. . . . . .

Section 4.2The Mean Value Theorem

V63.0121.006/016, Calculus I

March 25, 2010

AnnouncementsI Please resubmit your exams to be my 4:00pm TODAY for

regrading (it’s only one problem, and scores will either goup or stay the same)

I Quiz 3 in recitation Friday, April 2. Covers §§2.6–3.5.Contact me if you have religious conflicts.

. . . . . .

Announcements

I Please resubmit your exams to be my 4:00pm TODAY forregrading (it’s only one problem, and scores will either goup or stay the same)

I Quiz 3 in recitation Friday, April 2. Covers §§2.6–3.5.Contact me if you have religious conflicts.

. . . . . .

Outline

Review: The Closed Interval Method

Rolle’s Theorem

The Mean Value TheoremApplications

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

. . . . . .

Flowchart for placing extremaThanks to Fermat

Suppose f is a continuous function on the closed, boundedinterval [a,b], and c is a global maximum point.

..start

.Is c an

endpoint?

. c = a orc = b

.c is a

local max

.Is f diff’ble

at c?

.f is notdiff at c

.f′(c) = 0

.no

.yes

.no

.yes

. . . . . .

The Closed Interval Method

This means to find the maximum value of f on [a,b], we needto:

I Evaluate f at the endpoints a and bI Evaluate f at the critical points x where either f′(x) = 0 or

f is not differentiable at x.I The points with the largest function value are the global

maximum pointsI The points with the smallest or most negative function

value are the global minimum points.

. . . . . .

Outline

Review: The Closed Interval Method

Rolle’s Theorem

The Mean Value TheoremApplications

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

. . . . . .

Heuristic Motivation for Rolle's Theorem

If you bike up a hill, then back down, at some point yourelevation was stationary.

.

.Image credit: SpringSun

. . . . . .

Mathematical Statement of Rolle's Theorem

Theorem (Rolle’s Theorem)

Let f be continuous on[a,b] and differentiableon (a,b). Supposef(a) = f(b). Then thereexists a point c in (a,b)such that f′(c) = 0. . .

.a..b

..c

. . . . . .

Mathematical Statement of Rolle's Theorem

Theorem (Rolle’s Theorem)

Let f be continuous on[a,b] and differentiableon (a,b). Supposef(a) = f(b). Then thereexists a point c in (a,b)such that f′(c) = 0. . .

.a..b

..c

. . . . . .

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

. . . . . .

Outline

Review: The Closed Interval Method

Rolle’s Theorem

The Mean Value TheoremApplications

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

. . . . . .

Heuristic Motivation for The Mean Value Theorem

If you drive between points A and B, at some time yourspeedometer reading was the same as your average speedover the drive.

.

.Image credit: ClintJCL

. . . . . .

The Mean Value Theorem

Theorem (The Mean Value Theorem)

Let f be continuous on[a,b] and differentiableon (a,b). Then thereexists a point c in (a,b)such that

f(b)− f(a)b− a

= f′(c). . ..a

..b

.c

. . . . . .

The Mean Value Theorem

Theorem (The Mean Value Theorem)

Let f be continuous on[a,b] and differentiableon (a,b). Then thereexists a point c in (a,b)such that

f(b)− f(a)b− a

= f′(c). . ..a

..b

.c

. . . . . .

The Mean Value Theorem

Theorem (The Mean Value Theorem)

Let f be continuous on[a,b] and differentiableon (a,b). Then thereexists a point c in (a,b)such that

f(b)− f(a)b− a

= f′(c). . ..a

..b

.c

. . . . . .

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.

. . . . . .

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.

. . . . . .

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 fis. Also g(a) = 0 and g(b) = 0 (check both) So by Rolle’sTheorem there exists a point c in (a,b) such that

0 = g′(c) = f′(c)− f(b)− f(a)b− a

.

. . . . . .

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 fis. Also g(a) = 0 and g(b) = 0 (check both) So by Rolle’sTheorem there exists a point c in (a,b) such that

0 = g′(c) = f′(c)− f(b)− f(a)b− a

.

. . . . . .

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 fis.

Also g(a) = 0 and g(b) = 0 (check both) So by Rolle’sTheorem there exists a point c in (a,b) such that

0 = g′(c) = f′(c)− f(b)− f(a)b− a

.

. . . . . .

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 fis. Also g(a) = 0 and g(b) = 0 (check both)

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

0 = g′(c) = f′(c)− f(b)− f(a)b− a

.

