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Section 4.2The Mean Value Theorem

V63.0121.021, Calculus I

New York University

November 11, 2010

AnnouncementsI Quiz 4 next week (November 16, 18, 19) on Sections 3.3, 3.4, 3.5,

3.7

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. . . . . .

Announcements

I Quiz 4 next week(November 16, 18, 19) onSections 3.3, 3.4, 3.5, 3.7

V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 2 / 29

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Objectives

I Understand and be able toexplain the statement ofRolle’s Theorem.

I Understand and be able toexplain the statement ofthe Mean Value Theorem.


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Outline

Rolle’s Theorem

The Mean Value TheoremApplications

Why the MVT is the MITCFunctions with derivatives that are zeroMVT and differentiability


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Heuristic Motivation for Rolle's Theorem

If you bike up a hill, then back down, at some point your elevation wasstationary.

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.Image credit: SpringSunV63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 5 / 29

http://flickr.com/photos/63543004@N00/

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Mathematical Statement of Rolle's Theorem

Theorem (Rolle’s Theorem)

Let f be continuous on [a,b]and differentiable on (a,b).Suppose f(a) = f(b). Thenthere exists a point c in(a,b) such that f′(c) = 0.

. ..a

..b

..c


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Mathematical Statement of Rolle's Theorem

Theorem (Rolle’s Theorem)

Let f be continuous on [a,b]and differentiable on (a,b).Suppose f(a) = f(b). Thenthere exists a point c in(a,b) such that f′(c) = 0.

. ..a

..b

..c


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Flowchart proof of Rolle's Theorem

.

.

..Let c be

the max pt.

.Let d bethe min pt

..endpointsare maxand min

.

..is c an

endpoint?.

.is d an

endpoint?.

.f is

constanton [a,b]

..f′(c) = 0 ..

f′(d) = 0 ..f′(x) ≡ 0on (a,b)

.no .no

.yes .yes


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Outline

Rolle’s Theorem




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Heuristic Motivation for The Mean Value Theorem

If you drive between points A and B, at some time your speedometerreading was the same as your average speed over the drive.

.

.Image credit: ClintJCLV63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 10 / 29

http://www.flickr.com/photos/clintjcl/2476598913/

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The Mean Value Theorem

Theorem (The Mean Value Theorem)

Let f be continuous on [a,b]and differentiable on (a,b).Then there exists a point cin (a,b) such that

f(b)− f(a)b− a

= f′(c). . ..a

..b

.c


. . . . . .




f(b)− f(a)b− a

= f′(c). . ..a

..b

.c


. . . . . .




f(b)− f(a)b− a

= f′(c). . ..a

..b

.c


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Rolle vs. MVT

f′(c) = 0f(b)− f(a)

b− a= f′(c)

. ..a

..b

..c

. ..a

..b

..c

If the x-axis is skewed the pictures look the same.


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Rolle vs. MVT

f′(c) = 0f(b)− f(a)

b− a= f′(c)

. ..a

..b

..c

. ..a

..b

..c

If the x-axis is skewed the pictures look the same.


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Proof of the Mean Value Theorem

Proof.The line connecting (a, f(a)) and (b, f(b)) has equation

y− f(a) =f(b)− f(a)

b− a(x− a)

Apply Rolle’s Theorem to the function

g(x) = f(x)− f(a)− f(b)− f(a)b− a

(x− a).

Then g is continuous on [a,b] and differentiable on (a,b) since f is.Also g(a) = 0 and g(b) = 0 (check both) So by Rolle’s Theorem thereexists a point c in (a,b) such that

0 = g′(c) = f′(c)− f(b)− f(a)b− a

.


. . . . . .



y− f(a) =f(b)− f(a)

b− a(x− a)


g(x) = f(x)− f(a)− f(b)− f(a)b− a

(x− a).


0 = g′(c) = f′(c)− f(b)− f(a)b− a

.


. . . . . .



y− f(a) =f(b)− f(a)

b− a(x− a)


g(x) = f(x)− f(a)− f(b)− f(a)b− a

(x− a).

Then g is continuous on [a,b] and differentiable on (a,b) since f is.

Also g(a) = 0 and g(b) = 0 (check both) So by Rolle’s Theorem thereexists a point c in (a,b) such that

0 = g′(c) = f′(c)− f(b)− f(a)b− a

.


. . . . . .



y− f(a) =f(b)− f(a)

b− a(x− a)


g(x) = f(x)− f(a)− f(b)− f(a)b− a

(x− a).

Then g is continuous on [a,b] and differentiable on (a,b) since f is.Also g(a) = 0 and g(b) = 0 (check both)

So by Rolle’s Theorem thereexists a point c in (a,b) such that

0 = g′(c) = f′(c)− f(b)− f(a)b− a

.


. . . . . .



y− f(a) =f(b)− f(a)

b− a(x− a)


g(x) = f(x)− f(a)− f(b)− f(a)b− a

(x− a).


0 = g′(c) = f′(c)− f(b)− f(a)b− a

.


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Using the MVT to count solutions

Example

Show that there is a unique solution to the equation x3 − x = 100 in theinterval [4,5].

