MA 105 Calculus: Lecture 6Maxima/Minima (Contd.)

How Derivatives Affect the Shape of a Graph

S. Sivaji Ganesh

Mathematics DepartmentIIT Bombay

August 4, 2009

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 1 / 30

What does f ′ say about f ?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I if

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I if

f (x1) > f (x2)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I if

f (x1) > f (x2) whenever x1, x2 are in I and

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I if

f (x1) > f (x2) whenever x1, x2 are in I and x1 < x2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I if

f (x1) > f (x2) whenever x1, x2 are in I and x1 < x2.

Say True/False

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I if

f (x1) > f (x2) whenever x1, x2 are in I and x1 < x2.

Say True/False1 If f is increasing on an interval, then

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I if

f (x1) > f (x2) whenever x1, x2 are in I and x1 < x2.

Say True/False1 If f is increasing on an interval, then f ′(x) > 0 on that interval.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I if

f (x1) > f (x2) whenever x1, x2 are in I and x1 < x2.

Say True/False1 If f is increasing on an interval, then f ′(x) > 0 on that interval.

False.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I if

f (x1) > f (x2) whenever x1, x2 are in I and x1 < x2.

Say True/False1 If f is increasing on an interval, then f ′(x) > 0 on that interval.

False. Can we correct this statement and make it correct?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I if

f (x1) > f (x2) whenever x1, x2 are in I and x1 < x2.

Say True/False1 If f is increasing on an interval, then f ′(x) > 0 on that interval.

False. Can we correct this statement and make it correct?2 If f is decreasing on an interval, then

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I if

f (x1) > f (x2) whenever x1, x2 are in I and x1 < x2.

Say True/False1 If f is increasing on an interval, then f ′(x) > 0 on that interval.

False. Can we correct this statement and make it correct?2 If f is decreasing on an interval, then f ′(x) < 0 on that interval.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I if

f (x1) > f (x2) whenever x1, x2 are in I and x1 < x2.

Say True/False1 If f is increasing on an interval, then f ′(x) > 0 on that interval.

False. Can we correct this statement and make it correct?2 If f is decreasing on an interval, then f ′(x) < 0 on that interval.

Must be False.S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I if

f (x1) > f (x2) whenever x1, x2 are in I and x1 < x2.

Say True/False1 If f is increasing on an interval, then f ′(x) > 0 on that interval.

False. Can we correct this statement and make it correct?2 If f is decreasing on an interval, then f ′(x) < 0 on that interval.

Must be False. why?S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

Definition1 A function f is called increasing on an interval I if

f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.

2 A function f is called decreasing on an interval I if

f (x1) > f (x2) whenever x1, x2 are in I and x1 < x2.

Say True/False1 If f is increasing on an interval, then f ′(x) > 0 on that interval.

False. Can we correct this statement and make it correct?2 If f is decreasing on an interval, then f ′(x) < 0 on that interval.

Must be False. why? because the previous one is false!S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30

What does f ′ say about f ?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

What does f ′ say about f ?

Increasing/Decreasing test

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

What does f ′ say about f ?

Increasing/Decreasing test

If f ′(x) > 0 on an interval, then

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

What does f ′ say about f ?

Increasing/Decreasing test

If f ′(x) > 0 on an interval, then f is increasing on that interval.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

What does f ′ say about f ?

Increasing/Decreasing test

If f ′(x) > 0 on an interval, then f is increasing on that interval.

If f ′(x) < 0 on an interval, then

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

What does f ′ say about f ?

Increasing/Decreasing test

If f ′(x) > 0 on an interval, then f is increasing on that interval.

If f ′(x) < 0 on an interval, then f is decreasing on that interval.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

What does f ′ say about f ?

Increasing/Decreasing test

If f ′(x) > 0 on an interval, then f is increasing on that interval.

If f ′(x) < 0 on an interval, then f is decreasing on that interval.

Proof Let x1, x2 be any two points in the interval

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

What does f ′ say about f ?

Increasing/Decreasing test

If f ′(x) > 0 on an interval, then f is increasing on that interval.

If f ′(x) < 0 on an interval, then f is decreasing on that interval.

Proof Let x1, x2 be any two points in the interval with x1 < x2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

What does f ′ say about f ?

Increasing/Decreasing test

If f ′(x) > 0 on an interval, then f is increasing on that interval.

If f ′(x) < 0 on an interval, then f is decreasing on that interval.

Proof Let x1, x2 be any two points in the interval with x1 < x2. By MVT,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

What does f ′ say about f ?

