SLecture5 Rolle's Theorem and Mean Value Theorem

MA 105 Calculus: Lecture 5Rolle’s Theorem and Mean Value Theorem

S. Sivaji Ganesh

Mathematics DepartmentIIT Bombay

August 3, 2009

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 1 / 20

Rolle’s theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

We want to draw a graph of a function f with domain [0, 1].

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Plot the values of f ,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Plot the values of f , given by f (0) = f (1).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Plot the values of f , given by f (0) = f (1). Choose this value to be

anything that you like.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Plot the values of f , given by f (0) = f (1). Choose this value to be

anything that you like.3 Draw any ‘nice’ continuous curve joining these two points

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Plot the values of f , given by f (0) = f (1). Choose this value to be

anything that you like.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Plot the values of f , given by f (0) = f (1). Choose this value to be

anything that you like.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

4 We have finished defining our function

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Plot the values of f , given by f (0) = f (1). Choose this value to be

anything that you like.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

4 We have finished defining our function5 Draw some more graphs

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Plot the values of f , given by f (0) = f (1). Choose this value to be

anything that you like.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

4 We have finished defining our function5 Draw some more graphs6 What are your observations?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Plot the values of f , given by f (0) = f (1). Choose this value to be

anything that you like.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

4 We have finished defining our function5 Draw some more graphs6 What are your observations?

There is a point at which

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Plot the values of f , given by f (0) = f (1). Choose this value to be

anything that you like.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

4 We have finished defining our function5 Draw some more graphs6 What are your observations?

There is a point at which tangent is parallel to x-axis.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Plot the values of f , given by f (0) = f (1). Choose this value to be

anything that you like.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

4 We have finished defining our function5 Draw some more graphs6 What are your observations?

There is a point at which tangent is parallel to x-axis. There may bemore than one.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 2 / 20

Rolle’s theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 3 / 20

Rolle’s theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 4 / 20

Rolle’s theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 5 / 20

Rolle’s theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 6 / 20

Rolle’s theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that f ′(c) = 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that f ′(c) = 0.

Note: All our examples illustrate Rolle’s theorem.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that f ′(c) = 0.

Note: All our examples illustrate Rolle’s theorem.Proof: is an application of EVT and Fermat’s theorem.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that f ′(c) = 0.

Note: All our examples illustrate Rolle’s theorem.Proof: is an application of EVT and Fermat’s theorem.

1 If f is a constant function

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that f ′(c) = 0.

Note: All our examples illustrate Rolle’s theorem.Proof: is an application of EVT and Fermat’s theorem.

1 If f is a constant function i.e., f (x) = f (a) for every x ∈ [a, b],

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that f ′(c) = 0.

Note: All our examples illustrate Rolle’s theorem.Proof: is an application of EVT and Fermat’s theorem.

1 If f is a constant function i.e., f (x) = f (a) for every x ∈ [a, b], sucha c exists.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that f ′(c) = 0.

Note: All our examples illustrate Rolle’s theorem.Proof: is an application of EVT and Fermat’s theorem.

1 If f is a constant function i.e., f (x) = f (a) for every x ∈ [a, b], sucha c exists. why?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that f ′(c) = 0.

Note: All our examples illustrate Rolle’s theorem.Proof: is an application of EVT and Fermat’s theorem.

1 If f is a constant function i.e., f (x) = f (a) for every x ∈ [a, b], sucha c exists. why?

2 If f is not a constant,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that f ′(c) = 0.

Note: All our examples illustrate Rolle’s theorem.Proof: is an application of EVT and Fermat’s theorem.

1 If f is a constant function i.e., f (x) = f (a) for every x ∈ [a, b], sucha c exists. why?

2 If f is not a constant, then at least one of the following holds.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that f ′(c) = 0.

Note: All our examples illustrate Rolle’s theorem.Proof: is an application of EVT and Fermat’s theorem.

1 If f is a constant function i.e., f (x) = f (a) for every x ∈ [a, b], sucha c exists. why?

2 If f is not a constant, then at least one of the following holds.(i) The graph of f goes above the line y = f (a)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that f ′(c) = 0.

Note: All our examples illustrate Rolle’s theorem.Proof: is an application of EVT and Fermat’s theorem.

