..

Sec on 4.2The Mean Value Theorem

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

April 6, 2011

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

Courant Lecture tomorrow

Persi Diaconis (Stanford)“The Search for Randomness”(General Audience Lecture)

Thursday, April 7, 2011, 3:30pmWarren Weaver Hall, room 109

Recep on to follow

Visit http://cims.nyu.edu/ for detailsand to RSVP

Objectives

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

I Understand and be ableto explain the statementof theMean ValueTheorem.

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

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

..Image credit: SpringSun

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

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

Flowchart proof of Rolle’s Theorem

..

..Let c bethe max pt

..Let d bethe min pt

..endpointsare maxand min

.

..is c anendpoint?

..is d anendpoint?

..f is

constanton [a, b]

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

.

no

.

no

.yes . yes

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

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.

..Image credit: ClintJCL

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

.

c

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

.

c

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

.

c

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.

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.

Proof of the Mean Value TheoremProof.The line connec ng (a, f(a)) and (b, f(b)) has equa on

L(x) = f(a) +f(b)− f(a)

b− a(x− a)

Proof of the Mean Value TheoremProof.The line connec ng (a, f(a)) and (b, f(b)) has equa on

L(x) = f(a) +f(b)− f(a)

b− a(x− a)

Apply Rolle’s Theorem to the func on

g(x) = f(x)− L(x) = f(x)− f(a)− f(b)− f(a)b− a

(x− a).

Proof of the Mean Value TheoremProof.The line connec ng (a, f(a)) and (b, f(b)) has equa on

L(x) = f(a) +f(b)− f(a)

b− a(x− a)

Apply Rolle’s Theorem to the func on

g(x) = f(x)− L(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.

Proof of the Mean Value TheoremProof.The line connec ng (a, f(a)) and (b, f(b)) has equa on

L(x) = f(a) +f(b)− f(a)

b− a(x− a)

Apply Rolle’s Theorem to the func on

g(x) = f(x)− L(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).

Proof of the Mean Value TheoremProof.

g(x) = f(x)− L(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

.

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.

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.

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.

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.

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|

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|

Using the MVT to estimate II

Example

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?

Using the MVT to estimate IISolu 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

..

(1, 3)

..

(4, 9)

..

(4, f(4))

Using the MVT to estimate IISolu 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

..

(1, 3)

..

(4, 9)

..

(4, f(4))

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).

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).

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

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

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

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

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

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.

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.

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.

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)

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)

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)

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)

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 constant

I This means f(x)− g(x) = C on (a, b)

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)

MVT and differentiability

Example

Let

f(x) =

{−x if x ≤ 0x2 if x ≥ 0

Is f differen able at 0?

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.

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.

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

.

f(x)

...

f′(x)

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

.

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.

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

.

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!

Differentiability FAIL

.. x.

f(x)

This func on is differen ableat 0.

.. x.

f′(x)

.

But the deriva ve is notcon nuous at 0!

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.

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.

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.

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.

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.