. . . . . .

Section 4.2The Mean Value Theorem

V63.0121, Calculus I

March 25–26, 2009

Announcements

I Quiz 4 next week: Sections 2.5–3.5I Thank you for your midterm evaluations!

..Image credit: Jimmywayne22

. . . . . .

Outline

Review: The Closed Interval Method

Rolle’s Theorem

The Mean Value TheoremApplications

Why the MVT is the MITC

. . . . . .

Flowchart for placing extremaThanks to Fermat

Suppose f is a continuous function on the closed, bounded interval[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 need to: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 MITC

. . . . . .

Heuristic Motivation for Rolle’s Theorem

If you bike up a hill, then back down, at some point your elevationwas stationary.

.

.Image credit: SpringSun

. . . . . .

Mathematical Statement of Rolle’s Theorem

Theorem (Rolle’s Theorem)Let f be continuous on [a, b] anddifferentiable on (a, b). Supposef(a) = f(b). Then there exists apoint c in (a, b) such thatf′(c) = 0. . .•

.a.•.b

.•.c

. . . . . .

Mathematical Statement of Rolle’s Theorem

Theorem (Rolle’s Theorem)Let f be continuous on [a, b] anddifferentiable on (a, b). Supposef(a) = f(b). Then there exists apoint c in (a, b) such thatf′(c) = 0. . .•

.a.•.b

.•.c

. . . . . .

Proof of Rolle’s Theorem

Proof.

I By the Extreme Value Theorem f must achieve its maximumvalue at a point c in [a, b].

I If c is in (a, b), great: it’s a local maximum and so by Fermat’sTheorem f′(c) = 0.

I On the other hand, if c = a or c = b, try with the minimum. Theminimum of f on [a, b] must be achieved at a point d in [a, b].

I If d is in (a, b), great: it’s a local minimum and so by Fermat’sTheorem f′(d) = 0. If not, d = a or d = b.

I If we still haven’t found a point in the interior, we have that themaximum and minimum values of f on [a, b] occur at bothendpoints. But we already know that f(a) = f(b). If these arethe maximum and minimum values, f is constant on [a, b] and anypoint x in (a, b) will have f′(x) = 0.

. . . . . .

Proof of Rolle’s Theorem

Proof.

I By the Extreme Value Theorem f must achieve its maximumvalue at a point c in [a, b].

I If c is in (a, b), great: it’s a local maximum and so by Fermat’sTheorem f′(c) = 0.

I On the other hand, if c = a or c = b, try with the minimum. Theminimum of f on [a, b] must be achieved at a point d in [a, b].

I If d is in (a, b), great: it’s a local minimum and so by Fermat’sTheorem f′(d) = 0. If not, d = a or d = b.

I If we still haven’t found a point in the interior, we have that themaximum and minimum values of f on [a, b] occur at bothendpoints. But we already know that f(a) = f(b). If these arethe maximum and minimum values, f is constant on [a, b] and anypoint x in (a, b) will have f′(x) = 0.

. . . . . .

Proof of Rolle’s Theorem

Proof.

I By the Extreme Value Theorem f must achieve its maximumvalue at a point c in [a, b].

I If c is in (a, b), great: it’s a local maximum and so by Fermat’sTheorem f′(c) = 0.

I On the other hand, if c = a or c = b, try with the minimum. Theminimum of f on [a, b] must be achieved at a point d in [a, b].

I If d is in (a, b), great: it’s a local minimum and so by Fermat’sTheorem f′(d) = 0. If not, d = a or d = b.

I If we still haven’t found a point in the interior, we have that themaximum and minimum values of f on [a, b] occur at bothendpoints. But we already know that f(a) = f(b). If these arethe maximum and minimum values, f is constant on [a, b] and anypoint x in (a, b) will have f′(x) = 0.

. . . . . .

Proof of Rolle’s Theorem

Proof.

I By the Extreme Value Theorem f must achieve its maximumvalue at a point c in [a, b].

I If c is in (a, b), great: it’s a local maximum and so by Fermat’sTheorem f′(c) = 0.

I On the other hand, if c = a or c = b, try with the minimum. Theminimum of f on [a, b] must be achieved at a point d in [a, b].

I If d is in (a, b), great: it’s a local minimum and so by Fermat’sTheorem f′(d) = 0. If not, d = a or d = b.

I If we still haven’t found a point in the interior, we have that themaximum and minimum values of f on [a, b] occur at bothendpoints. But we already know that f(a) = f(b). If these arethe maximum and minimum values, f is constant on [a, b] and anypoint x in (a, b) will have f′(x) = 0.

. . . . . .

Proof of Rolle’s Theorem

Proof.

I By the Extreme Value Theorem f must achieve its maximumvalue at a point c in [a, b].

I If c is in (a, b), great: it’s a local maximum and so by Fermat’sTheorem f′(c) = 0.