1 If f is a constant function i.e., f (x) = f (a) for every x ∈ [a, b], sucha c exists. why?

2 If f is not a constant, then at least one of the following holds.(i) The graph of f goes above the line y = f (a) i.e., f (x) > f (a) for

some x ∈ (a, b).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that f ′(c) = 0.

Note: All our examples illustrate Rolle’s theorem.Proof: is an application of EVT and Fermat’s theorem.

1 If f is a constant function i.e., f (x) = f (a) for every x ∈ [a, b], sucha c exists. why?

2 If f is not a constant, then at least one of the following holds.(i) The graph of f goes above the line y = f (a) i.e., f (x) > f (a) for

some x ∈ (a, b).(ii) The graph of f goes below the line y = f (a)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that f ′(c) = 0.

Note: All our examples illustrate Rolle’s theorem.Proof: is an application of EVT and Fermat’s theorem.

1 If f is a constant function i.e., f (x) = f (a) for every x ∈ [a, b], sucha c exists. why?

2 If f is not a constant, then at least one of the following holds.(i) The graph of f goes above the line y = f (a) i.e., f (x) > f (a) for

some x ∈ (a, b).(ii) The graph of f goes below the line y = f (a) i.e., f (x) < f (a) for

some x ∈ (a, b).S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

Theorem (Rolle’s theorem)

Let f be a function that satisfies the following three hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).3 f (a) = f (b).

Then there is a number c in the open interval (a, b) such that f ′(c) = 0.

Note: All our examples illustrate Rolle’s theorem.Proof: is an application of EVT and Fermat’s theorem.

1 If f is a constant function i.e., f (x) = f (a) for every x ∈ [a, b], sucha c exists. why?

2 If f is not a constant, then at least one of the following holds.(i) The graph of f goes above the line y = f (a) i.e., f (x) > f (a) for

some x ∈ (a, b).(ii) The graph of f goes below the line y = f (a) i.e., f (x) < f (a) for

some x ∈ (a, b).S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 7 / 20

Rolle’s theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it c.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it c.That is, global maximum on [a, b] is actually a local maximum

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it c.That is, global maximum on [a, b] is actually a local maximum ByFermat’s theorem f ′(c) = 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it c.That is, global maximum on [a, b] is actually a local maximum ByFermat’s theorem f ′(c) = 0.

2(ii) If the graph of f goes below the line y = f (a), then where is theglobal minimum attained? (EVT guarantees existence of such apoint)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it c.That is, global maximum on [a, b] is actually a local maximum ByFermat’s theorem f ′(c) = 0.

2(ii) If the graph of f goes below the line y = f (a), then where is theglobal minimum attained? (EVT guarantees existence of such apoint) It cannot be at a or b.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it c.That is, global maximum on [a, b] is actually a local maximum ByFermat’s theorem f ′(c) = 0.

2(ii) If the graph of f goes below the line y = f (a), then where is theglobal minimum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it c.That is, global maximum on [a, b] is actually a local maximum ByFermat’s theorem f ′(c) = 0.

2(ii) If the graph of f goes below the line y = f (a), then where is theglobal minimum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it c.That is, global maximum on [a, b] is actually a local maximum ByFermat’s theorem f ′(c) = 0.

2(ii) If the graph of f goes below the line y = f (a), then where is theglobal minimum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it d .

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it c.That is, global maximum on [a, b] is actually a local maximum ByFermat’s theorem f ′(c) = 0.

2(ii) If the graph of f goes below the line y = f (a), then where is theglobal minimum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it d .That is, global minimum on [a, b] is actually a local minimum

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it c.That is, global maximum on [a, b] is actually a local maximum ByFermat’s theorem f ′(c) = 0.

2(ii) If the graph of f goes below the line y = f (a), then where is theglobal minimum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it d .That is, global minimum on [a, b] is actually a local minimum ByFermat’s theorem f ′(d) = 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it c.That is, global maximum on [a, b] is actually a local maximum ByFermat’s theorem f ′(c) = 0.

2(ii) If the graph of f goes below the line y = f (a), then where is theglobal minimum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it d .That is, global minimum on [a, b] is actually a local minimum ByFermat’s theorem f ′(d) = 0.