Increasing/Decreasing test

If f ′(x) > 0 on an interval, then f is increasing on that interval.

If f ′(x) < 0 on an interval, then f is decreasing on that interval.

Proof Let x1, x2 be any two points in the interval with x1 < x2. By MVT,there is a point c in (x1, x2)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

What does f ′ say about f ?

Increasing/Decreasing test

If f ′(x) > 0 on an interval, then f is increasing on that interval.

If f ′(x) < 0 on an interval, then f is decreasing on that interval.

Proof Let x1, x2 be any two points in the interval with x1 < x2. By MVT,there is a point c in (x1, x2) such that

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

What does f ′ say about f ?

Increasing/Decreasing test

If f ′(x) > 0 on an interval, then f is increasing on that interval.

If f ′(x) < 0 on an interval, then f is decreasing on that interval.

Proof Let x1, x2 be any two points in the interval with x1 < x2. By MVT,there is a point c in (x1, x2) such that

f (x2) − f (x1) = f ′(c)(x2 − x1).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

What does f ′ say about f ?

Increasing/Decreasing test

If f ′(x) > 0 on an interval, then f is increasing on that interval.

If f ′(x) < 0 on an interval, then f is decreasing on that interval.

Proof Let x1, x2 be any two points in the interval with x1 < x2. By MVT,there is a point c in (x1, x2) such that

f (x2) − f (x1) = f ′(c)(x2 − x1).

The above two statements now follow.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

What does f ′ say about f ?

Increasing/Decreasing test

If f ′(x) > 0 on an interval, then f is increasing on that interval.

If f ′(x) < 0 on an interval, then f is decreasing on that interval.

Proof Let x1, x2 be any two points in the interval with x1 < x2. By MVT,there is a point c in (x1, x2) such that

f (x2) − f (x1) = f ′(c)(x2 − x1).

The above two statements now follow.Exercise Find where the function f (x) = 3x4 − 4x3 − 12x2 + 5 isincreasing and

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

What does f ′ say about f ?

Increasing/Decreasing test

If f ′(x) > 0 on an interval, then f is increasing on that interval.

If f ′(x) < 0 on an interval, then f is decreasing on that interval.

Proof Let x1, x2 be any two points in the interval with x1 < x2. By MVT,there is a point c in (x1, x2) such that

f (x2) − f (x1) = f ′(c)(x2 − x1).

The above two statements now follow.Exercise Find where the function f (x) = 3x4 − 4x3 − 12x2 + 5 isincreasing and where it is decreasing.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 3 / 30

Increasing/Decreasing

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 4 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 4 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 4 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note that

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 4 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 4 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2),

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 4 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2), thus f has a localmaximum at 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 4 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2), thus f has a localmaximum at 0. Note that f ′ change sign from +ve to -ve at 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 4 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2), thus f has a localmaximum at 0. Note that f ′ change sign from +ve to -ve at 0.f decreases on (0,2)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 4 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2), thus f has a localmaximum at 0. Note that f ′ change sign from +ve to -ve at 0.f decreases on (0,2) and increases on (2,3),

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 4 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2), thus f has a localmaximum at 0. Note that f ′ change sign from +ve to -ve at 0.f decreases on (0,2) and increases on (2,3), thus f has a localminimum at 2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 4 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2), thus f has a localmaximum at 0. Note that f ′ change sign from +ve to -ve at 0.f decreases on (0,2) and increases on (2,3), thus f has a localminimum at 2. Note that f ′ change sign from -ve to +ve at 2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 4 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2), thus f has a localmaximum at 0. Note that f ′ change sign from +ve to -ve at 0.f decreases on (0,2) and increases on (2,3), thus f has a localminimum at 2. Note that f ′ change sign from -ve to +ve at 2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 4 / 30

How to catch local Minima/Maxima points using First Derivat e?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 5 / 30

How to catch local Minima/Maxima points using First Derivat e?

We know, by Fermat’s theorem,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 5 / 30

How to catch local Minima/Maxima points using First Derivat e?

We know, by Fermat’s theorem, that

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 5 / 30

How to catch local Minima/Maxima points using First Derivat e?

We know, by Fermat’s theorem, that if f has a local extremum at c,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 5 / 30

How to catch local Minima/Maxima points using First Derivat e?

We know, by Fermat’s theorem, that if f has a local extremum at c,then c must be a critical point.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 5 / 30

How to catch local Minima/Maxima points using First Derivat e?

We know, by Fermat’s theorem, that if f has a local extremum at c,then c must be a critical point. We also know that

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 5 / 30

How to catch local Minima/Maxima points using First Derivat e?

