Mixing
Rolle's theorem: what's the right statement of the theorem?
Clash Royale CLAN TAG#URR8PPP
up vote
7
down vote
favorite
In the fourth edition of "Introduction to Real Analysis" by Bartle and Sherbert, theorem 6.2.3 (Rolle's theorem) states,
Suppose that f is continuous on a closed interval $I := [a, b]$, that
the derivative of $f$ exists at every point of the open interval $(a, b)$, and that $f(a) = f(b) = 0$.
Then there exists at least one point $c$ in $(a, b)$ such that the derivative of $f$ is zero at $c$.
Now, why are we taking $f(a)=0=f(b)$? Is $f(a)=f(b)$ not sufficient?
real-analysis definition rolles-theorem
edited 2 days ago
Brahadeesh
3,25731144
asked 2 days ago
U Know me
384
add a comment |Â
up vote
7
down vote
favorite
In the fourth edition of "Introduction to Real Analysis" by Bartle and Sherbert, theorem 6.2.3 (Rolle's theorem) states,
Suppose that f is continuous on a closed interval $I := [a, b]$, that
the derivative of $f$ exists at every point of the open interval $(a, b)$, and that $f(a) = f(b) = 0$.
Then there exists at least one point $c$ in $(a, b)$ such that the derivative of $f$ is zero at $c$.
Now, why are we taking $f(a)=0=f(b)$? Is $f(a)=f(b)$ not sufficient?
real-analysis definition rolles-theorem
edited 2 days ago
Brahadeesh
3,25731144
asked 2 days ago
U Know me
384
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
2 days ago
If by 'right statement' you mean 'most general', then no we don't need f(a),f(b) to equal zero.
– smci
18 hours ago
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
In the fourth edition of "Introduction to Real Analysis" by Bartle and Sherbert, theorem 6.2.3 (Rolle's theorem) states,
Suppose that f is continuous on a closed interval $I := [a, b]$, that
the derivative of $f$ exists at every point of the open interval $(a, b)$, and that $f(a) = f(b) = 0$.
Then there exists at least one point $c$ in $(a, b)$ such that the derivative of $f$ is zero at $c$.
Now, why are we taking $f(a)=0=f(b)$? Is $f(a)=f(b)$ not sufficient?
real-analysis definition rolles-theorem
edited 2 days ago
Brahadeesh
3,25731144
asked 2 days ago
U Know me
384
In the fourth edition of "Introduction to Real Analysis" by Bartle and Sherbert, theorem 6.2.3 (Rolle's theorem) states,
Suppose that f is continuous on a closed interval $I := [a, b]$, that
the derivative of $f$ exists at every point of the open interval $(a, b)$, and that $f(a) = f(b) = 0$.
Then there exists at least one point $c$ in $(a, b)$ such that the derivative of $f$ is zero at $c$.
Now, why are we taking $f(a)=0=f(b)$? Is $f(a)=f(b)$ not sufficient?
real-analysis definition rolles-theorem
edited 2 days ago
Brahadeesh
3,25731144
asked 2 days ago
U Know me
384
edited 2 days ago
Brahadeesh
3,25731144
edited 2 days ago
Brahadeesh
3,25731144
edited 2 days ago
Brahadeesh
3,25731144
3,25731144
asked 2 days ago
U Know me
384
asked 2 days ago
U Know me
384
asked 2 days ago
U Know me
384
384
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
2 days ago
If by 'right statement' you mean 'most general', then no we don't need f(a),f(b) to equal zero.
– smci
18 hours ago
add a comment |Â
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
2 days ago
If by 'right statement' you mean 'most general', then no we don't need f(a),f(b) to equal zero.
– smci
18 hours ago
1
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
2 days ago
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
2 days ago
If by 'right statement' you mean 'most general', then no we don't need f(a),f(b) to equal zero.
– smci
18 hours ago
If by 'right statement' you mean 'most general', then no we don't need f(a),f(b) to equal zero.
– smci
18 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
9
down vote
accepted
You are right, taking $f(a) = f(b)$ is sufficient.