This finishes the proof of Rolle’s theorem.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it c.That is, global maximum on [a, b] is actually a local maximum ByFermat’s theorem f ′(c) = 0.

2(ii) If the graph of f goes below the line y = f (a), then where is theglobal minimum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it d .That is, global minimum on [a, b] is actually a local minimum ByFermat’s theorem f ′(d) = 0.

This finishes the proof of Rolle’s theorem.Graph of any function satisfying the hypothesis of Rolle’s theorem beremembered as

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Rolle’s theorem

Proof of Rolle’s theorem (contd.):

2(i) If the graph of f goes above the line y = f (a), then where is theglobal maximum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it c.That is, global maximum on [a, b] is actually a local maximum ByFermat’s theorem f ′(c) = 0.

2(ii) If the graph of f goes below the line y = f (a), then where is theglobal minimum attained? (EVT guarantees existence of such apoint) It cannot be at a or b. So? it must inside (a, b). call it d .That is, global minimum on [a, b] is actually a local minimum ByFermat’s theorem f ′(d) = 0.

This finishes the proof of Rolle’s theorem.Graph of any function satisfying the hypothesis of Rolle’s theorem beremembered as a graph having geometry of Rolle’s theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 8 / 20

Mean Value Theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1].

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Mark the points (0, f (0)) and (1, f (1)).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Mark the points (0, f (0)) and (1, f (1)). Let f (0), f (1) be of your

choice.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Mark the points (0, f (0)) and (1, f (1)). Let f (0), f (1) be of your

choice.3 Draw any ‘nice’ continuous curve joining these two points

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Mark the points (0, f (0)) and (1, f (1)). Let f (0), f (1) be of your

choice.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Mark the points (0, f (0)) and (1, f (1)). Let f (0), f (1) be of your

choice.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

4 We have finished defining our function

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Mark the points (0, f (0)) and (1, f (1)). Let f (0), f (1) be of your

choice.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

4 We have finished defining our function5 Join the two points (0, f (0)) and (1, f (1)) by a straight line

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Mark the points (0, f (0)) and (1, f (1)). Let f (0), f (1) be of your

choice.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

4 We have finished defining our function5 Join the two points (0, f (0)) and (1, f (1)) by a straight line6 What are your observations?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Mark the points (0, f (0)) and (1, f (1)). Let f (0), f (1) be of your

choice.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

4 We have finished defining our function5 Join the two points (0, f (0)) and (1, f (1)) by a straight line6 What are your observations?7 Is there a point on the graph

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Mark the points (0, f (0)) and (1, f (1)). Let f (0), f (1) be of your

choice.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

4 We have finished defining our function5 Join the two points (0, f (0)) and (1, f (1)) by a straight line6 What are your observations?7 Is there a point on the graph where the tangent is

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Mark the points (0, f (0)) and (1, f (1)). Let f (0), f (1) be of your

choice.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

4 We have finished defining our function5 Join the two points (0, f (0)) and (1, f (1)) by a straight line6 What are your observations?7 Is there a point on the graph where the tangent is parallel to the

straight line which was drawn?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Mark the points (0, f (0)) and (1, f (1)). Let f (0), f (1) be of your

choice.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

4 We have finished defining our function5 Join the two points (0, f (0)) and (1, f (1)) by a straight line6 What are your observations?7 Is there a point on the graph where the tangent is parallel to the

straight line which was drawn?

There is such a point.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean Value Theorem

We want to draw a graph of a function f with domain [0, 1]. Do thefollowing:

1 Mark the domain on x-axis2 Mark the points (0, f (0)) and (1, f (1)). Let f (0), f (1) be of your

choice.3 Draw any ‘nice’ continuous curve joining these two points i.e.,

(0, f (0)) and (1, f (1))

4 We have finished defining our function5 Join the two points (0, f (0)) and (1, f (1)) by a straight line6 What are your observations?7 Is there a point on the graph where the tangent is parallel to the

straight line which was drawn?

