f .− . ↘ .. 0 .0. +. ↗ So f is decreasing on (−∞, 0) and increasing on (0, ∞). In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 15 / 32
  • 37. . . . . . . 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′ . f .− . ↘ .. 0 .0. +. ↗ . 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, ∞) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 15 / 32
  • 38. . . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + . ↗ .. − . ↘ .. + . ↗ V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 16 / 32
  • 39. . . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + . ↗ .. − . ↘ .. + . ↗ . max V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 16 / 32
  • 40. . . . . . . Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 3 x−1/3 (x + 2) + x2/3 = 1 3 x−1/3 (5x + 4) The critical points are 0 and and −4/5. .. x−1/3.. 0 .×.− . +. 5x + 4 .. −4/5 . 0 . − . + . f′ (x) . f(x) .. −4/5 . 0 .. 0 . × .. + . ↗ .. − . ↘ .. + . ↗ . max . min V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 16 / 32
  • 41. . . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 17 / 32
  • 42. . . . . . . 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”. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 18 / 32
  • 43. . . . . . . Synonyms for concavity Remark “concave up” = “concave upwards” = “convex” “concave down” = “concave downwards” = “concave” V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 19 / 32
  • 44. . . . . . . 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 down . concave up .. inflection point V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 20 / 32
  • 45. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 21 / 32
  • 46. . . . . . . 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). V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 21 / 32
  • 47. . . . . . . Finding Intervals of Concavity I Example Find the intervals of concavity for the graph of f(x) = x3 + x2 . V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32
  • 48. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32
  • 49. . . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32
  • 50. . . . . . . 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) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32
  • 51. . . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
  • 52. . . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
  • 53. . . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
  • 54. . . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × .. −− V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
  • 55. . . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × .. −− .. −− V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
  • 56. . . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × .. −− .. −− .. ++ V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
  • 57. . . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × .. −− .. −− .. ++ . ⌢ V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
  • 58. . . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × .. −− .. −− .. ++ . ⌢ . ⌢ V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
  • 59. . . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × .. −− .. −− .. ++ . ⌢ . ⌢ . ⌣ V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
  • 60. . . . . . . Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution We have f′′ (x) = 10 9 x−1/3 − 4 9 x−4/3 = 2 9 x−4/3 (5x − 2). .. x−4/3.. 0 .×.+ . +. 5x − 2 .. 2/5 . 0 . − . + . f′′ (x) . f(x) .. 2/5 . 0 .. 0 . × .. −− .. −− .. ++ . ⌢ . ⌢ . ⌣ . IP V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
  • 61. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 24 / 32
  • 62. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 24 / 32
  • 63. . . . . . . Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. .. f′′ = (f′ )′ . f′ ... c .+. f′ . f ... c . 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
  • 64. . . . . . . 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. .. f′′ = (f′ )′ . f′ ... c .+. f′ . f ... c . 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
  • 65. . . . . . . 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. .. f′′ = (f′ )′ . f′ ... c .+..+ . f′ . f ... c . 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
  • 66. . . . . . . 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. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. f′ . f ... c . 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
  • 67. . . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. f′ . f ... c . 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
  • 68. . . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . f′ . f ... c . 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
  • 69. . . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
  • 70. . . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
  • 71. . . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 .. − 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
  • 72. . . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 .. − .. + 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
  • 73. . . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 .. − .. + 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
  • 74. . . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 .. − .. + . ↘ 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
  • 75. . . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 .. − .. + . ↘ . ↗ 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
  • 76. . . . . . . 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. Since f′′ = (f′ )′ , we know f′ is increasing near c. .. f′′ = (f′ )′ . f′ ... c .+..+ .. +. ↗ . ↗ . f′ . f ... c . 0 .. − .. + . ↘ . ↗ . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
  • 77. . . . . . . Using the Second Derivative Test I Example Find the local extrema of f(x) = x3 + x2 . V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
  • 78. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
  • 79. . . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
  • 80. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
  • 81. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
  • 82. . . . . . . Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
  • 83. . . . . . . Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solution Remember f′ (x) = 1 3 x−1/3 (5x + 4) which is zero when x = −4/5 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
  • 84. . . . . . . Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solution Remember f′ (x) = 1 3 x−1/3 (5x + 4) which is zero when x = −4/5 Remember f′′ (x) = 10 9 x−4/3 (5x − 2), which is negative when x = −4/5 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
  • 85. . . . . . . Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solution Remember f′ (x) = 1 3 x−1/3 (5x + 4) which is zero when x = −4/5 Remember f′′ (x) = 10 9 x−4/3 (5x − 2), which is negative when x = −4/5 So x = −4/5 is a local maximum. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
  • 86. . . . . . . Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solution Remember f′ (x) = 1 3 x−1/3 (5x + 4) which is zero when x = −4/5 Remember f′′ (x) = 10 9 x−4/3 (5x − 2), which is negative when 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
  • 87. . . . . . . Using the Second Derivative Test II: Graph Graph of f(x) = x2/3 (x + 2): .. x. y .. (−4/5, 1.03413) .. (0, 0) .. (2/5, 1.30292) .. (−2, 0) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 28 / 32
  • 88. . . . . . . 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? V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 29 / 32
  • 89. . . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 29 / 32
  • 90. . . . . . . When first and second derivative are zero function derivatives graph type f(x) = x4 f′ (x) = 4x3, f′ (0) = 0 . min f′′ (x) = 12x2, f′′ (0) = 0 g(x) = −x4 g′(x) = −4x3, g′(0) = 0 . max g′′(x) = −12x2, g′′(0) = 0 h(x) = x3 h′ (x) = 3x2, h′ (0) = 0 . infl. h′′ (x) = 6x, h′′ (0) = 0 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 30 / 32
  • 91. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 31 / 32
  • 92. . . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 32 / 32