I On the other hand, if c = a or c = b, try with the minimum. Theminimum of f on [a, b] must be achieved at a point d in [a, b].

I If d is in (a, b), great: it’s a local minimum and so by Fermat’sTheorem f′(d) = 0. If not, d = a or d = b.

I If we still haven’t found a point in the interior, we have that themaximum and minimum values of f on [a, b] occur at bothendpoints.

But we already know that f(a) = f(b). If these arethe maximum and minimum values, f is constant on [a, b] and anypoint x in (a, b) will have f′(x) = 0.

. . . . . .

Proof of Rolle’s Theorem

Proof.

I By the Extreme Value Theorem f must achieve its maximumvalue at a point c in [a, b].

I If c is in (a, b), great: it’s a local maximum and so by Fermat’sTheorem f′(c) = 0.

I On the other hand, if c = a or c = b, try with the minimum. Theminimum of f on [a, b] must be achieved at a point d in [a, b].

I If d is in (a, b), great: it’s a local minimum and so by Fermat’sTheorem f′(d) = 0. If not, d = a or d = b.

I If we still haven’t found a point in the interior, we have that themaximum and minimum values of f on [a, b] occur at bothendpoints. But we already know that f(a) = f(b).

If these arethe maximum and minimum values, f is constant on [a, b] and anypoint x in (a, b) will have f′(x) = 0.

. . . . . .

Proof of Rolle’s Theorem

Proof.

I By the Extreme Value Theorem f must achieve its maximumvalue at a point c in [a, b].

I If c is in (a, b), great: it’s a local maximum and so by Fermat’sTheorem f′(c) = 0.

I On the other hand, if c = a or c = b, try with the minimum. Theminimum of f on [a, b] must be achieved at a point d in [a, b].

I If d is in (a, b), great: it’s a local minimum and so by Fermat’sTheorem f′(d) = 0. If not, d = a or d = b.

I If we still haven’t found a point in the interior, we have that themaximum and minimum values of f on [a, b] occur at bothendpoints. But we already know that f(a) = f(b). If these arethe maximum and minimum values, f is constant on [a, b] and anypoint x in (a, b) will have f′(x) = 0.

. . . . . .

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

Review: The Closed Interval Method

Rolle’s Theorem

The Mean Value TheoremApplications

Why the MVT is the MITC

. . . . . .

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: ClintJCL

. . . . . .

The Mean Value Theorem

Theorem (The Mean ValueTheorem)Let f be continuous on [a, b] anddifferentiable on (a, b). Thenthere exists 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 ValueTheorem)Let f be continuous on [a, b] anddifferentiable on (a, b). Thenthere exists 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 ValueTheorem)Let f be continuous on [a, b] anddifferentiable on (a, b). Thenthere exists a point c in (a, b)such that

f(b) − f(a)b − a

= f′(c). . .•.a

.•.b

.•.c

. . . . . .

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 Theoremthere 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 f is.Also g(a) = 0 and g(b) = 0 (check both). So by Rolle’s Theoremthere 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 f is.

Also g(a) = 0 and g(b) = 0 (check both). So by Rolle’s Theoremthere 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 f is.Also g(a) = 0 and g(b) = 0 (check both).

So by Rolle’s Theoremthere 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 f is.Also g(a) = 0 and g(b) = 0 (check both). So by Rolle’s Theoremthere 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

ExampleShow that there is a unique solution to the equation x3 − x = 100 inthe interval [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.

. . . . . .

Using the MVT to count solutions

ExampleShow that there is a unique solution to the equation x3 − x = 100 inthe interval [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.

. . . . . .

Using the MVT to count solutions

ExampleShow that there is a unique solution to the equation x3 − x = 100 inthe interval [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.

. . . . . .

Using the MVT to count solutions

ExampleShow that there is a unique solution to the equation x3 − x = 100 inthe interval [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.

. . . . . .

ExampleWe 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|

. . . . . .

ExampleWe 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|

. . . . . .

QuestionA driver travels along the New Jersey Turnpike using EZ-Pass. Thesystem takes note of the time and place the driver enters and exitsthe Turnpike. A week after his trip, the driver gets a speeding ticketin the mail. Which of the following best describes the situation?

(a) EZ-Pass cannot prove that the driver was speeding

(b) EZ-Pass 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 EZ-Pass. Thesystem takes note of the time and place the driver enters and exitsthe Turnpike. A week after his trip, the driver gets a speeding ticketin the mail. Which of the following best describes the situation?

(a) EZ-Pass cannot prove that the driver was speeding

(b) EZ-Pass 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 MITC

. . . . . .

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 informationabout 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 determine informationabout 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 determine informationabout 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 determine informationabout 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 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.

. . . . . .

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.

. . . . . .

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.

. . . . . .

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

. . . . . .

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