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Lesson 21: Derivatives and the Shapes of Curves
 

Lesson 21: Derivatives and the Shapes of Curves

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We discuss the ideas of monotonicity (increasing or decreasing) and concavity (up or down) of a function. Because of the Mean Value Theorem, we can determine these characteristics using derivatives.

We discuss the ideas of monotonicity (increasing or decreasing) and concavity (up or down) of a function. Because of the Mean Value Theorem, we can determine these characteristics using derivatives.

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    Lesson 21: Derivatives and the Shapes of Curves Lesson 21: Derivatives and the Shapes of Curves Presentation Transcript

    • Section 4.2 Derivatives and the Shapes of Curves V63.0121.034, Calculus I November 11, 2009 Announcements Final Exam Friday, December 18, 2:00–3:50pm Wednesday, November 25 is a regular class day . . Image credit: cobalt123 . . . . . .
    • 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 . . . . . .
    • 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). a . b−a . . . . . .
    • 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). a . b−a . . . . . .
    • Recall: The Mean Value Theorem Theorem (The Mean Value c .. 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). a . b−a . . . . . .
    • Why the MVT is the MITC Most Important Theorem In Calculus! Theorem Let f′ = 0 on an interval (a, b). Then f is constant on (a, b). Proof. Pick any points x and y in (a, b) with x < y. Then f is continuous on [x, y] and differentiable on (x, y). By MVT there exists a point z in (x, y) such that f(y) − f(x) = f′ (z) = 0. y−x So f(y) = f(x). Since this is true for all x and y in (a, b), then f is constant. . . . . . .
    • 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 . . . . . .
    • 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. . . . . . .
    • 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 . . . . . .
    • 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). . . . . . .
    • The Increasing/Decreasing Test Theorem (The Increasing/Decreasing Test) If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b), then f is decreasing on (a, b). Proof. It works the same as the last theorem. Pick two points x and y in (a, b) with x < y. We must show f(x) < f(y). By MVT there exists a point c in (x, y) such that f(y) − f(x) = f′ (c) > 0. y−x So f(y) − f(x) = f′ (c)(y − x) > 0. . . . . . .
    • Finding intervals of monotonicity I Example Find the intervals of monotonicity of f(x) = 2x − 5. . . . . . .
    • 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 (−∞, ∞). . . . . . .
    • 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). . . . . . .
    • 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 . . . . . .
    • Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. . . . . . .
    • 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. . . . . . .
    • 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 . . . . . . .
    • 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 . . . . . . .
    • 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, ∞). . . . . . .
    • 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, ∞) . . . . . .
    • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). . . . . . .
    • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − . × .. . + . −1/3 x 0 . − . 0 .. . + .x+4 5 − . 4/5 . . . . . .
    • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − . × .. . + . −1/3 x 0 . − . 0 .. . + .x+4 5 − . 4/5 0 .. × .. .′ (x) f − . 4/5 0 . f .(x) . . . . . .
    • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − . × .. . + . −1/3 x 0 . − . 0 .. . + .x+4 5 − . 4/5 . + 0 − × .. . . . . + .′ (x) f ↗ . − ↘ . . 4/5 . 0 ↗ . f .(x) . . . . . .
    • 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. . . . . . .
    • 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, ∞) . . . . . .
    • 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 . m . in So f is decreasing on (−∞, 0) and increasing on (0, ∞). In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞) . . . . . .
    • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − . × .. . + . −1/3 x 0 . − . 0 .. . + .x+4 5 − . 4/5 . + 0 − × .. . . . . + .′ (x) f ↗ . − ↘ . . 4/5 . 0 ↗ . f .(x) . . . . . .
    • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − . × .. . + . −1/3 x 0 . − . 0 .. . + .x+4 5 − . 4/5 . + 0 − × .. . . . . + .′ (x) f ↗ . − ↘ . . 4/5 . 0 ↗ . f .(x) m . ax . in m . . . . . .
    • 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 . . . . . .
    • Definition The graph of f is called concave up on and interval I if it lies above all its tangents on I. The graph of f is called concave down on I if it lies below all its tangents on I. . . concave up concave down We sometimes say a concave up graph “holds water” and a concave down graph “spills water”. . . . . . .
    • Definition A point P on a curve y = f(x) is called an inflection point if f is continuous there and the curve changes from concave upward to concave downward at P (or vice versa). .concave up i .nflection point . . . concave down . . . . . .