There is such a point.There may be more than one.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 9 / 20

Mean value theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 10 / 20

Mean value theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 11 / 20

Mean value theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 12 / 20

Mean value theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 13 / 20

Mean Value Theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 14 / 20

Mean Value Theorem

This theorem is due to J.-L. Lagrange.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 14 / 20

Mean Value Theorem

This theorem is due to J.-L. Lagrange.

Theorem (Mean value theorem)

Let f be a function that satisfies the following hypotheses:

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 14 / 20

Mean Value Theorem

This theorem is due to J.-L. Lagrange.

Theorem (Mean value theorem)

Let f be a function that satisfies the following hypotheses:1 f is continuous on the closed interval [a, b].

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 14 / 20

Mean Value Theorem

This theorem is due to J.-L. Lagrange.

Theorem (Mean value theorem)

Let f be a function that satisfies the following hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 14 / 20

Mean Value Theorem

This theorem is due to J.-L. Lagrange.

Theorem (Mean value theorem)

Let f be a function that satisfies the following hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).

Then

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 14 / 20

Mean Value Theorem

This theorem is due to J.-L. Lagrange.

Theorem (Mean value theorem)

Let f be a function that satisfies the following hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).

Then there is a number c in the open interval (a, b)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 14 / 20

Mean Value Theorem

This theorem is due to J.-L. Lagrange.

Theorem (Mean value theorem)

Let f be a function that satisfies the following hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).

Then there is a number c in the open interval (a, b) such that

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 14 / 20

Mean Value Theorem

This theorem is due to J.-L. Lagrange.

Theorem (Mean value theorem)

Let f be a function that satisfies the following hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).

Then there is a number c in the open interval (a, b) such that

f ′(c) =f (b) − f (a)

b − a.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 14 / 20

Mean Value Theorem

This theorem is due to J.-L. Lagrange.

Theorem (Mean value theorem)

Let f be a function that satisfies the following hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).

Then there is a number c in the open interval (a, b) such that

f ′(c) =f (b) − f (a)

b − a.

or, equivalently,f (b)− f (a) = f ′(c)(b − a).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 14 / 20

Mean Value Theorem

This theorem is due to J.-L. Lagrange.

Theorem (Mean value theorem)

Let f be a function that satisfies the following hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).

Then there is a number c in the open interval (a, b) such that

f ′(c) =f (b) − f (a)

b − a.

or, equivalently,f (b)− f (a) = f ′(c)(b − a).

Note: Our examples illustrate Mean value theorem.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 14 / 20

Mean Value Theorem

This theorem is due to J.-L. Lagrange.

Theorem (Mean value theorem)

Let f be a function that satisfies the following hypotheses:1 f is continuous on the closed interval [a, b].2 f is differentiable on the open interval (a, b).

Then there is a number c in the open interval (a, b) such that

f ′(c) =f (b) − f (a)

b − a.

or, equivalently,f (b)− f (a) = f ′(c)(b − a).

Note: Our examples illustrate Mean value theorem.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 14 / 20

Mean Value Theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 15 / 20

Mean Value Theorem

Questions:

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 15 / 20

Mean Value Theorem

Questions:1 Can we prove Rolle’s theorem using the Mean value theorem?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 15 / 20

Mean Value Theorem

Questions:1 Can we prove Rolle’s theorem using the Mean value theorem?

yes!

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 15 / 20

Mean Value Theorem

Questions:1 Can we prove Rolle’s theorem using the Mean value theorem?

yes! Easy proof.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 15 / 20

Mean Value Theorem

Questions:1 Can we prove Rolle’s theorem using the Mean value theorem?

yes! Easy proof.2 We deduce Mean value theorem from Rolle’s theorem here.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 15 / 20

Mean Value Theorem

Questions:1 Can we prove Rolle’s theorem using the Mean value theorem?

yes! Easy proof.2 We deduce Mean value theorem from Rolle’s theorem here.3 Therefore, the statements of the two theorems are “one and the

same” or “equivalent”.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 15 / 20

Mean Value Theorem

Questions:1 Can we prove Rolle’s theorem using the Mean value theorem?

yes! Easy proof.2 We deduce Mean value theorem from Rolle’s theorem here.3 Therefore, the statements of the two theorems are “one and the

same” or “equivalent”.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 15 / 20

Mean Value Theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 16 / 20

Mean Value Theorem

Proof of Mean value theorem:

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 16 / 20

Mean Value Theorem

Proof of Mean value theorem: The strategy is to define a newfunction φ(x)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 16 / 20

Mean Value Theorem

Proof of Mean value theorem: The strategy is to define a newfunction φ(x) satisfying the hypothesis of Rolle’s theorem.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 16 / 20

Mean Value Theorem

Proof of Mean value theorem: The strategy is to define a newfunction φ(x) satisfying the hypothesis of Rolle’s theorem. Theconclusion of Rolle’s theorem for φ should yield the conclusion of MVTfor f .

