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

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

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 (slides) Lesson 20: Derivatives and the Shapes of Curves (slides) Presentation Transcript

  • Sec on 4.3 Deriva ves and the Shapes of Curves V63.0121.001: Calculus I Professor Ma hew Leingang New York University April 11, 2011.
  • 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
  • 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.
  • Objectives Use the second deriva ve of a func on to determine the intervals along which the graph of the func on is concave up or concave down (The Concavity Test) Use the first and second deriva ve of a func on to classify cri cal points as local maxima or local minima, when applicable (The Second Deriva ve Test)
  • 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
  • Recall: The Mean Value Theorem Theorem (The Mean Value Theorem) 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
  • Recall: The Mean Value Theorem Theorem (The Mean Value Theorem) 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
  • 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
  • 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 MITCMost 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.
  • 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
  • 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.
  • 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
  • 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.
  • 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).
  • 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.
  • 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).
  • 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. Solu on f′ (x) = 2 is always posi ve, so f is increasing on (−∞, ∞).
  • 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).
  • 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
  • Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solu on .
  • 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. .
  • 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
  • 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
  • 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, ∞).
  • 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, ∞)
  • 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). Solu on . f′ (x) = 2 x−1/3 (x + 2) + x2/3 3 = 3 x−1/3 (5x + 4) 1
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on . f′ (x) = 2 x−1/3 (x + 2) + x2/3 3 = 3 x−1/3 (5x + 4) 1 The cri cal points are 0 and and −4/5.
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on . x−1/3 f′ (x) = 2 x−1/3 (x + 2) + x2/3 3 = 3 x−1/3 (5x + 4) 1 The cri cal points are 0 and and −4/5.
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on ×. ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 = 3x (5x + 4) The cri cal points are 0 and and −4/5.
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 = 3x (5x + 4) The cri cal points are 0 and and −4/5.
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 = 3x (5x + 4) The cri cal points are 0 and and −4/5.
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 = 3x (5x + 4) 5x + 4 The cri cal points are 0 and and −4/5.
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 0 = 3x (5x + 4) 5x + 4 −4/5 The cri cal points are 0 and and −4/5.
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 − 0 = 3x (5x + 4) 5x + 4 −4/5 The cri cal points are 0 and and −4/5.
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 − 0 + = 3x (5x + 4) 5x + 4 −4/5 The cri cal points are 0 and and −4/5.
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 − 0 + = 3x (5x + 4) 5x + 4 −4/5 f′ (x) The cri cal points are 0 and and −4/5. f(x)
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 − 0 + = 3x (5x + 4) 5x + 4 −4/5 f′ (x) 0 The cri cal points are 0 and and −4/5. −4/5 f(x)
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 − 0 + = 3x (5x + 4) 5x + 4 −4/5 × f′ (x) 0 The cri cal points are 0 and and −4/5. −4/5 0 f(x)
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 − 0 + = 3x (5x + 4) 5x + 4 + −0/5 f′ (x) 4 × The cri cal points are 0 and and −4/5. −4/5 0 f(x)
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 − 0 + = 3x (5x + 4) 5x + 4 + −0/5− × f′ (x) 4 The cri cal points are 0 and and −4/5. −4/5 0 f(x)
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 − 0 + = 3x (5x + 4) 5x + 4 + −0/5− × f′ (x) 4 + The cri cal points are 0 and and −4/5. −4/5 0 f(x)
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 − 0 + = 3x (5x + 4) 5x + 4 + −0/5− × f′ (x) 4 + The cri cal points are 0 and and −4/5. ↗ −4/5 0 f(x)
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 − 0 + = 3x (5x + 4) 5x + 4 + −0/5− × f′ (x) 4 + The cri cal points are 0 and and −4/5. ↗ −4/5 0 ↘ f(x)
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 − 0 + = 3x (5x + 4) 5x + 4 + −0/5− × f′ (x) 4 + The cri cal points are 0 and and −4/5. ↗ −4/5 0 ↘ ↗ f(x)
  • 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.
  • 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, ∞)
  • Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solu on − 0 + f′ . ↘ 0 ↗ f min So f is decreasing on (−∞, 0) and increasing on (0, ∞). In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞)
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 − 0 + = 3x (5x + 4) 5x + 4 + −0/5− × f′ (x) 4 + The cri cal points are 0 and and −4/5. ↗ −4/5 0 ↘ ↗ f(x)
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 − 0 + = 3x (5x + 4) 5x + 4 + −0/5− × f′ (x) 4 + The cri cal points are 0 and and −4/5. ↗ −4/5 0 ↘ ↗ f(x) max
  • Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on − ×. + ′ f (x) = 2 −1/3 3x (x + 2) + x2/3 x−1/3 0 1 −1/3 − 0 + = 3x (5x + 4) 5x + 4 + −0/5− × f′ (x) 4 + The cri cal points are 0 and and −4/5. ↗ −4/5 0 ↘ ↗ f(x) max min
  • 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
  • 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.
  • 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
  • Concavity Defini on . . concave up concave down We some mes say a concave up graph “holds water” and a concave down graph “spills water”.
