Lesson 19: The Mean Value Theorem

Section 4.2The Mean Value Theorem

V63.0121.002.2010Su, Calculus I

New York University

June 8, 2010

Announcements

I Exams not graded yet

I Assignment 4 is on the website

I Quiz 3 on Thursday covering 3.3, 3.4, 3.5, 3.7

Announcements

I Exams not graded yet

I Assignment 4 is on thewebsite

I Quiz 3 on Thursday covering3.3, 3.4, 3.5, 3.7

V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 2 / 28

Objectives

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

I Understand and be able toexplain the statement of theMean Value Theorem.


Outline

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 your elevation wasstationary.

Image credit: SpringSunV63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 5 / 28

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

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


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


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 isconstanton [a, b]

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

no no

yes yes


Outline

Rolle’s Theorem




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: ClintJCLV63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 10 / 28

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

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 c in(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


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 f is. Alsog(a) = 0 and g(b) = 0 (check both) So by Rolle’s Theorem there exists apoint c in (a, b) such that

0 = g ′(c) = f ′(c)− f (b)− f (a)

b − a.



Proof.


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.



Proof.


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.

Alsog(a) = 0 and g(b) = 0 (check both) So by Rolle’s Theorem there exists apoint c in (a, b) such that

0 = g ′(c) = f ′(c)− f (b)− f (a)

b − a.



Proof.


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. Alsog(a) = 0 and g(b) = 0 (check both)

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

0 = g ′(c) = f ′(c)− f (b)− f (a)

b − a.



Proof.


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.


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 − x musttake 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 this isimpossible.



Example


Solution






Example


Solution






Example


Solution





Using the MVT to estimate

Example

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

Solution

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

sin x − sin 0

x − 0= cos(c)

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

x

∣∣∣∣ ≤ 1 =⇒ |sin x | ≤ |x |


Using the MVT to estimate

Example

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

Solution

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

sin x − sin 0

x − 0= cos(c)

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

x

∣∣∣∣ ≤ 1 =⇒ |sin x | ≤ |x |


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))


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))


Food for Thought

Question

A driver travels along the New Jersey Turnpike using E-ZPass. The systemtakes note of the time and place the driver enters and exits the Turnpike.A week after his trip, the driver gets a speeding ticket in the mail. Whichof 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).


Food for Thought

Question

A driver travels along the New Jersey Turnpike using E-ZPass. The systemtakes note of the time and place the driver enters and exits the Turnpike.A week after his trip, the driver gets a speeding ticket in the mail. Whichof 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).


Outline

Rolle’s Theorem




Functions with derivatives that are zero

Fact

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

I The limit of difference quotients must be 0

I The tangent line to a line is that line, and a constant function’s graphis a horizontal line, which has slope 0.

I Implied by the power rule since c = cx0

Question

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

I It seems true

I But so far no theorem (that we have proven) uses information aboutthe derivative of a function to determine information about thefunction itself



Fact





Question


I It seems true




Fact





Question


I It seems true




Fact





Question


I It seems true



Why the MVT is the MITCMost Important Theorem In Calculus!

Theorem

Let 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.



Theorem

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

Proof.


f (y)− f (x)

y − x= f ′(z) = 0.




Theorem

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

Proof.


f (y)− f (x)

y − x= f ′(z) = 0.



Functions with the same derivative

Theorem

Suppose 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 C suchthat 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 derivative

Theorem

Suppose 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 C suchthat 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)


MVT and differentiability

Example

Let

f (x) =

{−x if x ≤ 0

x2 if x ≥ 0

Is f differentiable at 0?



Example

Let

f (x) =

{−x if x ≤ 0

x2 if x ≥ 0


Solution (from the definition)

We have

limx→0−

f (x)− f (0)

x − 0= lim

x→0−

−x

x= −1

limx→0+

f (x)− f (0)

x − 0= lim

x→0+

x2

x= lim

x→0+x = 0

Since these limits disagree, f is not differentiable at 0.



Example

Let

f (x) =

{−x if x ≤ 0

x2 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.


Why only “sort of”?

I This solution is valid but lessdirect.

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)


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 6= 0

0 if x = 0. Then when x 6= 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= lim

x→0

x2 sin(1/x)

x= lim

x→0x sin(1/x) = 0

So f ′(0) = 0. Hence f is differentiable for all x , but f ′ is not continuous at0!


Differentiability FAIL

x

f (x)

This function is differentiable at0.

x

f ′(x)

But the derivative is notcontinuous at 0!


MVT to the rescue

Lemma

Suppose f is continuous on [a, b] and limx→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= lim

x→a+f ′(cx) = lim

x→a+f ′(x) = m.


MVT to the rescue

Lemma

Suppose f is continuous on [a, b] and limx→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= lim

x→a+f ′(cx) = lim

x→a+f ′(x) = m.


Theorem

Supposelim

x→a−f ′(x) = m1 and lim

x→a+f ′(x) = m2

If m1 = m2, then f is differentiable at a. If m1 6= m2, then f is notdifferentiable at a.

Proof.

We know by the lemma that

limx→a−

f (x)− f (a)

x − a= lim

x→a−f ′(x)

limx→a+

f (x)− f (a)

x − a= lim

x→a+f ′(x)

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


Theorem

Supposelim

x→a−f ′(x) = m1 and lim

x→a+f ′(x) = m2

If m1 = m2, then f is differentiable at a. If m1 6= m2, then f is notdifferentiable at a.

Proof.

We know by the lemma that

limx→a−

f (x)− f (a)

x − a= lim

x→a−f ′(x)

limx→a+

f (x)− f (a)

x − a= lim

x→a+f ′(x)

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


Summary

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

I Mean Value Theorem: under suitable conditions, functions must havean instantaneous rate of change equal to the average rate of change.

I A function whose derivative is identically zero on an interval must beconstant on that interval.

I E-ZPass is kinder than we realized.


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Lesson 19: The Mean Value Theorem

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