We know, by Fermat’s theorem, that if f has a local extremum at c,then c must be a critical point. We also know that we may not have alocal extremum at every critical point.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 5 / 30

How to catch local Minima/Maxima points using First Derivat e?

We know, by Fermat’s theorem, that if f has a local extremum at c,then c must be a critical point. We also know that we may not have alocal extremum at every critical point.

Question: At what critical points we have a local extremum?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 5 / 30

How to catch local Minima/Maxima points using First Derivat e?

We know, by Fermat’s theorem, that if f has a local extremum at c,then c must be a critical point. We also know that we may not have alocal extremum at every critical point.

Question: At what critical points we have a local extremum? At whatcritical points we do not have a local extremum?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 5 / 30

How to catch local Minima/Maxima points using First Derivat e?

We know, by Fermat’s theorem, that if f has a local extremum at c,then c must be a critical point. We also know that we may not have alocal extremum at every critical point.

Question: At what critical points we have a local extremum? At whatcritical points we do not have a local extremum?

The next result answers these questions.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 5 / 30

What does f ′ say about f ?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative Test

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

1 If f ′ changes sign from positive to negative at c,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

1 If f ′ changes sign from positive to negative at c, then f has a localmaximum at c.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

1 If f ′ changes sign from positive to negative at c, then f has a localmaximum at c.

2 If f ′ changes sign from negative to positive at c,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

1 If f ′ changes sign from positive to negative at c, then f has a localmaximum at c.

2 If f ′ changes sign from negative to positive at c, then f has a localminimum at c.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

1 If f ′ changes sign from positive to negative at c, then f has a localmaximum at c.

2 If f ′ changes sign from negative to positive at c, then f has a localminimum at c.

3 If f ′ does not change sign at c,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

1 If f ′ changes sign from positive to negative at c, then f has a localmaximum at c.

2 If f ′ changes sign from negative to positive at c, then f has a localminimum at c.

3 If f ′ does not change sign at c, then f has no local extremum at c.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

1 If f ′ changes sign from positive to negative at c, then f has a localmaximum at c.

2 If f ′ changes sign from negative to positive at c, then f has a localminimum at c.

3 If f ′ does not change sign at c, then f has no local extremum at c.

Proof follows from Increasing/Decreasing test.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

1 If f ′ changes sign from positive to negative at c, then f has a localmaximum at c.

2 If f ′ changes sign from negative to positive at c, then f has a localminimum at c.

3 If f ′ does not change sign at c, then f has no local extremum at c.

Proof follows from Increasing/Decreasing test.Remark/Questions:

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

1 If f ′ changes sign from positive to negative at c, then f has a localmaximum at c.

2 If f ′ changes sign from negative to positive at c, then f has a localminimum at c.

3 If f ′ does not change sign at c, then f has no local extremum at c.

Proof follows from Increasing/Decreasing test.Remark/Questions:

1 Can we drop the hypothesis

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

1 If f ′ changes sign from positive to negative at c, then f has a localmaximum at c.

2 If f ′ changes sign from negative to positive at c, then f has a localminimum at c.

3 If f ′ does not change sign at c, then f has no local extremum at c.

Proof follows from Increasing/Decreasing test.Remark/Questions:

1 Can we drop the hypothesis of continuity on f ?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

1 If f ′ changes sign from positive to negative at c, then f has a localmaximum at c.

2 If f ′ changes sign from negative to positive at c, then f has a localminimum at c.

3 If f ′ does not change sign at c, then f has no local extremum at c.

Proof follows from Increasing/Decreasing test.Remark/Questions:

1 Can we drop the hypothesis of continuity on f ? No.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

1 If f ′ changes sign from positive to negative at c, then f has a localmaximum at c.

2 If f ′ changes sign from negative to positive at c, then f has a localminimum at c.

3 If f ′ does not change sign at c, then f has no local extremum at c.

Proof follows from Increasing/Decreasing test.Remark/Questions:

1 Can we drop the hypothesis of continuity on f ? No.2 We require f to be differentiable

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

1 If f ′ changes sign from positive to negative at c, then f has a localmaximum at c.

2 If f ′ changes sign from negative to positive at c, then f has a localminimum at c.

3 If f ′ does not change sign at c, then f has no local extremum at c.

Proof follows from Increasing/Decreasing test.Remark/Questions:

1 Can we drop the hypothesis of continuity on f ? No.2 We require f to be differentiable in an open interval containing c,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

What does f ′ say about f ?

The First Derivative TestLet f be a continuous function and c be a critical point of f .