  • Synonyms for concavity Remark “concave up” = “concave upwards” = “convex” “concave down” = “concave downwards” = “concave”
  • 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
  • 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.
  • Testing for Concavity Proof. Suppose f′′ (x) > 0 on the interval I (which could be infinite). This means f′ is increasing on I.
  • 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)
  • 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)
  • 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).
  • Finding Intervals of Concavity I Example Find the intervals of concavity for the graph of f(x) = x3 + x2 .
  • 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.
  • 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
  • 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)
  • Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
  • 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 + ×. + x−4/3 10 −1/3 4 −4/3 0 f′′ (x) = x − x − 0 + 9 9 5x − 2 2/5 2 −4/3 = x (5x − 2) 9
  • 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 + ×. + x−4/3 10 −1/3 4 −4/3 0 f′′ (x) = x − x − 0 + 9 9 5x − 2 2 −4/3 × 2/5 0 f′′ (x) = x (5x − 2) 9 0 2/5 f(x)
  • 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 + ×. + x−4/3 10 −1/3 4 −4/3 0 f′′ (x) = x − x − 0 + 9 9 5x − 2 2 −4/3 −− × 2/5 0 f′′ (x) = x (5x − 2) 9 0 2/5 f(x)
  • 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 + ×. + x−4/3 10 −1/3 4 −4/3 0 f′′ (x) = x − x − 0 + 9 9 5x − 2 2 −4/3 −− 2 × /5 −− 0 f′′ (x) = x (5x − 2) 9 0 2/5 f(x)
  • 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 + ×. + x−4/3 10 −1/3 4 −4/3 0 f′′ (x) = x − x − 0 + 9 9 5x − 2 2 −4/3 −− 2 × /5 −− 0 ++ f′′ (x) = x (5x − 2) 9 0 2/5 f(x)
  • 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 + ×. + x−4/3 10 −1/3 4 −4/3 0 f′′ (x) = x − x − 0 + 9 9 5x − 2 2 −4/3 −− 2 × /5 −− 0 ++ f′′ (x) = x (5x − 2) ⌢ 9 0 2/5 f(x)
  • 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 + ×. + x−4/3 10 −1/3 4 −4/3 0 f′′ (x) = x − x − 0 + 9 9 5x − 2 2 −4/3 −− 2 × /5 −− 0 ++ f′′ (x) = x (5x − 2) ⌢ 9 0⌢ 2/5 f(x)
  • 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 + ×. + x−4/3 10 −1/3 4 −4/3 0 f′′ (x) = x − x − 0 + 9 9 5x − 2 2 −4/3 −− 2 × /5 −− 0 ++ f′′ (x) = x (5x − 2) ⌢ 9 0⌢ 2/5 ⌣ f(x)
  • 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 + ×. + x−4/3 10 −1/3 4 −4/3 0 f′′ (x) = x − x − 0 + 9 9 5x − 2 2 −4/3 −− 2 × /5 −− 0 ++ f′′ (x) = x (5x − 2) ⌢ 9 0⌢ 2/5 ⌣ f(x) IP
  • 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.
  • 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.
  • Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. + f′′ = (f′ )′ . c f′ 0 f′ c f
  • 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′ 0 f′ c f
  • 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′ 0 f′ c f
  • 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′ 0 f′ c f
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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 f
  • 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 f
  • 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 ↗ f
  • 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
  • Using the Second Derivative Test I Example Find the local extrema of f(x) = x3 + x2 .
  • 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.
  • 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
  • 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.
  • 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.
  • Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2)
  • Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solu on 1 Remember f′ (x) = x−1/3 (5x + 4) which is zero when x = −4/5 3
  • Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solu on 1 Remember f′ (x) = x−1/3 (5x + 4) which is zero when x = −4/5 3 10 Remember f′′ (x) = x−4/3 (5x − 2), which is nega ve when 9 x = −4/5
  • Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solu on 1 Remember f′ (x) = x−1/3 (5x + 4) which is zero when x = −4/5 3 10 Remember f′′ (x) = x−4/3 (5x − 2), which is nega ve when 9 x = −4/5 So x = −4/5 is a local maximum.
  • Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solu on 1 Remember f′ (x) = x−1/3 (5x + 4) which is zero when x = −4/5 3 10 Remember f′′ (x) = x−4/3 (5x − 2), which is nega ve when 9 x = −4/5 So x = −4/5 is a local maximum. No ce the Second Deriva ve Test doesn’t catch the local minimum x = 0 since f is not differen able there.
  • Using the Second Derivative Test IIGraph 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 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?
