Section 5.5 The Intermediate Value Theorem Rolle’s Theorem The Mean Value Theorem 3.6.
Lesson 19: The Mean Value Theorem (Section 021 slides)
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Transcript of Lesson 19: The Mean Value Theorem (Section 021 slides)
Section 4.2The Mean Value Theorem
V63.0121.021, Calculus I
New York University
November 11, 2010
AnnouncementsI Quiz 4 next week (November 16, 18, 19) on Sections 3.3, 3.4, 3.5,
3.7
. . . . . .
. . . . . .
Announcements
I Quiz 4 next week(November 16, 18, 19) onSections 3.3, 3.4, 3.5, 3.7
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 2 / 29
. . . . . .
Objectives
I Understand and be able toexplain the statement ofRolle’s Theorem.
I Understand and be able toexplain the statement ofthe Mean Value Theorem.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 3 / 29
. . . . . .
Outline
Rolle’s Theorem
The Mean Value TheoremApplications
Why the MVT is the MITCFunctions with derivatives that are zeroMVT and differentiability
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 4 / 29
. . . . . .
Heuristic Motivation for Rolle's Theorem
If you bike up a hill, then back down, at some point your elevation wasstationary.
.
.Image credit: SpringSunV63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 5 / 29
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 6 / 29
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 6 / 29
. . . . . .
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
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 8 / 29
. . . . . .
Outline
Rolle’s Theorem
The Mean Value TheoremApplications
Why the MVT is the MITCFunctions with derivatives that are zeroMVT and differentiability
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 9 / 29
. . . . . .
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: ClintJCLV63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 10 / 29
. . . . . .
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 cin (a,b) such that
f(b)− f(a)b− a
= f′(c). . ..a
..b
.c
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 11 / 29
. . . . . .
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 cin (a,b) such that
f(b)− f(a)b− a
= f′(c). . ..a
..b
.c
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 11 / 29
. . . . . .
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 cin (a,b) such that
f(b)− f(a)b− a
= f′(c). . ..a
..b
.c
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 11 / 29
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 12 / 29
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 12 / 29
. . . . . .
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 Theorem thereexists a point c in (a,b) such that
0 = g′(c) = f′(c)− f(b)− f(a)b− a
.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 13 / 29
. . . . . .
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 Theorem thereexists a point c in (a,b) such that
0 = g′(c) = f′(c)− f(b)− f(a)b− a
.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 13 / 29
. . . . . .
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 Theorem thereexists a point c in (a,b) such that
0 = g′(c) = f′(c)− f(b)− f(a)b− a
.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 13 / 29
. . . . . .
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 Theorem thereexists a point c in (a,b) such that
0 = g′(c) = f′(c)− f(b)− f(a)b− a
.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 13 / 29
. . . . . .
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 Theorem thereexists a point c in (a,b) such that
0 = g′(c) = f′(c)− f(b)− f(a)b− a
.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 13 / 29
. . . . . .
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 − 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, 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 thisis impossible.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 14 / 29
. . . . . .
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 − 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, 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 thisis impossible.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 14 / 29
. . . . . .
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 − 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, 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 thisis impossible.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 14 / 29
. . . . . .
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 − 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, 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 thisis impossible.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 14 / 29
. . . . . .
Using the MVT to estimate
Example
We know that |sin x| ≤ 1 for all x, and that sin x ≈ x for small x. Showthat |sin x| ≤ |x| for all 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|
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 15 / 29
. . . . . .
Using the MVT to estimate
Example
We know that |sin x| ≤ 1 for all x, and that sin x ≈ x for small x. Showthat |sin x| ≤ |x| for all 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|
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 15 / 29
. . . . . .
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))
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 16 / 29
. . . . . .
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))
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 16 / 29
. . . . . .
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))
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 16 / 29
. . . . . .
Food for Thought
QuestionA driver travels along the New Jersey Turnpike using E-ZPass. Thesystem takes note of the time and place the driver enters and exits theTurnpike. A week after his trip, the driver gets a speeding ticket in themail. Which of 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).
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 17 / 29
. . . . . .
Food for Thought
QuestionA driver travels along the New Jersey Turnpike using E-ZPass. Thesystem takes note of the time and place the driver enters and exits theTurnpike. A week after his trip, the driver gets a speeding ticket in themail. Which of 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).
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 17 / 29
. . . . . .
Outline
Rolle’s Theorem
The Mean Value TheoremApplications
Why the MVT is the MITCFunctions with derivatives that are zeroMVT and differentiability
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 18 / 29
. . . . . .
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 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 information aboutthe function itself
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 19 / 29
. . . . . .
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 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 information aboutthe function itself
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 19 / 29
. . . . . .
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 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 information aboutthe function itself
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 19 / 29
. . . . . .
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 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 information aboutthe function itself
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 19 / 29
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 20 / 29
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 20 / 29
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 20 / 29
. . . . . .
Functions with the same derivative
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 Csuch 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)
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 21 / 29
. . . . . .
Functions with the same derivative
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 Csuch 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)
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 21 / 29
. . . . . .
Functions with the same derivative
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 Csuch 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)
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 21 / 29
. . . . . .
Functions with the same derivative
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 Csuch 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)
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 21 / 29
. . . . . .
Functions with the same derivative
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 Csuch 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)
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 21 / 29
. . . . . .
Functions with the same derivative
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 Csuch 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)
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 21 / 29
. . . . . .
MVT and differentiability
Example
Let
f(x) =
{−x if x ≤ 0x2 if x ≥ 0
Is f differentiable at 0?
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 22 / 29
. . . . . .
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.V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 22 / 29
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 22 / 29
. . . . . .
Why only “sort of"?
I This solution is valid butless direct.
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)
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 24 / 29
. . . . . .
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 ̸= 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 not continuousat 0!
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 25 / 29
. . . . . .
Differentiability FAIL
. .x
.f(x)
This function is differentiable at0.
. .x
.f′(x)
.
But the derivative is notcontinuous at 0!
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 26 / 29
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 27 / 29
. . . . . .
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.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 27 / 29
. . . . . .
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-hand sidesagree.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 28 / 29
. . . . . .
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-hand sidesagree.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 28 / 29
. . . . . .
Summary
I Rolle’s Theorem: under suitable conditions, functions must havecritical points.
I Mean Value Theorem: under suitable conditions, functions musthave an instantaneous rate of change equal to the average rate ofchange.
I A function whose derivative is identically zero on an interval mustbe constant on that interval.
I E-ZPass is kinder than we realized.
V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 29 / 29