MA 105: Calculus Lecture 6 . Prof. B.V. Limaye IIT Bombay - - PowerPoint PPT Presentation

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MA 105: Calculus Lecture 6

• Prof. B.V. Limaye

IIT Bombay Friday, 20 January 2017

B.V. Limaye, IITB MA 105: Lec-06

. Theorem (Second derivative test for a local extremum) . . Let D ⊂ R, and let c be an interior point of D. Suppose f : D → R is twice differentiable at c, and f ′(c) = 0. (i) If f ′′(c) < 0, then f has a local maximum at c. (ii) If f ′′(c) > 0, then f has a local minimum at c. Proof: (i) Let f ′′(c) < 0. Since f ′′(c) exists, there is r > 0 such that f is differentiable on (c − r, c + r). Now lim

x→c

f ′(x) x − c = lim

x→c

f ′(x) − f ′(c) x − c = f ′′(c) < 0. Hence there is δ > 0 such that f ′(x)/(x − c) < 0 for all x ∈ (c − δ, c) ∪ (c, c + δ). Now x ∈ (c − δ, c) = ⇒ f ′(x) > 0 and x ∈ (c, c + δ) = ⇒ f ′(x) < 0. By the first derivative test, f has a local maximum at c. A similar argument for (ii).

B.V. Limaye, IITB MA 105: Lec-06

Examples: (i) Let f (x) := x4 − 2x2 for x ∈ R. Then f ′(x) = 4x3−4x = 4(x +1)x(x −1) = 0 ⇐ ⇒ x ∈ {−1, 0, 1}. f is twice differentiable, and f ′′(x) = 4(3x2 − 1) for x ∈ R. Since f ′′(−1) = 8 > 0, f ′′(0) = −4 < 0 and f ′′(1) = 8 > 0, f has a local maximum at 0, and has a local minimum at ±1. (ii) Let f (x) := x4 for x ∈ (−1, 1). Then f ′(x) = 4x3 and f ′′(x) = 12x2 for x ∈ (−1, 1), and so f (0) = f ′(0) = f ′′(0) = 0. Hence the second derivative test is not applicable at 0. But since f ′ < 0 on (−1, 0) and f ′ > 0 on (0, 1), the first derivative test shows that f has a local minimum at 0. The first derivative test is more general than the second.

B.V. Limaye, IITB MA 105: Lec-06

. . Point of Inflection

Let I be an interval, and let f : I → R. An interior point of I at which convexity of f changes to concavity, or the other way round, is called a point of inflection for f . More precisely, . Definition . . An interior point c of I is called a point of inflection for f if there is δ > 0 such that either f is convex on (c − δ, c) and concave on (c, c + δ),

• r f is concave on (c − δ, c) and convex on (c, c + δ).

Example: Let f (x) := x3, x ∈ R. Then 0 is a point of inflection for f . We give characterizations of a point of inflection of a differentiable (and of a twice differentiable) function.

B.V. Limaye, IITB MA 105: Lec-06

. . Point of Inflection and Derivatives

. Theorem (Derivative tests for a point of inflection) . . Let c be an interior point of I, and let f : I → R. (i) Suppose there is δ > 0 such that f is differentiable on (c − δ, c) ∪ (c, c + δ). Then c is point of inflection for f ⇐ ⇒ f ′ is increasing on (c − δ, c) and f ′ is decreasing on (c, c + δ), or vice versa. (ii) Suppose there is δ > 0 such that f is twice differentiable

• n (c − δ, c) ∪ (c, c + δ).Thenc is point of inflection for f

⇐ ⇒ f ′′ ≥ 0 on (c − δ, c) and f ′′ ≤ 0 on (c, c + δ),

• r vice versa.

Proof: See characterizations of convexity and concavity. Thumb Rule: f ′′ changes sign at c ⇐ ⇒ c is a point of inflection for f .

B.V. Limaye, IITB MA 105: Lec-06

.(Necessary condition for a point of inflection) . . Let c be an interior point of I, and let f : I → R. Suppose f is twice differentiable at c. If c is point of inflection for f , then f ′′(c) = 0. Proof: Since f ′′(c) exists, there is r > 0 such that f is differentiable on (c − r, c + r). Apply (i) of the previous test. Convex to concave: There is δ > 0 such that f ′ is increasing

• n (c − δ, c) and f ′ is decreasing on (c, c + δ). Let g := f ′ on

(c − δ, c + δ). Now g is continuous at c, g ′ ≥ 0 on (c − δ, c) and g ′ ≤ 0 on (c, c + δ). By the first derivative test, there is a local maximum of g at c, and so f ′′(c) = g ′(c) = 0. Concave to convex: There is δ > 0 such that f ′ is decreasing

• n (c − δ, c) and f ′ is increasing on (c, c + δ). As above, by

the first derivative test for g := f ′ on (c − δ, c + δ), there is a local minimum of g at c, and so f ′′(c) = g ′(c) = 0.