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 16 / 20

Mean Value Theorem

Proof of Mean value theorem: The strategy is to define a newfunction φ(x) satisfying the hypothesis of Rolle’s theorem. Theconclusion of Rolle’s theorem for φ should yield the conclusion of MVTfor f .

1 Define φ on [a, b] by

φ(x) = f (x) − f (a) − f (b) − f (a)

b − a(x − a).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 16 / 20

Mean Value Theorem

Proof of Mean value theorem: The strategy is to define a newfunction φ(x) satisfying the hypothesis of Rolle’s theorem. Theconclusion of Rolle’s theorem for φ should yield the conclusion of MVTfor f .

1 Define φ on [a, b] by

φ(x) = f (x) − f (a) − f (b) − f (a)

b − a(x − a).

What is the graph of φ?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 16 / 20

Mean Value Theorem

Proof of Mean value theorem: The strategy is to define a newfunction φ(x) satisfying the hypothesis of Rolle’s theorem. Theconclusion of Rolle’s theorem for φ should yield the conclusion of MVTfor f .

1 Define φ on [a, b] by

φ(x) = f (x) − f (a) − f (b) − f (a)

b − a(x − a).

What is the graph of φ? Does it have Rolle’s theorem geometry?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 16 / 20

Mean Value Theorem

Proof of Mean value theorem: The strategy is to define a newfunction φ(x) satisfying the hypothesis of Rolle’s theorem. Theconclusion of Rolle’s theorem for φ should yield the conclusion of MVTfor f .

1 Define φ on [a, b] by

φ(x) = f (x) − f (a) − f (b) − f (a)

b − a(x − a).

What is the graph of φ? Does it have Rolle’s theorem geometry?2 Can we apply Rolle’s theorem to φ

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 16 / 20

Mean Value Theorem

Proof of Mean value theorem: The strategy is to define a newfunction φ(x) satisfying the hypothesis of Rolle’s theorem. Theconclusion of Rolle’s theorem for φ should yield the conclusion of MVTfor f .

1 Define φ on [a, b] by

φ(x) = f (x) − f (a) − f (b) − f (a)

b − a(x − a).

What is the graph of φ? Does it have Rolle’s theorem geometry?2 Can we apply Rolle’s theorem to φ on [a, b]?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 16 / 20

Mean Value Theorem

Proof of Mean value theorem: The strategy is to define a newfunction φ(x) satisfying the hypothesis of Rolle’s theorem. Theconclusion of Rolle’s theorem for φ should yield the conclusion of MVTfor f .

1 Define φ on [a, b] by

φ(x) = f (x) − f (a) − f (b) − f (a)

b − a(x − a).

What is the graph of φ? Does it have Rolle’s theorem geometry?2 Can we apply Rolle’s theorem to φ on [a, b]? Yes!

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 16 / 20

Mean Value Theorem

Proof of Mean value theorem: The strategy is to define a newfunction φ(x) satisfying the hypothesis of Rolle’s theorem. Theconclusion of Rolle’s theorem for φ should yield the conclusion of MVTfor f .

1 Define φ on [a, b] by

φ(x) = f (x) − f (a) − f (b) − f (a)

b − a(x − a).

What is the graph of φ? Does it have Rolle’s theorem geometry?2 Can we apply Rolle’s theorem to φ on [a, b]? Yes!3 Rolle’s theorem asserts the existence of c ∈ (a, b)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 16 / 20

Mean Value Theorem

Proof of Mean value theorem: The strategy is to define a newfunction φ(x) satisfying the hypothesis of Rolle’s theorem. Theconclusion of Rolle’s theorem for φ should yield the conclusion of MVTfor f .

