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

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

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

1. 1. Section 4.2 Derivatives and the Shapes of Curves V63.0121.041, Calculus I New York University November 15, 2010 Announcements Quiz this week in recitation on 3.3, 3.4, 3.5, 3.7 There is class on November 24 Announcements Quiz 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 / 31 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 / 31 Notes Notes Notes 1 Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
2. 2. 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 / 31 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 / 31 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 / 31 Notes Notes Notes 2 Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
3. 3. 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 / 31 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 / 31 What does it mean for a function to be increasing? Deﬁnition A function f is increasing on (a, b) if f (x) < f (y) whenever x and y are two points in (a, b) with x < y. An increasing function “preserves order.” Write your own deﬁnition (mutatis mutandis) of decreasing, nonincreasing, nondecreasing V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 9 / 31 Notes Notes Notes 3 Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
4. 4. The Increasing/Decreasing Test Theorem (The Increasing/Decreasing Test) If f > 0 on (a, b), then f is increasing on (a, b). If f < 0 on (a, b), then f is decreasing on (a, b). Proof. It works the same as the last theorem. Pick two points x and y in (a, b) with x < y. We must show f (x) < f (y). By MVT there exists a point c in (x, y) such that f (y) − f (x) = f (c)(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 / 31 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 / 31 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 / 31 Notes Notes Notes 4 Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
5. 5. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f (x) = x2/3 (x + 2). Solution V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 31 The First Derivative Test Theorem (The First Derivative Test) Let f be continuous on [a, b] and c a critical point of f in (a, b). If f (x) > 0 on (a, c) and f (x) < 0 on (c, b), then c is a local maximum. If f (x) < 0 on (a, c) and f (x) > 0 on (c, b), then c is a local minimum. If f (x) has the same sign on (a, c) and (c, b), then c is not a local extremum. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 14 / 31 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 / 31 Notes Notes Notes 5 Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
6. 6. Concavity Deﬁnition The graph of f is called concave up on an interval I if it lies above all its tangents on I. The graph of f is called concave down on I if it lies below all its tangents on I. concave up concave down We sometimes say a concave up graph “holds water” and a concave down graph “spills water”. V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 18 / 31 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 19 / 31 Theorem (Concavity Test) If f (x) > 0 for all x in an interval I, then the graph of f is concave upward on I. If f (x) < 0 for all x in I, then the graph of f is concave downward on I. Proof. Suppose f (x) > 0 on I. This means f is increasing on I. Let a and x be in I. The tangent line through (a, f (a)) is the graph of L(x) = f (a) + f (a)(x − a) 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 20 / 31 Notes Notes Notes 6 Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
7. 7. 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 (−∞, −1/3), concave up on (−1/3, ∞), and has an inﬂection point at (−1/3, 2/27) V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 21 / 31 Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f (x) = x2/3 (x + 2). Solution V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 31 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 23 / 31 Notes Notes Notes 7 Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
8. 8. 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. 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 24 / 31 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 25 / 31 Using the Second Derivative Test II Example Find the local extrema of f (x) = x2/3 (x + 2) Solution V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 31 Notes Notes Notes 8 Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
9. 9. Using the Second Derivative Test II: Graph V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 31 When the second derivative is zero At inﬂection points c, if f is diﬀerentiable at c, then f (c) = 0 Is it necessarily true, though? Consider these examples: f (x) = 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 28 / 31 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 29 / 31 Notes Notes Notes 9 Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
10. 10. When the second derivative is zero At inﬂection points c, if f is diﬀerentiable at c, then f (c) = 0 Is it necessarily true, though? Consider these examples: f (x) = 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 30 / 31 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 31 / 31 Notes Notes Notes 10 Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010