1 If f ′ changes sign from positive to negative at c, then f has a localmaximum at c.

2 If f ′ changes sign from negative to positive at c, then f has a localminimum at c.

3 If f ′ does not change sign at c, then f has no local extremum at c.

Proof follows from Increasing/Decreasing test.Remark/Questions:

1 Can we drop the hypothesis of continuity on f ? No.2 We require f to be differentiable in an open interval containing c,

and we do not insist on differentiability of f at c.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 6 / 30

Increasing/Decreasing

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 7 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 7 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5, critical points are

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 7 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5, critical points are −1,0,2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 7 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5, critical points are −1,0,2.

Note that

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 7 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5, critical points are −1,0,2.

Note thatf ′ changes sign from +ve to -ve at 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 7 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5, critical points are −1,0,2.

Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 7 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5, critical points are −1,0,2.

Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0) and decreases on (0,2),

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 7 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5, critical points are −1,0,2.

Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0) and decreases on (0,2), thus f has a local maximum at 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 7 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5, critical points are −1,0,2.

Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0) and decreases on (0,2), thus f has a local maximum at 0.f ′ changes sign from -ve to +ve at 2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 7 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5, critical points are −1,0,2.

Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0) and decreases on (0,2), thus f has a local maximum at 0.f ′ changes sign from -ve to +ve at 2. Note that f decreases on(0,2)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 7 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5, critical points are −1,0,2.

Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0) and decreases on (0,2), thus f has a local maximum at 0.f ′ changes sign from -ve to +ve at 2. Note that f decreases on(0,2) and increases on (2,3),

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 7 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5, critical points are −1,0,2.

Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0) and decreases on (0,2), thus f has a local maximum at 0.f ′ changes sign from -ve to +ve at 2. Note that f decreases on(0,2) and increases on (2,3), thus f has a local minimum at 2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 7 / 30

Increasing/Decreasing

Graph of f (x) = 3x4 − 4x3 − 12x2 + 5, critical points are −1,0,2.

Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0) and decreases on (0,2), thus f has a local maximum at 0.f ′ changes sign from -ve to +ve at 2. Note that f decreases on(0,2) and increases on (2,3), thus f has a local minimum at 2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 7 / 30

The First Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 8 / 30

The First Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 8 / 30

The First Derivative Test Illustrated

2 is a critical point,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 8 / 30

The First Derivative Test Illustrated

2 is a critical point, f ′ changes sign from

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 8 / 30

The First Derivative Test Illustrated

2 is a critical point, f ′ changes sign from +ve to -ve at 2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 8 / 30

The First Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 9 / 30

The First Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 9 / 30

The First Derivative Test Illustrated

1 is a critical point,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 9 / 30

The First Derivative Test Illustrated

1 is a critical point, f ′ changes sign from

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 9 / 30

The First Derivative Test Illustrated

1 is a critical point, f ′ changes sign from -ve to +ve at 1.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 9 / 30

The First Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 10 / 30

The First Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 10 / 30

The First Derivative Test Illustrated

2 is a critical point,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 10 / 30

The First Derivative Test Illustrated

2 is a critical point, f ′ keeps

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 10 / 30

The First Derivative Test Illustrated

2 is a critical point, f ′ keeps +ve sign around x = 2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 10 / 30

The First Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 11 / 30

The First Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 11 / 30

The First Derivative Test Illustrated

2 is a critical point,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 11 / 30

The First Derivative Test Illustrated

2 is a critical point, f ′ keeps

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 11 / 30

The First Derivative Test Illustrated

2 is a critical point, f ′ keeps -ve sign around x = 2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 11 / 30

The First Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 12 / 30

The First Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 12 / 30

The First Derivative Test Illustrated

Observe

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 12 / 30

The First Derivative Test Illustrated

Observe how

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 12 / 30

The First Derivative Test Illustrated

Observe how f ′ changes its sign.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 12 / 30

What does f ′′ say about f ?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c),

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents on an interval I,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I. That is,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I. That is,

f (x) > lc(x)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I. That is,

f (x) > lc(x) for all x ∈ I \ {c}

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I. That is,

f (x) > lc(x) for all x ∈ I \ {c} and for all c ∈ I.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I. That is,

f (x) > lc(x) for all x ∈ I \ {c} and for all c ∈ I.

3 If the graph of f

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I. That is,

f (x) > lc(x) for all x ∈ I \ {c} and for all c ∈ I.

3 If the graph of f lies below all of its tangents

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I. That is,

f (x) > lc(x) for all x ∈ I \ {c} and for all c ∈ I.