Solution

I By the Intermediate Value Theorem, the function f(x) = x3 − xmust take the value 100 at some point on c in (4,5).

I If there were two points c1 and c2 with f(c1) = f(c2) = 100, thensomewhere between them would be a point c3 between them withf′(c3) = 0.

I However, f′(x) = 3x2 − 1, which is positive all along (4,5). So thisis impossible.


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Example


Solution





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Example


Solution





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Example


Solution





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Using the MVT to estimate

Example

We know that |sin x| ≤ 1 for all x, and that sin x ≈ x for small x. Showthat |sin x| ≤ |x| for all x.

SolutionApply the MVT to the function f(t) = sin t on [0, x]. We get

sin x− sin 0x− 0

= cos(c)

for some c in (0, x). Since |cos(c)| ≤ 1, we get∣∣∣∣sin xx∣∣∣∣ ≤ 1 =⇒ |sin x| ≤ |x|


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Using the MVT to estimate

Example

We know that |sin x| ≤ 1 for all x, and that sin x ≈ x for small x. Showthat |sin x| ≤ |x| for all x.

SolutionApply the MVT to the function f(t) = sin t on [0, x]. We get

sin x− sin 0x− 0

= cos(c)

for some c in (0, x). Since |cos(c)| ≤ 1, we get∣∣∣∣sin xx∣∣∣∣ ≤ 1 =⇒ |sin x| ≤ |x|


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Using the MVT to estimate II

Example

Let f be a differentiable function with f(1) = 3 and f′(x) < 2 for all x in[0,5]. Could f(4) ≥ 9?

Solution

By MVT

f(4)− f(1)4− 1

= f′(c) < 2

for some c in (1,4). Therefore

f(4) = f(1) + f′(c)(3) < 3+ 2 · 3 = 9.

So no, it is impossible that f(4) ≥ 9.. .x

.y

..(1, 3)

..(4,9)

..(4, f(4))


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Example


Solution

By MVT

f(4)− f(1)4− 1

= f′(c) < 2


f(4) = f(1) + f′(c)(3) < 3+ 2 · 3 = 9.

So no, it is impossible that f(4) ≥ 9.

. .x

.y

..(1,3)

..(4,9)

..(4, f(4))


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Example


Solution

By MVT

f(4)− f(1)4− 1

= f′(c) < 2


f(4) = f(1) + f′(c)(3) < 3+ 2 · 3 = 9.

So no, it is impossible that f(4) ≥ 9.. .x

.y

..(1,3)

..(4,9)

..(4, f(4))


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Food for Thought

QuestionA driver travels along the New Jersey Turnpike using E-ZPass. Thesystem takes note of the time and place the driver enters and exits theTurnpike. A week after his trip, the driver gets a speeding ticket in themail. Which of the following best describes the situation?(a) E-ZPass cannot prove that the driver was speeding(b) E-ZPass can prove that the driver was speeding(c) The driver’s actual maximum speed exceeds his ticketed speed(d) Both (b) and (c).


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Food for Thought

QuestionA driver travels along the New Jersey Turnpike using E-ZPass. Thesystem takes note of the time and place the driver enters and exits theTurnpike. A week after his trip, the driver gets a speeding ticket in themail. Which of the following best describes the situation?(a) E-ZPass cannot prove that the driver was speeding(b) E-ZPass can prove that the driver was speeding(c) The driver’s actual maximum speed exceeds his ticketed speed(d) Both (b) and (c).


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Outline

Rolle’s Theorem




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Functions with derivatives that are zero

FactIf f is constant on (a,b), then f′(x) = 0 on (a,b).

I The limit of difference quotients must be 0I The tangent line to a line is that line, and a constant function’s

graph is a horizontal line, which has slope 0.I Implied by the power rule since c = cx0

QuestionIf f′(x) = 0 is f necessarily a constant function?

I It seems trueI But so far no theorem (that we have proven) uses information

about the derivative of a function to determine information aboutthe function itself


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Why the MVT is the MITCMost Important Theorem In Calculus!

TheoremLet f′ = 0 on an interval (a,b).

Then f is constant on (a,b).

Proof.Pick any points x and y in (a,b) with x < y. Then f is continuous on[x, y] and differentiable on (x, y). By MVT there exists a point z in (x, y)such that

f(y)− f(x)y− x

= f′(z) = 0.

So f(y) = f(x). Since this is true for all x and y in (a,b), then f isconstant.


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TheoremLet f′ = 0 on an interval (a,b). Then f is constant on (a,b).


f(y)− f(x)y− x

= f′(z) = 0.



. . . . . .


TheoremLet f′ = 0 on an interval (a,b). Then f is constant on (a,b).


f(y)− f(x)y− x

= f′(z) = 0.



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Functions with the same derivative

TheoremSuppose f and g are two differentiable functions on (a,b) with f′ = g′.Then f and g differ by a constant. That is, there exists a constant Csuch that f(x) = g(x) + C.

Proof.