. . . . . .

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 fis. Also g(a) = 0 and g(b) = 0 (check both) So by Rolle’sTheorem 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 solutions

Example

Show that there is a unique solution to the equationx3 − x = 100 in the interval [4,5].

Solution

I By the Intermediate Value Theorem, the functionf(x) = x3 − x must 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 c3between them with f′(c3) = 0.

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

. . . . . .

Using the MVT to count solutions

Example

Show that there is a unique solution to the equationx3 − x = 100 in the interval [4,5].

Solution

I By the Intermediate Value Theorem, the functionf(x) = x3 − x must 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 c3between them with f′(c3) = 0.

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

. . . . . .

Using the MVT to count solutions

Example

Show that there is a unique solution to the equationx3 − x = 100 in the interval [4,5].

Solution

I By the Intermediate Value Theorem, the functionf(x) = x3 − x must 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 c3between them with f′(c3) = 0.

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

. . . . . .

Using the MVT to count solutions

Example

Show that there is a unique solution to the equationx3 − x = 100 in the interval [4,5].

Solution

I By the Intermediate Value Theorem, the functionf(x) = x3 − x must 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 c3between them with f′(c3) = 0.

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

. . . . . .

Example

We know that |sin x| ≤ 1 for all x. Show that |sin x| ≤ |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|

. . . . . .

Example

We know that |sin x| ≤ 1 for all x. Show that |sin x| ≤ |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|

. . . . . .

Example

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

SolutionBy 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 thatf(4) ≥ 9. . .x

.y

..(1,3)

..(4,9)

..(4, f(4))

. . . . . .

Example

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

SolutionBy 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 thatf(4) ≥ 9. . .x

.y

..(1,3)

..(4,9)

..(4, f(4))

. . . . . .

QuestionA driver travels along the New Jersey Turnpike using E-ZPass.The system takes note of the time and place the driver entersand exits the Turnpike. A week after his trip, the driver gets aspeeding ticket in the mail. Which of the following bestdescribes 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).Be prepared to justify your answer.

. . . . . .

QuestionA driver travels along the New Jersey Turnpike using E-ZPass.The system takes note of the time and place the driver entersand exits the Turnpike. A week after his trip, the driver gets aspeeding ticket in the mail. Which of the following bestdescribes 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).Be prepared to justify your answer.

. . . . . .

Outline

Review: The Closed Interval Method

Rolle’s Theorem

The Mean Value TheoremApplications

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

. . . . . .

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 determineinformation about the function itself

. . . . . .

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 determineinformation about the function itself

. . . . . .

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 determineinformation about the function itself

. . . . . .

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 determineinformation about the function itself

. . . . . .

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 continuouson [x, y] and differentiable on (x, y). By MVT there exists a pointz 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 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 continuouson [x, y] and differentiable on (x, y). By MVT there exists a pointz 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 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 continuouson [x, y] and differentiable on (x, y). By MVT there exists a pointz 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.

. . . . . .

TheoremSuppose f and g are two differentiable functions 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)

. . . . . .

TheoremSuppose f and g are two differentiable functions 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 differentiable at 0?

. . . . . .

MVT and differentiability

Example

Let

f(x) =

{−x if x ≤ 0x2 if x ≥ 0

Is f differentiable at 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.

. . . . . .

MVT and differentiability

Example

Let

f(x) =

{−x if x ≤ 0x2 if x ≥ 0

Is f differentiable at 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.

. . . . . .

Why only “sort of"?

I This solution is valid butless direct.

I We seem to be usingthe following fact: Iflimx→a

f′(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 nottrue!

. .x

.y .f(x)

.

.

.f′(x)

. . . . . .

Differentiable with discontinuous derivative

It is possible for a function f to be differentiable at a even iflimx→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 notcontinuous at 0!

. . . . . .

Differentiability FAIL

. .x

.f(x)

This function is differentiableat 0.

. .x

.f′(x)

But the derivative is notcontinuous at 0!

. . . . . .

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.

. . . . . .

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.

. . . . . .

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-handsides agree.

. . . . . .

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-handsides agree.

. . . . . .

What have we learned today?

I Rolle’s Theorem: under suitable conditions, functions musthave critical points.

I Mean Value Theorem: under suitable conditions, functionsmust have an instantaneous rate of change equal to theaverage rate of change.

I A function whose derivative is identically zero on aninterval must be constant on that interval.

I E-ZPass is kinder than we realized.