3 If the graph of f lies below all of its tangents on an interval I,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I. That is,

f (x) > lc(x) for all x ∈ I \ {c} and for all c ∈ I.

3 If the graph of f lies below all of its tangents on an interval I, then itis called concave downward (CD) on I.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I. That is,

f (x) > lc(x) for all x ∈ I \ {c} and for all c ∈ I.

3 If the graph of f lies below all of its tangents on an interval I, then itis called concave downward (CD) on I. That is,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I. That is,

f (x) > lc(x) for all x ∈ I \ {c} and for all c ∈ I.

3 If the graph of f lies below all of its tangents on an interval I, then itis called concave downward (CD) on I. That is,

f (x) < lc(x)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I. That is,

f (x) > lc(x) for all x ∈ I \ {c} and for all c ∈ I.

3 If the graph of f lies below all of its tangents on an interval I, then itis called concave downward (CD) on I. That is,

f (x) < lc(x) for all x ∈ I \ {c}

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

What does f ′′ say about f ?

DefinitionLet f be differentiable on an interval I, and c ∈ I.

1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .

2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I. That is,

f (x) > lc(x) for all x ∈ I \ {c} and for all c ∈ I.

3 If the graph of f lies below all of its tangents on an interval I, then itis called concave downward (CD) on I. That is,

f (x) < lc(x) for all x ∈ I \ {c} and for all c ∈ I.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 13 / 30

On the definition of CU and CD

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.4 Now have a look at this picture.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.4 Now have a look at this picture.

Then,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.4 Now have a look at this picture.

Then,

f is (CU) on I if

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.4 Now have a look at this picture.

Then,

f is (CU) on I if graph of f lies above all those lines drawn in (2)and (3) above.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.4 Now have a look at this picture.

Then,

f is (CU) on I if graph of f lies above all those lines drawn in (2)and (3) above.

f is (CD) on I if

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.4 Now have a look at this picture.

Then,

f is (CU) on I if graph of f lies above all those lines drawn in (2)and (3) above.

f is (CD) on I if graph of f lies below all those lines drawn in (2)and (3) above.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.4 Now have a look at this picture.

Then,

f is (CU) on I if graph of f lies above all those lines drawn in (2)and (3) above.

f is (CD) on I if graph of f lies below all those lines drawn in (2)and (3) above.

Caution: Some books call

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.4 Now have a look at this picture.

Then,

f is (CU) on I if graph of f lies above all those lines drawn in (2)and (3) above.

f is (CD) on I if graph of f lies below all those lines drawn in (2)and (3) above.

Caution: Some books call concave upward as convex ,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.4 Now have a look at this picture.

Then,

f is (CU) on I if graph of f lies above all those lines drawn in (2)and (3) above.

f is (CD) on I if graph of f lies below all those lines drawn in (2)and (3) above.

Caution: Some books call concave upward as convex , and

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.4 Now have a look at this picture.

Then,

f is (CU) on I if graph of f lies above all those lines drawn in (2)and (3) above.

f is (CD) on I if graph of f lies below all those lines drawn in (2)and (3) above.

Caution: Some books call concave upward as convex , and concavedownward as concave .

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.4 Now have a look at this picture.

Then,

f is (CU) on I if graph of f lies above all those lines drawn in (2)and (3) above.

f is (CD) on I if graph of f lies below all those lines drawn in (2)and (3) above.

Caution: Some books call concave upward as convex , and concavedownward as concave . We defined the concepts of CU and CD

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

On the definition of CU and CD

Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.

1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.4 Now have a look at this picture.

Then,

f is (CU) on I if graph of f lies above all those lines drawn in (2)and (3) above.

f is (CD) on I if graph of f lies below all those lines drawn in (2)and (3) above.

Caution: Some books call concave upward as convex , and concavedownward as concave . We defined the concepts of CU and CD onlyon an interval.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 14 / 30

CU, CD Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 15 / 30

CU, CD Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 15 / 30

CU, CD Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 16 / 30

CU, CD Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 16 / 30

CU, CD Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 17 / 30

CU, CD Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 17 / 30

What does f ′′ say about f ?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is increasing.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is increasing.2 How do we prove f is CU on I?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is increasing.2 How do we prove f is CU on I? we prove graph of f lies above

tangent line at any point.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is increasing.2 How do we prove f is CU on I? we prove graph of f lies above

tangent line at any point.3 Let a ∈ I.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is increasing.2 How do we prove f is CU on I? we prove graph of f lies above

tangent line at any point.3 Let a ∈ I. let x > a.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is increasing.2 How do we prove f is CU on I? we prove graph of f lies above

tangent line at any point.3 Let a ∈ I. let x > a. Then f (x) − f (a) = f ′(c)(x − a) for some

c ∈ [a, x ]

