. 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 f ′ ( x ) f ′ ( x ) − f ′ ( c ) lim x − c = lim = f ′′ ( c ) < 0 . x − c x → c x → c 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 ) := x 4 − 2 x 2 for x ∈ R . Then f ′ ( x ) = 4 x 3 − 4 x = 4( x +1) x ( x − 1) = 0 ⇐ ⇒ x ∈ {− 1 , 0 , 1 } . f is twice differentiable, and f ′′ ( x ) = 4(3 x 2 − 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 ) := x 4 for x ∈ ( − 1 , 1). Then f ′ ( x ) = 4 x 3 and f ′′ ( x ) = 12 x 2 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 ), or f is concave on ( c − δ, c ) and convex on ( c , c + δ ). . Example: Let f ( x ) := x 3 , 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 on ( c − δ, c ) ∪ ( c , c + δ ).Then c is point of inflection for f ⇒ f ′′ ≥ 0 on ( c − δ, c ) and f ′′ ≤ 0 on ( c , c + δ ), ⇐ or 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 on ( 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 on ( 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 ) := x 4 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 f ′′ ( x ) f ′′ ( x ) − f ′′ ( c ) = f ′′′ ( c ) < 0 . lim x − c = lim x − c x → c x → c 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 ) := x 5 , 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 ) := x 4 − 4 x 3 for x ∈ R . Then f ′ ( x ) = 4 x 3 − 12 x 2 = 4 x 2 ( x − 3) = 0 ⇐ ⇒ x ∈ { 0 , 3 } , f ′′ ( x ) = 12 x 2 − 24 x = 12 x ( 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 ( a n ) be a sequence of real numbers. Then a n → ∞ if for every α > 0, there exists n 0 ∈ N such that a n > α for all n ≥ n 0 . a n → −∞ if for every β < 0, there exists n 0 ∈ N such that a n < β for all n ≥ n 0 . Examples: Let a n := n 2 for n ∈ N . Then a n → ∞ : Given α > 0, n 2 > α if n > α 1 / 2 . Take n 0 := [ α 1 / 2 ] + 1. Let a n := − n 1 / 3 for n ∈ N . Then a n → −∞ : Given β < 0, − n 1 / 3 < β if n > − β 3 . Take n 0 := [ − β 3 ] + 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 ( x n ) is a sequence in D and x n → ∞ = ⇒ f ( x n ) → ℓ . . Notation: f ( x ) → ℓ as x → ∞ or lim x →∞ f ( x ) = ℓ . Similarly, we define f ( x ) → ℓ as x → −∞ or x →−∞ f ( x ) = ℓ . lim 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 ( x n ) is a sequence in D , x n ̸ = c and x n → c = ⇒ f ( x n ) → ∞ . . 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
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