1 Define φ on [a, b] by

φ(x) = f (x) − f (a) − f (b) − f (a)

b − a(x − a).

What is the graph of φ? Does it have Rolle’s theorem geometry?2 Can we apply Rolle’s theorem to φ on [a, b]? Yes!3 Rolle’s theorem asserts the existence of c ∈ (a, b) such that

φ′(c) = 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 16 / 20

Mean Value Theorem

Proof of Mean value theorem: The strategy is to define a newfunction φ(x) satisfying the hypothesis of Rolle’s theorem. Theconclusion of Rolle’s theorem for φ should yield the conclusion of MVTfor f .

1 Define φ on [a, b] by

φ(x) = f (x) − f (a) − f (b) − f (a)

b − a(x − a).

What is the graph of φ? Does it have Rolle’s theorem geometry?2 Can we apply Rolle’s theorem to φ on [a, b]? Yes!3 Rolle’s theorem asserts the existence of c ∈ (a, b) such that

φ′(c) = 0.4 This finishes the proof of MVT.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 16 / 20

On the proof of Mean value theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 17 / 20

On the proof of Mean value theorem

What is φ in the following graph?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 17 / 20

On the proof of Mean value theorem

What is φ in the following graph?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 17 / 20

On the proof of Mean value theorem

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 18 / 20

On the proof of Mean value theorem

What is φ in the following graph?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 18 / 20

On the proof of Mean value theorem

What is φ in the following graph?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 18 / 20

Mean value theorem: Applications

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b)

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b) and x1 < x2.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b) and x1 < x2. ApplyMVT for f on [x1, x2].

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b) and x1 < x2. ApplyMVT for f on [x1, x2]. We get f (x1) = f (x2).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b) and x1 < x2. ApplyMVT for f on [x1, x2]. We get f (x1) = f (x2). Since this is true forany x1 and x2 in (a, b) with x1 < x2,

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b) and x1 < x2. ApplyMVT for f on [x1, x2]. We get f (x1) = f (x2). Since this is true forany x1 and x2 in (a, b) with x1 < x2, we conclude f is a constantfunction.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b) and x1 < x2. ApplyMVT for f on [x1, x2]. We get f (x1) = f (x2). Since this is true forany x1 and x2 in (a, b) with x1 < x2, we conclude f is a constantfunction.

2 If f ′(x) = g′(x) for all x ∈ (a, b), then

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b) and x1 < x2. ApplyMVT for f on [x1, x2]. We get f (x1) = f (x2). Since this is true forany x1 and x2 in (a, b) with x1 < x2, we conclude f is a constantfunction.

2 If f ′(x) = g′(x) for all x ∈ (a, b), then f − g is constant on (a, b).

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b) and x1 < x2. ApplyMVT for f on [x1, x2]. We get f (x1) = f (x2). Since this is true forany x1 and x2 in (a, b) with x1 < x2, we conclude f is a constantfunction.

2 If f ′(x) = g′(x) for all x ∈ (a, b), then f − g is constant on (a, b).That is, f (x) = g(x) + C for some real number C.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b) and x1 < x2. ApplyMVT for f on [x1, x2]. We get f (x1) = f (x2). Since this is true forany x1 and x2 in (a, b) with x1 < x2, we conclude f is a constantfunction.

2 If f ′(x) = g′(x) for all x ∈ (a, b), then f − g is constant on (a, b).That is, f (x) = g(x) + C for some real number C.

Let f be a function defined by f (x) = x|x| for x 6= 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b) and x1 < x2. ApplyMVT for f on [x1, x2]. We get f (x1) = f (x2). Since this is true forany x1 and x2 in (a, b) with x1 < x2, we conclude f is a constantfunction.

2 If f ′(x) = g′(x) for all x ∈ (a, b), then f − g is constant on (a, b).That is, f (x) = g(x) + C for some real number C.

Let f be a function defined by f (x) = x|x| for x 6= 0. What is f ′?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b) and x1 < x2. ApplyMVT for f on [x1, x2]. We get f (x1) = f (x2). Since this is true forany x1 and x2 in (a, b) with x1 < x2, we conclude f is a constantfunction.