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is increasing.2 How do we prove f is CU on I? we prove graph of f lies above

tangent line at any point.3 Let a ∈ I. let x > a. Then f (x) − f (a) = f ′(c)(x − a) for some

c ∈ [a, x ] by MVT.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is increasing.2 How do we prove f is CU on I? we prove graph of f lies above

tangent line at any point.3 Let a ∈ I. let x > a. Then f (x) − f (a) = f ′(c)(x − a) for some

c ∈ [a, x ] by MVT.4 f (x) − f (a) = f ′(c)(x − a) > f ′(a)(x − a).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is increasing.2 How do we prove f is CU on I? we prove graph of f lies above

tangent line at any point.3 Let a ∈ I. let x > a. Then f (x) − f (a) = f ′(c)(x − a) for some

c ∈ [a, x ] by MVT.4 f (x) − f (a) = f ′(c)(x − a) > f ′(a)(x − a). QED+5 repeat arguments in (3) and (4),

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is increasing.2 How do we prove f is CU on I? we prove graph of f lies above

tangent line at any point.3 Let a ∈ I. let x > a. Then f (x) − f (a) = f ′(c)(x − a) for some

c ∈ [a, x ] by MVT.4 f (x) − f (a) = f ′(c)(x − a) > f ′(a)(x − a). QED+5 repeat arguments in (3) and (4), for x < a.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is increasing.2 How do we prove f is CU on I? we prove graph of f lies above

tangent line at any point.3 Let a ∈ I. let x > a. Then f (x) − f (a) = f ′(c)(x − a) for some

c ∈ [a, x ] by MVT.4 f (x) − f (a) = f ′(c)(x − a) > f ′(a)(x − a). QED+5 repeat arguments in (3) and (4), for x < a. QED

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 18 / 30

What does f ′′ say about f ?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 19 / 30

What does f ′′ say about f ?

DefinitionA point P on a curve y = f (x) is called an inflection point if

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 19 / 30

What does f ′′ say about f ?

DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 19 / 30

What does f ′′ say about f ?

DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there and

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 19 / 30

What does f ′′ say about f ?

DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there and the curve changes from concave upward toconcave downward

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 19 / 30

What does f ′′ say about f ?

DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there and the curve changes from concave upward toconcave downward or

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 19 / 30

What does f ′′ say about f ?

DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there and the curve changes from concave upward toconcave downward or concave downward to concave upward at P.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 19 / 30

What does f ′′ say about f ?

DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there and the curve changes from concave upward toconcave downward or concave downward to concave upward at P.

Interpretation: If f (t) is the distance travelled by the car, secondderivative is

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 19 / 30

What does f ′′ say about f ?

DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there and the curve changes from concave upward toconcave downward or concave downward to concave upward at P.

Interpretation: If f (t) is the distance travelled by the car, secondderivative is its acceleration/deceleration.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 19 / 30

What does f ′′ say about f ?

DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there and the curve changes from concave upward toconcave downward or concave downward to concave upward at P.

Interpretation: If f (t) is the distance travelled by the car, secondderivative is its acceleration/deceleration. An inflection point is the timeat which car changes from acceleration to deceleration or vice versa.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 19 / 30

What does f ′′ say about f ?

DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there and the curve changes from concave upward toconcave downward or concave downward to concave upward at P.

Interpretation: If f (t) is the distance travelled by the car, secondderivative is its acceleration/deceleration. An inflection point is the timeat which car changes from acceleration to deceleration or vice versa.

Question: If f ′′(c) = 0, then is it true that (c, f (c)) is an inflection point?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 19 / 30

What does f ′′ say about f ?

DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there and the curve changes from concave upward toconcave downward or concave downward to concave upward at P.

Interpretation: If f (t) is the distance travelled by the car, secondderivative is its acceleration/deceleration. An inflection point is the timeat which car changes from acceleration to deceleration or vice versa.

Question: If f ′′(c) = 0, then is it true that (c, f (c)) is an inflection point?No.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 19 / 30

What does f ′′ say about f ?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 20 / 30

What does f ′′ say about f ?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 20 / 30

What does f ′′ say about f ?

Graph of f (x) = x4.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 20 / 30

What does f ′′ say about f ?

Graph of f (x) = x4. Is x = 0 an inflection point?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 20 / 30

What does f ′′ say about f ?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 21 / 30

What does f ′′ say about f ?