B.V. Limaye, IITB MA 105: Lec-06

The condition ‘f ′′(c) = 0’ is not sufficient. Example: Let f (x) := x4 for x ∈ R. Then f is twice differentiable at 0 and f ′′(0) = 0. But 0 is not a point of inflection for f . .(Sufficient condition for a point of inflection) . . Let c be an interior point of I, and let f : I → R. Suppose f is thrice differentiable at c. If f ′′(c) = 0 and f ′′′(c) ̸= 0, then c is point of inflection for f . Proof: Without loss of generality, suppose f ′′′(c) < 0. Then lim

x→c

f ′′(x) x − c = lim

x→c

f ′′(x) − f ′′(c) x − c = f ′′′(c) < 0. Hence there is δ > 0 such that f ′′(x)/(x − c) < 0 for all x ∈ (c − δ, c) ∪ (c, c + δ). Now f ′′ > 0 on (c − δ, c), and f ′′ < 0 on (c, c + δ). Hence c is point of inflection for f .

B.V. Limaye, IITB MA 105: Lec-06

The condition ‘f ′′′(c) ̸= 0’ is not necessary. Example: Let f (x) := x5, x ∈ R. Then 0 is a point of inflection for f , but f ′′′(0) = 0. To conclude this topic, let us consider the following example. Let f (x) := x4 − 4x3 for x ∈ R. Then f ′(x) = 4x3 − 12x2 = 4x2(x − 3) = 0 ⇐ ⇒ x ∈ {0, 3}, f ′′(x) = 12x2 − 24x = 12x(x − 2) = 0 ⇐ ⇒ x ∈ {0, 2}. Also, f ′′′(x) = 24(x − 1) for x ∈ R. Since f ′′(0) = 0 = f ′′(2) and f ′′′(0) = −f ′′′(2) = −24 ̸= 0, we see that 0 and 2 are points of inflection for f . Since f ′ < 0 on (−∞, 3), it is strictly decreasing near 0. Hence f does not have a local extremum at 0. Further, since f ′(3) = 0 and f ′′(3) = 36 > 0, f has a local minimum at 3.

B.V. Limaye, IITB MA 105: Lec-06

. . Geometric properties of a function

Let us recapitualte. Let I be an interval. We have considered the following geometric properties of a function from I to R. Boundedness, and attaining the bounds Intermediate value property Monotonicity Convexity/Concavity Absolute Extrema, local extrema Points of inflection Further, we have given analytical conditions on the function (such as continuity, nonnegativity of the derivatives, the vanishing of the derivatives etc.) which imply the above-mentioned geometric properties.

B.V. Limaye, IITB MA 105: Lec-06

. . Towards Curve Sketching

A function f : (−a, a) → R is called even if f (−x) = f (x). It is called odd if f (−x) = −f (x). If f : R → R and f (x) = f (x + p) for a fixed p > 0 and all x, then f is called periodic with period p.

B.V. Limaye, IITB MA 105: Lec-06

. . Towards asymptotes

Let (an) be a sequence of real numbers. Then an → ∞ if for every α > 0, there exists n0 ∈ N such that an > α for all n ≥ n0. an → −∞ if for every β < 0, there exists n0 ∈ N such that an < β for all n ≥ n0. Examples: Let an := n2 for n ∈ N. Then an → ∞: Given α > 0, n2 > α if n > α1/2. Take n0 := [α1/2] + 1. Let an := −n1/3 for n ∈ N. Then an → −∞: Given β < 0, −n1/3 < β if n > −β3. Take n0 := [−β3] + 1.

B.V. Limaye, IITB MA 105: Lec-06

. . Limits as x → ∞ and as x → −∞

. Definition . . Suppose D ⊂ R is such that (a, ∞) ⊂ D for some a ∈ R. For a function f : D → R, we say that a limit of f (x) exists as x → ∞ if there is ℓ ∈ R such that (xn) is a sequence in D and xn → ∞ = ⇒ f (xn) → ℓ. Notation: f (x) → ℓ as x → ∞ or lim

x→∞ f (x) = ℓ.

Similarly, we define f (x) → ℓ as x → −∞ or lim

x→−∞ f (x) = ℓ.

Examples: Let f (x) := 1 x for x > 0. Then f (x) → 0 as x → ∞. Let f (x) := x + 1 x − 1 for x < 1. Then f (x) → 1 as x → −∞.