2 If f ′(x) = g′(x) for all x ∈ (a, b), then f − g is constant on (a, b).That is, f (x) = g(x) + C for some real number C.

Let f be a function defined by f (x) = x|x| for x 6= 0. What is f ′?

Constant function 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b) and x1 < x2. ApplyMVT for f on [x1, x2]. We get f (x1) = f (x2). Since this is true forany x1 and x2 in (a, b) with x1 < x2, we conclude f is a constantfunction.

2 If f ′(x) = g′(x) for all x ∈ (a, b), then f − g is constant on (a, b).That is, f (x) = g(x) + C for some real number C.

Let f be a function defined by f (x) = x|x| for x 6= 0. What is f ′?

Constant function 0. Is f constant?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b) and x1 < x2. ApplyMVT for f on [x1, x2]. We get f (x1) = f (x2). Since this is true forany x1 and x2 in (a, b) with x1 < x2, we conclude f is a constantfunction.

2 If f ′(x) = g′(x) for all x ∈ (a, b), then f − g is constant on (a, b).That is, f (x) = g(x) + C for some real number C.

Let f be a function defined by f (x) = x|x| for x 6= 0. What is f ′?

Constant function 0. Is f constant? No.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Mean value theorem: Applications

1 If f ′(x) = 0 for all x ∈ (a, b), then f is constant on (a, b).Proof Let x1 and x2 be any two points in (a, b) and x1 < x2. ApplyMVT for f on [x1, x2]. We get f (x1) = f (x2). Since this is true forany x1 and x2 in (a, b) with x1 < x2, we conclude f is a constantfunction.

2 If f ′(x) = g′(x) for all x ∈ (a, b), then f − g is constant on (a, b).That is, f (x) = g(x) + C for some real number C.

Let f be a function defined by f (x) = x|x| for x 6= 0. What is f ′?

Constant function 0. Is f constant? No. Does it contradict (1) above?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 19 / 20

Problems

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 20 / 20

Problems

1 Does there exist a function f such that f (0) = −1, f (2) = 4, andf ′(x) ≤ 2 for all x?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 20 / 20

Problems

1 Does there exist a function f such that f (0) = −1, f (2) = 4, andf ′(x) ≤ 2 for all x?

2 Show that the equation 2x − 1 − sin x = 0 has exactly one root.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 20 / 20

Problems

1 Does there exist a function f such that f (0) = −1, f (2) = 4, andf ′(x) ≤ 2 for all x?

2 Show that the equation 2x − 1 − sin x = 0 has exactly one root.3 Show that the equation x3 − 15x + c = 0 has at most one root in

[−2, 2].

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 20 / 20

Problems

1 Does there exist a function f such that f (0) = −1, f (2) = 4, andf ′(x) ≤ 2 for all x?

2 Show that the equation 2x − 1 − sin x = 0 has exactly one root.3 Show that the equation x3 − 15x + c = 0 has at most one root in

[−2, 2].4 Show that the equation x4 + 4x + c = 0 has at most two real roots.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 20 / 20

Problems

1 Does there exist a function f such that f (0) = −1, f (2) = 4, andf ′(x) ≤ 2 for all x?

2 Show that the equation 2x − 1 − sin x = 0 has exactly one root.3 Show that the equation x3 − 15x + c = 0 has at most one root in

[−2, 2].4 Show that the equation x4 + 4x + c = 0 has at most two real roots.5 If f (1) = 10, and f ′(x) ≥ 2 for 1 ≤ x ≤ 4, how small can f (4)

possibly be?

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 20 / 20

Problems

1 Does there exist a function f such that f (0) = −1, f (2) = 4, andf ′(x) ≤ 2 for all x?

2 Show that the equation 2x − 1 − sin x = 0 has exactly one root.3 Show that the equation x3 − 15x + c = 0 has at most one root in

[−2, 2].4 Show that the equation x4 + 4x + c = 0 has at most two real roots.5 If f (1) = 10, and f ′(x) ≥ 2 for 1 ≤ x ≤ 4, how small can f (4)

possibly be?6 Show that

√1 + x < 1 + 1

2x if x > 0.

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 5 August 3, 2009 20 / 20