Lesson 20: Derivatives and the Shapes of Curves (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 Shapes of Curves (handout)

  1. 1. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Sec on 4.2 Deriva ves and the Shapes of Curves V63.0121.001: Calculus I Professor Ma hew Leingang New York University April 11, 2011 . . Notes Announcements Quiz 4 on Sec ons 3.3, 3.4, 3.5, and 3.7 this week (April 14/15) Quiz 5 on Sec ons 4.1–4.4 April 28/29 Final Exam Thursday May 12, 2:00–3:50pm . . Notes Objectives Use the deriva ve of a func on to determine the intervals along which the func on is increasing or decreasing (The Increasing/Decreasing Test) Use the First Deriva ve Test to classify cri cal points of a func on as local maxima, local minima, or neither. . . . 1.
  2. 2. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Outline Recall: The Mean Value Theorem Monotonicity The Increasing/Decreasing Test Finding intervals of monotonicity The First Deriva ve Test Concavity Defini ons Tes ng for Concavity The Second Deriva ve Test . . Notes Recall: The Mean Value Theorem Theorem (The Mean Value Theorem) c Let f be con nuous on [a, b] and differen able on (a, b). Then there exists a point c in (a, b) such that b f(b) − f(a) . = f′ (c). a b−a Another way to put this is that there exists a point c such that f(b) = f(a) + f′ (c)(b − a) . . Why the MVT is the MITC Notes 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 con nuous on [x, y] and differen able 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. . . . 2.
  3. 3. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Outline Recall: The Mean Value Theorem Monotonicity The Increasing/Decreasing Test Finding intervals of monotonicity The First Deriva ve Test Concavity Defini ons Tes ng for Concavity The Second Deriva ve Test . . Notes Increasing Functions Defini on A func on 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 func on “preserves order.” I could be bounded or infinite, open, closed, or half-open/half-closed. Write your own defini on (muta s mutandis) of decreasing, nonincreasing, nondecreasing . . Notes 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). . . . 3.
  4. 4. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Finding intervals of monotonicity I Example Find the intervals of monotonicity of f(x) = 2x − 5. Solu on f′ (x) = 2 is always posi ve, so f is increasing on (−∞, ∞). Example Describe the monotonicity of f(x) = arctan(x). Solu on 1 Since f′ (x) = is always posi ve, f(x) is always increasing. . 1 + x2 . Notes Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solu on f′ (x) = 2x, which is posi ve when x > 0 and nega ve when x is. We can draw a number line: − 0 + f′ . ↘ 0 ↗ f . . Notes Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solu on − 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, ∞) . . . 4.
  5. 5. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on . . Notes The First Derivative Test Theorem (The First Deriva ve Test) Let f be con nuous on [a, b] and c a cri cal point of f in (a, b). If f′ changes from posi ve to nega ve at c, then c is a local maximum. If f′ changes from nega ve to posi ve 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. . . Notes Outline Recall: The Mean Value Theorem Monotonicity The Increasing/Decreasing Test Finding intervals of monotonicity The First Deriva ve Test Concavity Defini ons Tes ng for Concavity The Second Deriva ve Test . . . 5.
  6. 6. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Concavity Defini on 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 some mes say a concave up graph “holds water” and a concave down graph “spills water”. . Notes Synonyms for concavity Remark “concave up” = “concave upwards” = “convex” “concave down” = “concave downwards” = “concave” . . Notes Inflection points mean change in concavity Defini on A point P on a curve y = f(x) is called an inflec on point if f is con nuous at P and the curve changes from concave upward to concave downward at P (or vice versa). concave up inflec on point . concave down . . . 6.
  7. 7. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Testing for Concavity 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. . . Notes Testing for Concavity 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). . . Notes Finding Intervals of Concavity I Example Find the intervals of concavity for the graph of f(x) = x3 + x2 . Solu on We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2. This is nega ve when x < −1/3, posi ve 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 inflec on point at the point (−1/3, 2/27) . . . 7.
  8. 8. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solu on We have 10 −1/3 4 −4/3 f′′ (x) = x − x 9 9 2 −4/3 = x (5x − 2) 9 . . Notes The Second Derivative Test Theorem (The Second Deriva ve Test) Let f, f′ , and f′′ be con nuous 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 of f. If f′′ (c) > 0, then c is a local minimum of f. Remarks If f′′ (c) = 0, the second deriva ve test is inconclusive We look for zeroes of f′ and plug them into f′′ to determine if their f values are local extreme values. . . Notes Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is con nuous, f′′ (x) > 0 for all x + + + f′′ = (f′ )′ sufficiently close to c. . ↗ c ↗ f′ Since f′′ = (f′ )′ , we know 0 f′ f′ is increasing near c. c f . . . 8.
  9. 9. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′ (c) = 0 and f′ is increasing, f′ (x) < 0 for x + + + f′′ = (f′ )′ . close to c and less than c, c ↗ ↗ f′ and f′ (x) > 0 for x close − 0 + f′ to c and more than c. c f . . Notes Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. This means f′ changes sign from nega ve to + + + f′′ = (f′ )′ . posi ve at c, which ↗ c ↗ f′ means (by the First − 0 + f′ Deriva ve Test) that f has a local minimum at c. c ↘ min ↗ f . . Notes Using the Second Derivative Test I Example Find the local extrema of f(x) = x3 + x2 . Solu on 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. . . . 9.
  10. 10. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solu on . . Using the Second Derivative Test II Notes Graph Graph of f(x) = x2/3 (x + 2): y . x . . Notes When the second derivative is zero Remark At inflec on points c, if f′ is differen able at c, then f′′ (c) = 0 If f′′ (c) = 0, must f have an inflec on point at c? Consider these examples: f(x) = x4 g(x) = −x4 h(x) = x3 All of them have cri cal points at zero with a second deriva ve of zero. But the first has a local min at 0, the second has a local max at 0, and the third has an inflec on point at 0. This is why we say 2DT has nothing to say when f′′ (c) = 0. . . . 10.
  11. 11. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes When first and second derivative are zero func on deriva ves graph type ′ 3 ′ f (x) = 4x , f (0) = 0 f(x) = x4 . min f (x) = 12x2 , f′′ (0) = 0 ′′ . g′ (x) = − 4x3 , g′ (0) = 0 g(x) = −x4 max g′′ (x) = − 12x2 , g′′ (0) = 0 h′ (x) = 3x2 , h′ (0) = 0 h(x) = x3 . infl. h′′ (x) = 6x, h′′ (0) = 0 . . Notes Summary Concepts: Mean Value Theorem, monotonicity, concavity Facts: deriva ves can detect monotonicity and concavity Techniques for drawing curves: the Increasing/Decreasing Test and the Concavity Test Techniques for finding extrema: the First Deriva ve Test and the Second Deriva ve Test . . Notes . . . 11.

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