    • 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 . . . . . .
    • Theorem (Concavity Test) If f′′ (x) > 0 for all x in an interval I, then the graph of f is concave upward on I If f′′ (x) < 0 for all x in I, then the graph of f is concave downward on I Proof. Suppose f′′ (x) > 0 on I. This means f′ is increasing on I. Let a and x be in I. The tangent line through (a, f(a)) is the graph of L(x) = f(a) + f′ (a)(x − a) f(x) − f(a) By MVT, there exists a c between a and x with = f′ (c). x−a So f(x) = f(a) + f′ (c)(x − a) ≥ f(a) + f′ (a)(x − a) = L(x) . . . . . .
    • Example Find the intervals of concavity for the graph of f(x) = x3 + x2 . . . . . . .
    • 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. . . . . . .
    • 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 . . . . . .
    • 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) . . . . . .
    • Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). . . . . . .
    • Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution 10 −1/3 4 −4/3 2 −4/3 f′′ (x) = x − x = x (5x − 2) 9 9 9 . . . . . .
    • Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution 10 −1/3 4 −4/3 2 −4/3 f′′ (x) = x − x = x (5x − 2) 9 9 9 The second derivative f′′ (x) is not defined at 0 . . . . . .
    • Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution 10 −1/3 4 −4/3 2 −4/3 f′′ (x) = x − x = x (5x − 2) 9 9 9 The second derivative f′′ (x) is not defined at 0 Otherwise, x−4/3 is always positive, so the concavity is determined by the 5x − 2 factor . . . . . .
    • Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution 10 −1/3 4 −4/3 2 −4/3 f′′ (x) = x − x = x (5x − 2) 9 9 9 The second derivative f′′ (x) is not defined at 0 Otherwise, x−4/3 is always positive, so the concavity is determined by the 5x − 2 factor So f is concave down on (−∞, 0], concave down on [0, 2/5), concave up on (2/5, ∞), and has an inflection point when x = 2/5 . . . . . .
    • The Second Derivative Test Theorem (The Second Derivative Test) Let f, f′ , and f′′ be continuous on [a, b]. Let c be be a point in (a, b) with f′ (c) = 0. If f′′ (c) < 0, then f(c) is a local maximum. If f′′ (c) > 0, then f(c) is a local minimum. . . . . . .
    • The Second Derivative Test Theorem (The Second Derivative Test) Let f, f′ , and f′′ be continuous on [a, b]. Let c be be a point in (a, b) with f′ (c) = 0. If f′′ (c) < 0, then f(c) is a local maximum. If f′′ (c) > 0, then f(c) is a local minimum. 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. . . . . . .
    • Example Find the local extrema of f(x) = x3 + x2 . . . . . . .
    • 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. . . . . . .
    • 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 . . . . . .
    • 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. . . . . . .
    • 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. . . . . . .
    • Example Find the local extrema of f(x) = x2/3 (x + 2) . . . . . .
    • 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 3 x = −4/5 . . . . . .
    • 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 3 x = −4/5 10 −4/3 Remember f′′ (x) = x (5x − 2), which is negative when 9 x = −4/5 . . . . . .
    • 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 3 x = −4/5 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. . . . . . .
    • 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 3 x = −4/5 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. . . . . . .
    • Graph Graph of f(x) = x2/3 (x + 2): y . . −4/5, 1.03413) ( . . . 2/5, 1.30292) ( . . x . . −2, 0) ( . 0, 0) ( . . . . . .
    • 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? . . . . . .
    • When the second derivative is zero At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0 Is it necessarily true, though? Consider these examples: f (x ) = x 4 g(x) = −x4 h(x) = x3 . . . . . .
    • When first and second derivative are zero function derivatives graph type f′ (x) = 4x3 , f′ (0) = 0 f (x ) = x 4 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 . . . . . . .
    • When the second derivative is zero At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0 Is it necessarily true, though? Consider these examples: f (x ) = x 4 g(x) = −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. . . . . . .
    • What have we learned today? Concepts: Mean Value Theorem, monotonicity, concavity Facts: derivatives can detect monotonicity and concavity Techniques for drawing curves: the Increasing/Decreasing Test and the Concavity Test Techniques for finding extrema: the First Derivative Test and the Second Derivative Test . . . . . .
    • What have we learned today? Concepts: Mean Value Theorem, monotonicity, concavity Facts: derivatives can detect monotonicity and concavity Techniques for drawing curves: the Increasing/Decreasing Test and the Concavity Test Techniques for finding extrema: the First Derivative Test and the Second Derivative Test Next week: Graphing functions! . . . . . .