lesson 19: the mean value theorem (slides)
DESCRIPTION
The Mean Value Theorem is the most important theorem in calculus. It is the first theorem which allows us to infer information about a function from information about its derivative. From the MVT we can derive tests for the monotonicity (increase or decrease) and concavity of a function.TRANSCRIPT
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..
Sec on 4.2The Mean Value Theorem
V63.0121.011: Calculus IProfessor Ma hew Leingang
New York University
April 6, 2011
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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
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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
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Objectives
I Understand and be ableto explain the statementof Rolle’s Theorem.
I Understand and be ableto explain the statementof theMean ValueTheorem.
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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
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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
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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
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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
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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
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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
![Page 11: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/11.jpg)
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
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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
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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
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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
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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.
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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.
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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)
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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).
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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.
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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).
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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
.
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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.
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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.
![Page 24: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/24.jpg)
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.
![Page 25: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/25.jpg)
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.
![Page 26: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/26.jpg)
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|
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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|
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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?
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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))
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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))
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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).
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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).
![Page 33: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/33.jpg)
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
![Page 34: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/34.jpg)
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
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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
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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
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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
![Page 38: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/38.jpg)
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.
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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.
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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.
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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)
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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)
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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)
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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)
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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)
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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)
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MVT and differentiability
Example
Let
f(x) =
{−x if x ≤ 0x2 if x ≥ 0
Is f differen able at 0?
![Page 48: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/48.jpg)
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.
![Page 49: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/49.jpg)
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.
![Page 50: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/50.jpg)
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)
![Page 51: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/51.jpg)
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.
![Page 52: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/52.jpg)
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!
![Page 53: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/53.jpg)
Differentiability FAIL
.. x.
f(x)
This func on is differen ableat 0.
.. x.
f′(x)
.
But the deriva ve is notcon nuous at 0!
![Page 54: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/54.jpg)
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.
![Page 55: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/55.jpg)
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.
![Page 56: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/56.jpg)
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.
![Page 57: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/57.jpg)
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.
![Page 58: Lesson 19: The Mean Value Theorem (slides)](https://reader035.vdocuments.site/reader035/viewer/2022062308/558c69b8d8b42afb508b4732/html5/thumbnails/58.jpg)
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.