I Let h(x) = f(x)− g(x)I Then h′(x) = f′(x)− g′(x) = 0 on (a,b)I So h(x) = C, a constantI This means f(x)− g(x) = C on (a,b)


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Proof.



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Proof.

I Let h(x) = f(x)− g(x)

I Then h′(x) = f′(x)− g′(x) = 0 on (a,b)I So h(x) = C, a constantI This means f(x)− g(x) = C on (a,b)


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Proof.

I Let h(x) = f(x)− g(x)I Then h′(x) = f′(x)− g′(x) = 0 on (a,b)

I So h(x) = C, a constantI This means f(x)− g(x) = C on (a,b)


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Proof.

I Let h(x) = f(x)− g(x)I Then h′(x) = f′(x)− g′(x) = 0 on (a,b)I So h(x) = C, a constant

I This means f(x)− g(x) = C on (a,b)


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Proof.



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MVT and differentiability

Example

Let

f(x) =

{−x if x ≤ 0x2 if x ≥ 0

Is f differentiable at 0?


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Example

Let

f(x) =

{−x if x ≤ 0x2 if x ≥ 0


Solution (from the definition)

We have

limx→0−

f(x)− f(0)x− 0

= limx→0−

−xx

= −1

limx→0+

f(x)− f(0)x− 0

= limx→0+

x2

x= lim

x→0+x = 0

Since these limits disagree, f is not differentiable at 0.V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 22 / 29

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Example

Let

f(x) =

{−x if x ≤ 0x2 if x ≥ 0


Solution (Sort of)

If x < 0, then f′(x) = −1. If x > 0, then f′(x) = 2x. Since

limx→0+

f′(x) = 0 and limx→0−

f′(x) = −1,

the limit limx→0

f′(x) does not exist and so f is not differentiable at 0.


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Why only “sort of"?

I This solution is valid butless direct.

I We seem to be using thefollowing fact: If lim

x→af′(x)

does not exist, then f is notdifferentiable at a.

I equivalently: If f isdifferentiable at a, thenlimx→a

f′(x) exists.

I But this “fact” is not true!

. .x

.y .f(x)

.

.

.f′(x)


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Differentiable with discontinuous derivative

It is possible for a function f to be differentiable at a even if limx→a

f′(x)does not exist.

Example

Let f′(x) =

{x2 sin(1/x) if x ̸= 00 if x = 0

. Then when x ̸= 0,

f′(x) = 2x sin(1/x) + x2 cos(1/x)(−1/x2) = 2x sin(1/x)− cos(1/x),

which has no limit at 0. However,

f′(0) = limx→0

f(x)− f(0)x− 0

= limx→0

x2 sin(1/x)x

= limx→0

x sin(1/x) = 0

So f′(0) = 0. Hence f is differentiable for all x, but f′ is not continuousat 0!


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Differentiability FAIL

. .x

.f(x)

This function is differentiable at0.

. .x

.f′(x)

.

But the derivative is notcontinuous at 0!


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MVT to the rescue

LemmaSuppose f is continuous on [a,b] and lim

x→a+f′(x) = m. Then

limx→a+

f(x)− f(a)x− a

= m.

Proof.Choose x near a and greater than a. Then

f(x)− f(a)x− a

= f′(cx)

for some cx where a < cx < x. As x → a, cx → a as well, so:

limx→a+

f(x)− f(a)x− a

= limx→a+

f′(cx) = limx→a+

f′(x) = m.


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MVT to the rescue

LemmaSuppose f is continuous on [a,b] and lim

x→a+f′(x) = m. Then

limx→a+

f(x)− f(a)x− a

= m.

Proof.Choose x near a and greater than a. Then

f(x)− f(a)x− a

= f′(cx)

for some cx where a < cx < x. As x → a, cx → a as well, so:

limx→a+

f(x)− f(a)x− a

= limx→a+

f′(cx) = limx→a+

f′(x) = m.


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TheoremSuppose

limx→a−

f′(x) = m1 and limx→a+

f′(x) = m2

If m1 = m2, then f is differentiable at a. If m1 ̸= m2, then f is notdifferentiable at a.

Proof.We know by the lemma that

limx→a−

f(x)− f(a)x− a

= limx→a−

f′(x)

limx→a+

f(x)− f(a)x− a

= limx→a+

f′(x)

The two-sided limit exists if (and only if) the two right-hand sidesagree.


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TheoremSuppose

limx→a−

f′(x) = m1 and limx→a+

f′(x) = m2

If m1 = m2, then f is differentiable at a. If m1 ̸= m2, then f is notdifferentiable at a.

Proof.We know by the lemma that

limx→a−

f(x)− f(a)x− a

= limx→a−

f′(x)

limx→a+

f(x)− f(a)x− a

= limx→a+

f′(x)

The two-sided limit exists if (and only if) the two right-hand sidesagree.


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Summary

I Rolle’s Theorem: under suitable conditions, functions must havecritical points.

I Mean Value Theorem: under suitable conditions, functions musthave an instantaneous rate of change equal to the average rate ofchange.

I A function whose derivative is identically zero on an interval mustbe constant on that interval.

I E-ZPass is kinder than we realized.