Sketch a possible graph of a function f that satisfies the followingconditions:

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 21 / 30

What does f ′′ say about f ?

Sketch a possible graph of a function f that satisfies the followingconditions:

1 f ′(x) > 0 on (−∞,1), f ′(x) < 0 on (1,∞).2 f ′′(x) > 0 on (−∞,−2) and (2,∞), f ′′(x) < 0 on (−2,2)

3 limx→−∞

f (x) = −2, limx→∞

f (x) = 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 21 / 30

What does f ′′ say about f ?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 22 / 30

How to catch local Minima/Maxima points using Second Deriva te?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 23 / 30

How to catch local Minima/Maxima points using Second Deriva te?

Second Derivative Test

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 23 / 30

How to catch local Minima/Maxima points using Second Deriva te?

Second Derivative Test1 If f ′(c) = 0 and f ′′(c) > 0,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 23 / 30

How to catch local Minima/Maxima points using Second Deriva te?

Second Derivative Test1 If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 23 / 30

How to catch local Minima/Maxima points using Second Deriva te?

Second Derivative Test1 If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.2 If f ′(c) = 0 and f ′′(c) < 0,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 23 / 30

How to catch local Minima/Maxima points using Second Deriva te?

Second Derivative Test1 If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.2 If f ′(c) = 0 and f ′′(c) < 0, then f has a local maximum at c.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 23 / 30

How to catch local Minima/Maxima points using Second Deriva te?

Second Derivative Test1 If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.2 If f ′(c) = 0 and f ′′(c) < 0, then f has a local maximum at c.

Remark: The Second derivative test is silent if

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 23 / 30

How to catch local Minima/Maxima points using Second Deriva te?

Second Derivative Test1 If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.2 If f ′(c) = 0 and f ′′(c) < 0, then f has a local maximum at c.

Remark: The Second derivative test is silent if f ′′(c) = 0

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 23 / 30

How to catch local Minima/Maxima points using Second Deriva te?

Second Derivative Test1 If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.2 If f ′(c) = 0 and f ′′(c) < 0, then f has a local maximum at c.

Remark: The Second derivative test is silent if f ′′(c) = 0Proof is a consequence of Concavity test .

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 23 / 30

Why Second Derivate Test could be true?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 24 / 30

Why Second Derivate Test could be true?

Let’s prove:

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 24 / 30

Why Second Derivate Test could be true?

Let’s prove: If f ′(c) = 0 and f ′′(c) > 0,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 24 / 30

Why Second Derivate Test could be true?

Let’s prove: If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 24 / 30

Why Second Derivate Test could be true?

Let’s prove: If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.Proof: Recall

TheoremSuppose f is differentiable at a.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 24 / 30

Why Second Derivate Test could be true?

Let’s prove: If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.Proof: Recall

TheoremSuppose f is differentiable at a. Then there exists a function φ suchthat

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 24 / 30

Why Second Derivate Test could be true?

Let’s prove: If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.Proof: Recall

TheoremSuppose f is differentiable at a. Then there exists a function φ suchthat

f (x) = f (a) + (x − a)f ′(a) + (x − a)φ(x),

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 24 / 30

Why Second Derivate Test could be true?

Let’s prove: If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.Proof: Recall

TheoremSuppose f is differentiable at a. Then there exists a function φ suchthat

f (x) = f (a) + (x − a)f ′(a) + (x − a)φ(x),

andlimx→a

φ(x) = 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 24 / 30

Why Second Derivate Test could be true?

Let’s prove: If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.Proof: Recall

TheoremSuppose f is differentiable at a. Then there exists a function φ suchthat

f (x) = f (a) + (x − a)f ′(a) + (x − a)φ(x),

andlimx→a

φ(x) = 0.

A similar result is true:

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 24 / 30

Why Second Derivate Test could be true? (contd.)

TheoremSuppose f is twice differentiable at a.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 25 / 30

Why Second Derivate Test could be true? (contd.)

TheoremSuppose f is twice differentiable at a. Then there exists a function ψsuch that

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 25 / 30

Why Second Derivate Test could be true? (contd.)

TheoremSuppose f is twice differentiable at a. Then there exists a function ψsuch that

f (x) = f (a) + (x − a)f ′(a) +(x − a)2

2f ′′(a) +

(x − a)2

2ψ(x),

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 25 / 30

Why Second Derivate Test could be true? (contd.)

TheoremSuppose f is twice differentiable at a. Then there exists a function ψsuch that

f (x) = f (a) + (x − a)f ′(a) +(x − a)2

2f ′′(a) +

(x − a)2

2ψ(x),

andlimx→a

ψ(x) = 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 25 / 30

Why Second Derivate Test could be true? (contd.)