  • 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
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = , f′ (0) = f(x) = x f′′ (x) = , f′′ (0) = g′ (x) = , g′ (0) = g(x) = −x 4 g′′ (x) = , g′′ (0) = 3 h′ (x) = , h′ (0) = h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = f(x) = x f′′ (x) = , f′′ (0) = g′ (x) = , g′ (0) = g(x) = −x 4 g′′ (x) = , g′′ (0) = 3 h′ (x) = , h′ (0) = h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x f′′ (x) = , f′′ (0) = g′ (x) = , g′ (0) = g(x) = −x 4 g′′ (x) = , g′′ (0) = 3 h′ (x) = , h′ (0) = h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x f′′ (x) = 12x2 , f′′ (0) = g′ (x) = , g′ (0) = g(x) = −x 4 g′′ (x) = , g′′ (0) = 3 h′ (x) = , h′ (0) = h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x f′′ (x) = 12x2 , f′′ (0) = 0 g′ (x) = , g′ (0) = g(x) = −x 4 g′′ (x) = , g′′ (0) = 3 h′ (x) = , h′ (0) = h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x . f′′ (x) = 12x2 , f′′ (0) = 0 g′ (x) = , g′ (0) = g(x) = −x 4 g′′ (x) = , g′′ (0) = 3 h′ (x) = , h′ (0) = h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x . min ′′ 2 ′′ f (x) = 12x , f (0) = 0 g′ (x) = , g′ (0) = g(x) = −x 4 g′′ (x) = , g′′ (0) = 3 h′ (x) = , h′ (0) = h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x . min ′′ 2 ′′ f (x) = 12x , f (0) = 0 g′ (x) = − 4x3 , g′ (0) = g(x) = −x 4 g′′ (x) = , g′′ (0) = 3 h′ (x) = , h′ (0) = h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x . min ′′ 2 ′′ f (x) = 12x , f (0) = 0 g′ (x) = − 4x3 , g′ (0) = 0 g(x) = −x 4 g′′ (x) = , g′′ (0) = 3 h′ (x) = , h′ (0) = h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x . min ′′ 2 ′′ f (x) = 12x , f (0) = 0 g′ (x) = − 4x3 , g′ (0) = 0 g(x) = −x 4 g′′ (x) = − 12x2 , g′′ (0) = 3 h′ (x) = , h′ (0) = h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x . min ′′ 2 ′′ f (x) = 12x , f (0) = 0 g′ (x) = − 4x3 , g′ (0) = 0 g(x) = −x 4 g′′ (x) = − 12x2 , g′′ (0) = 0 3 h′ (x) = , h′ (0) = h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x . min ′′ 2 ′′ f (x) = 12x , f (0) = 0 . g′ (x) = − 4x3 , g′ (0) = 0 g(x) = −x4 g′′ (x) = − 12x2 , g′′ (0) = 0 3 h′ (x) = , h′ (0) = h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x . min ′′ 2 ′′ f (x) = 12x , f (0) = 0 . g′ (x) = − 4x3 , g′ (0) = 0 g(x) = −x4 max ′′ ′′ g (x) = − 12x , g (0) = 0 2 3 h′ (x) = , h′ (0) = h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x . min ′′ 2 ′′ f (x) = 12x , f (0) = 0 . g′ (x) = − 4x3 , g′ (0) = 0 g(x) = −x4 max ′′ ′′ g (x) = − 12x , g (0) = 0 2 3 h′ (x) = 3x2 , h′ (0) = h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x . min ′′ 2 ′′ f (x) = 12x , f (0) = 0 . g′ (x) = − 4x3 , g′ (0) = 0 g(x) = −x4 max ′′ ′′ g (x) = − 12x , g (0) = 0 2 3 h′ (x) = 3x2 , h′ (0) = 0 h(x) = x h′′ (x) = , h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x . min ′′ 2 ′′ f (x) = 12x , f (0) = 0 . g′ (x) = − 4x3 , g′ (0) = 0 g(x) = −x4 max ′′ ′′ g (x) = − 12x , g (0) = 0 2 3 h′ (x) = 3x2 , h′ (0) = 0 h(x) = x h′′ (x) = 6x, h′′ (0) =
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x . min ′′ 2 ′′ f (x) = 12x , f (0) = 0 . g′ (x) = − 4x3 , g′ (0) = 0 g(x) = −x4 max ′′ ′′ g (x) = − 12x , g (0) = 0 2 3 h′ (x) = 3x2 , h′ (0) = 0 h(x) = x h′′ (x) = 6x, h′′ (0) = 0
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x . min ′′ 2 ′′ f (x) = 12x , f (0) = 0 . g′ (x) = − 4x3 , g′ (0) = 0 g(x) = −x4 max ′′ ′′ g (x) = − 12x , g (0) = 0 2 3 h′ (x) = 3x2 , h′ (0) = 0 h(x) = x . h′′ (x) = 6x, h′′ (0) = 0
  • When first and second derivative are zero func on deriva ves graph type 4 f′ (x) = 4x3 , f′ (0) = 0 f(x) = x . min ′′ 2 ′′ f (x) = 12x , f (0) = 0 . g′ (x) = − 4x3 , g′ (0) = 0 g(x) = −x4 max ′′ ′′ g (x) = − 12x , g (0) = 02 3 h′ (x) = 3x2 , h′ (0) = 0 h(x) = x . infl. ′′ ′′ h (x) = 6x, h (0) = 0
  • 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.
  • 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