B.V. Limaye, IITB MA 105: Lec-06

. . f (x) → ∞ and f (x) → −∞

. Definition . . Let f : D → R, c ∈ R such that there is r > 0 with (c − r, c) ∪ (c, c + r) ⊂ D. We say that f (x) tends to ∞ as x → c if (xn) is a sequence in D, xn ̸= c and xn → c = ⇒ f (xn) → ∞. Notation: f (x) → ∞ as x → c. Similarly, we define f (x) → −∞ as x → c. We shall not write lim

x→c f (x) = ∞ or lim x→c f (x) = −∞.

Example: Let f (x) := 1/x for x ∈ R \ {0}. Then f (x) → ∞ as x → 0+ and f (x) → −∞ as x → 0−.

B.V. Limaye, IITB MA 105: Lec-06

. . Asymptotes

Let D ⊂ R, and let f : D → R. There are three possible types

• f asymptotes: horizontal, oblique and vertical. They all imply

nearness of the curve y = f (x) to a straight line. A straight line y = b is called a horizontal asymptote

• f the curve y = f (x) if

lim

x→∞(f (x) − b) = 0 or

lim

x→−∞(f (x) − b) = 0.

A straight line y = ax + b, where a ̸= 0, is called an

• blique asymptote of the curve y = f (x) if

lim

x→∞(f (x) − ax − b) = 0 or

lim

x→−∞(f (x) − ax − b) = 0.

A line x = a is called a vertical asymptote of the curve y = f (x) if f (x) → ∞ as x → a−, or f (x) → −∞ as x → a−, or f (x) → ∞ as x → a+, or f (x) → −∞ as x → a+.

B.V. Limaye, IITB MA 105: Lec-06

. . Illustrations of asymptotes

B.V. Limaye, IITB MA 105: Lec-06

. . Examples of asymptotes

Example: Consider f : (−∞, 0) ∪ (1, ∞) → R defined by f (x) = { (3x2 + 4x + 1)/x if x < 0, (2x − 1)/(x − 1) if x > 1. For x > 1, we have f (x) = 2 + 1/(x − 1), and so lim

x→∞ f (x) = 2. Hence the straight line y = 2 is a

horizontal asymptote of the curve y = f (x). f (x) → ∞ as x → 1+. Hence the straight line x = 1 is a vertical asymptote of the curve y = f (x). For x < 0, f (x) = 3x + 4 + (1/x), and so lim

x→−∞ (f (x) − (3x + 4)) = 0. Hence the straight line

y = 3x + 4 is an oblique asymptote of the curve y = f (x). f (x) → −∞ as x → 0−. Hence the straight line x = 0 is a vertical asymptote of the curve y = f (x). Exercise: Draw the graph of f .

B.V. Limaye, IITB MA 105: Lec-06

. . Tips for Curve Sketching

Look for symmetries of the given function f : Is f even/ odd/ periodic? Look for intercepts, that is, the intersections of the curve with co-ordinate axes. Look for points (if any) where f is not defined. If such a point exists, study the behaviour of f near that point. Locate critical points, local maxima, local minima of f . Find intervals on which f is increasing or decreasing. Find intervals on which f is convex or concave. Locate points of inflection of f . Determine the behaviour of f (x) as x becomes large through positive values or through negative values. Find the asymptotes, if any.

B.V. Limaye, IITB MA 105: Lec-06

. . Curve Sketching: Example

Let f (x) := (x2 − 1)/(x + 2), x ̸= −2. f is not even, not odd, not periodic. Intercepts of f : (0, −1/2), (1, 0), (−1, 0). f ′(x) = (x + 2 + √ 3)(x + 2 − √ 3)/(x + 2)2 and f ′′(x) = 6/(x + 2)3 for x ̸= −2. f is increasing on (−∞, −2 − √ 3) and on (−2 + √ 3, ∞). f is decreasing on (−2 − √ 3, −2) and on (−2, −2 + √ 3). f is concave on (−∞, −2) and convex on (−2, ∞). Since f ′(−2 − √ 3) = 0 = f ′(−2 + √ 3) and f ′′(−2 − √ 3) = −2/ √ 3 < 0, f ′′(−2 + √ 3) = 2/ √ 3 > 0, f has a local maximum at −2 − √ 3 and f has a local minimum at −2 + √

• 3. Note: f (−2 ±

√ 3) = −4 ± 2 √ 3. There is no point of inflection for f . The straight line x = −2 is a vertical asymptote. Since f (x) = x − 2 + 3/(x + 2) for x ̸= −2, y = x − 2 is an oblique asymptote.

B.V. Limaye, IITB MA 105: Lec-06