Lesson 19: The Mean Value Theorem (handout)

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Sec on 4.2The Mean Value Theorem

V63.0121.001: Calculus IProfessor Ma hew Leingang

New York University

April 6, 2011

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Announcements

I Quiz 4 on Sec ons 3.3,3.4, 3.5, and 3.7 nextweek (April 14/15)

I Quiz 5 on Sec ons4.1–4.4 April 28/29

I Final Exam Thursday May12, 2:00–3:50pm

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Objectives

I Understand and be ableto explain the statementof Rolle’s Theorem.

I Understand and be ableto explain the statementof theMean ValueTheorem.

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. Sec on 4.2: The Mean Value Theorem. V63.0121.001: Calculus I . April 6, 2011

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Outline

Rolle’s Theorem

The Mean Value TheoremApplica ons

Why the MVT is the MITCFunc ons with deriva ves that are zeroMVT and differen ability

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Heuristic Motivation for Rolle’s TheoremIf you bike up a hill, then back down, at some point your eleva onwas sta onary.

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Image credit: SpringSun

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Mathematical Statement of Rolle’sTheorem

Theorem (Rolle’s Theorem)

Let f be con nuous on [a, b]and differen able 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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http://flickr.com/photos/63543004@N00/

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Flowchart proof of Rolle’s Theorem

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..Let c bethe max pt

..Let d bethe min pt

..endpointsare maxand min

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..is c anendpoint?

..is d anendpoint?

..f is

constanton [a, b]

..f′(c) = 0 ..f′(d) = 0 ..f′(x) ≡ 0on (a, b)

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no

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no

.yes . yes

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Outline

Rolle’s Theorem



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Heuristic Motivation for The Mean Value TheoremIf you drive between points A and B, at some me your speedometerreading was the same as your average speed over the drive.

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Image credit: ClintJCL

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

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The Mean Value TheoremTheorem (The Mean Value Theorem)

Let f be con nuous on[a, b] and differen able on(a, b). Then there exists apoint c in (a, b) such that

f(b)− f(a)b− a

= f′(c). ...a

..b

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c

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Rolle vs. MVTf′(c) = 0

f(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 TheoremProof.The line connec ng (a, f(a)) and (b, f(b)) has equa on

y− f(a) =f(b)− f(a)

b− a(x− a)

Apply Rolle’s Theorem to the func on

g(x) = f(x)− f(a)− f(b)− f(a)b− a

(x− a).

Then g is con nuous on [a, b] and differen able on (a, b) since f is.Also g(a) = 0 and g(b) = 0 (check both)

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Proof of the Mean Value TheoremProof.

g(x) = f(x)− f(a)− f(b)− f(a)b− a

(x− a).

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

0 = g′(c) = f′(c)− f(b)− f(a)b− a

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Using the MVT to count solutionsExampleShow that there is a unique solu on to the equa on x3 − x = 100 in theinterval [4, 5].

Solu onI By the Intermediate Value Theorem, the func on 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,then somewhere between them would be a point c3 betweenthem with f′(c3) = 0.

I However, f′(x) = 3x2 − 1, which is posi ve all along (4, 5). Sothis is impossible.

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Using the MVT to estimateExample

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

Solu onApply the MVT to the func onf(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 IIExample

Let f be a differen able func on with f(1) = 3 and f′(x) < 2 for all xin [0, 5]. Could f(4) ≥ 9?

Solu on

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

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(1, 3)

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(4, 9)

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(4, f(4))

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Food for ThoughtQues on

A driver travels along the New Jersey Turnpike using E-ZPass. Thesystem takes note of the me and place the driver enters and exitsthe Turnpike. A week a er his trip, the driver gets a speeding cketin the mail. Which of the following best describes the situa on?(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 cketed 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 quo ents must be 0I The tangent line to a line is that line, and a constant func on’sgraph is a horizontal line, which has slope 0.

I Implied by the power rule since c = cx0

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Functions with derivatives that are zero

Ques on

If f′(x) = 0 is f necessarily a constant func on?

I It seems trueI But so far no theorem (that we have proven) uses informa onabout the deriva ve of a func on to determine informa onabout the func on itself

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Why the MVT is the MITC(Most 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 con nuous on[x, y] and differen able 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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Functions with the same derivativeTheoremSuppose f and g are two differen able func ons on (a, b) withf′ = g′.

Then f and g differ by a constant. That is, there exists aconstant C such 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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MVT and differentiability

Example

Let

f(x) =

{−x if x ≤ 0x2 if x ≥ 0

Is f differen able at 0?

Solu on (from the defini on)

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 notdifferen able at 0.

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MVT and differentiability

Example

Let

f(x) =

{−x if x ≤ 0x2 if x ≥ 0

Is f differen able at 0?

Solu on (Sort of)

If x < 0, then f′(x) = −1. If x > 0, thenf′(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 isnot differen able at 0.

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Why only “sort of”?I This solu on is valid but lessdirect.

I We seem to be using thefollowing fact: If lim

x→af′(x) does

not exist, then f is notdifferen able at a.

I equivalently: If f is differen ableat a, then lim

x→af′(x) exists.

I But this “fact” is not true!

.. x.

y

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f(x)

...

f′(x)

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Differentiable with discontinuous derivativeIt is possible for a func on f to be differen able at a even if lim

x→af′(x)

does not exist.Example

Let f′(x) =

{x2 sin(1/x) if x ̸= 00 if x = 0

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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.

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Differentiable with discontinuous derivativeIt is possible for a func on f to be differen able at a even if lim

x→af′(x)

does not exist.Example

Let f′(x) =

{x2 sin(1/x) if x ̸= 00 if x = 0

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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 differen able for all x, but f′ is notcon nuous at 0!

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Differentiability FAIL

.. x.

f(x)

This func on is differen ableat 0.

.. x.

f′(x)

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But the deriva ve is notcon nuous at 0!

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MVT to the rescue

LemmaSuppose f is con nuous on [a, b] and lim

x→a+f′(x) = m. Then

limx→a+

f(x)− f(a)x− a

= m.

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MVT to the rescueProof.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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Using the MVT to find limits

TheoremSuppose

limx→a−

f′(x) = m1 and limx→a+

f′(x) = m2

If m1 = m2, then f is differen able at a. If m1 ̸= m2, then f is notdifferen able at a.

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Using the MVT to find limitsProof.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 condi ons, func ons musthave cri cal points.

I Mean Value Theorem: under suitable condi ons, func onsmust have an instantaneous rate of change equal to theaverage rate of change.

I A func on whose deriva ve is iden cally zero on an intervalmust be constant on that interval.

I E-ZPass is kinder than we realized.

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Lesson 19: The Mean Value Theorem (handout)