TheoremSuppose f is twice differentiable at a. Then there exists a function ψsuch that

f (x) = f (a) + (x − a)f ′(a) +(x − a)2

2f ′′(a) +

(x − a)2

2ψ(x),

andlimx→a

ψ(x) = 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 25 / 30

Why Second Derivate Test could be true? (contd.)

Now, at the critical point c,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 26 / 30

Why Second Derivate Test could be true? (contd.)

Now, at the critical point c,

f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 26 / 30

Why Second Derivate Test could be true? (contd.)

Now, at the critical point c,

f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 26 / 30

Why Second Derivate Test could be true? (contd.)

Now, at the critical point c,

f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.

The required conclusion follows by looking at the three terms on RHSof the first equation:

1 first term is 0

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 26 / 30

Why Second Derivate Test could be true? (contd.)

Now, at the critical point c,

f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.

The required conclusion follows by looking at the three terms on RHSof the first equation:

1 first term is 02 The third term is negligible,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 26 / 30

Why Second Derivate Test could be true? (contd.)

Now, at the critical point c,

f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.

The required conclusion follows by looking at the three terms on RHSof the first equation:

1 first term is 02 The third term is negligible, compared to the second term,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 26 / 30

Why Second Derivate Test could be true? (contd.)

Now, at the critical point c,

f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.

The required conclusion follows by looking at the three terms on RHSof the first equation:

1 first term is 02 The third term is negligible, compared to the second term, when

we are near c.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 26 / 30

Why Second Derivate Test could be true? (contd.)

Now, at the critical point c,

f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.

The required conclusion follows by looking at the three terms on RHSof the first equation:

1 first term is 02 The third term is negligible, compared to the second term, when

we are near c.3 Thus the sign of RHS is determined by that of

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 26 / 30

Why Second Derivate Test could be true? (contd.)

Now, at the critical point c,

f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.

The required conclusion follows by looking at the three terms on RHSof the first equation:

1 first term is 02 The third term is negligible, compared to the second term, when

we are near c.3 Thus the sign of RHS is determined by that of f ′′(c).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 26 / 30

Why Second Derivate Test could be true? (contd.)

Now, at the critical point c,

f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.

The required conclusion follows by looking at the three terms on RHSof the first equation:

1 first term is 02 The third term is negligible, compared to the second term, when

we are near c.3 Thus the sign of RHS is determined by that of f ′′(c). QED

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 26 / 30

The Second Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 27 / 30

The Second Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 27 / 30

The Second Derivative Test Illustrated

2 is a critical point,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 27 / 30

The Second Derivative Test Illustrated

2 is a critical point, f ′′(2) < 0

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 27 / 30

The Second Derivative Test Illustrated

2 is a critical point, f ′′(2) < 0 and f has a local maximum at 2

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 27 / 30

The Second Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 28 / 30

The Second Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 28 / 30

The Second Derivative Test Illustrated

1 is a critical point,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 28 / 30

The Second Derivative Test Illustrated

1 is a critical point, f ′′(1) > 0

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 28 / 30

The Second Derivative Test Illustrated

1 is a critical point, f ′′(1) > 0 and f has a local maximum at 1

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 28 / 30

The Second Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 29 / 30

The Second Derivative Test Illustrated

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 29 / 30

The Second Derivative Test Illustrated

2 is a critical point,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 29 / 30

The Second Derivative Test Illustrated

2 is a critical point, f ′′(2) = 0

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 29 / 30

The Second Derivative Test Illustrated

2 is a critical point, f ′′(2) = 0 and f does not have a local extremum at2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 29 / 30

Problems

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 30 / 30

Problems

1 Sketch a possible graph of a function f that satisfies the followingconditions:

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 30 / 30

Problems

1 Sketch a possible graph of a function f that satisfies the followingconditions:

1 f ′(0) = f ′(2) = f ′(4) = 0,2 f ′(x) > 0 on (−∞, 0) and (2, 4),3 f ′(x) < 0 on (0, 2) and (4,∞),4 f ′′(x) > 0 on (1, 3),5 f ′′(x) < 0 on (−∞, 1) and (3,∞).

2 Show that the function g(x) = x |x | has an inflection point at (0,0)but g′′(0) does not exist.

3 Prove that if (c, f (c)) is an inflection point of the graph of f andf ′′(c) exists in an open interval that contains c, then f ′′(c) = 0.

4 Show that tan x > x for 0 < x < π

2 .

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 30 / 30