But, one can prove the theorem in this general scenario using the theorem for the case $f(a) = 0 = f(b)$, as follows:
Assume Rolle's theorem as stated in the question details is true. Let $f$ be a function satisfying the same hypotheses, except that $f(a) = f(b) = k$, where $k$ is not necessarily equal to zero. Then, the function $g(x) = f(x) - k$ satisfies the hypotheses of Rolle's theorem, and so there is a point $c$ such that $g'(c) = 0$. But $g'(c) = f'(c)$, so we are done.
So, it doesn't really matter which one we use, as both versions are seen to be equivalent to each other.
answered 2 days ago
Brahadeesh
3,25731144
If by 'right statement' OP means 'most general', then no we don't need f(a),f(b) to equal zero. Better to state in the more general. And Rolle's Theorem not always about roots (f(a)=f(b)=0). The version I was taught was simply a formalization that between two points with equal y-value, there is a peak with zero derivative.
– smci
17 hours ago
This could be an answer on its own.
– Brahadeesh
13 hours ago
add a comment |Â
up vote
8
down vote
Yes, you are right: $f(a)=f(b)$ is enough. However, it is usual to add the condition $=0$ after $f(a)=f(b)$ so that the theorem becomes: between two roots of $f$, there's a root of $f'$, which is the original theorem due Michel Rolle (who stated it for polynomials only).
Anyway, this is a moot point, since both statements are just a particular case of the Mean Value Theorem.
answered 2 days ago
José Carlos Santos
111k1695172
Thank you. It was really helpful.
– U Know me
2 days ago
I'm glad I could help.
– José Carlos Santos
2 days ago
3
OTOH, the usual way to prove the Mean Value Theorem is by applying the theorem of Rolle. Depending on how much one 'groks' working with derivatives, the step from the general ($f(a) = f(b)$ but not necessarily $=0$) version of Rolle to the Mean Value Theorem is not much bigger than from the "$f(a) = f(b)=0$" version of Rolle to the general one.
– Ingix
2 days ago
2
+1 for the history of these theorems.
– Paramanand Singh
yesterday
add a comment |Â
up vote
4
down vote
The two versions are clearly equivalent. Suppose just $f(a)=f(b)$. Let $g(t)=f(t)-f(a)$. Then $g(a)=g(b)=0$, so the formally weaker version shows $g'(c)=0$, hence $f'(c)=0$ for some $c$.
answered 2 days ago
David C. Ullrich
53.7k33481
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
You are right, taking $f(a) = f(b)$ is sufficient.
But, one can prove the theorem in this general scenario using the theorem for the case $f(a) = 0 = f(b)$, as follows:
Assume Rolle's theorem as stated in the question details is true. Let $f$ be a function satisfying the same hypotheses, except that $f(a) = f(b) = k$, where $k$ is not necessarily equal to zero. Then, the function $g(x) = f(x) - k$ satisfies the hypotheses of Rolle's theorem, and so there is a point $c$ such that $g'(c) = 0$. But $g'(c) = f'(c)$, so we are done.
So, it doesn't really matter which one we use, as both versions are seen to be equivalent to each other.
answered 2 days ago
Brahadeesh
3,25731144
If by 'right statement' OP means 'most general', then no we don't need f(a),f(b) to equal zero. Better to state in the more general. And Rolle's Theorem not always about roots (f(a)=f(b)=0). The version I was taught was simply a formalization that between two points with equal y-value, there is a peak with zero derivative.
– smci
17 hours ago
This could be an answer on its own.
– Brahadeesh
13 hours ago
add a comment |Â
up vote
9
down vote
accepted
You are right, taking $f(a) = f(b)$ is sufficient.
But, one can prove the theorem in this general scenario using the theorem for the case $f(a) = 0 = f(b)$, as follows:
Assume Rolle's theorem as stated in the question details is true. Let $f$ be a function satisfying the same hypotheses, except that $f(a) = f(b) = k$, where $k$ is not necessarily equal to zero. Then, the function $g(x) = f(x) - k$ satisfies the hypotheses of Rolle's theorem, and so there is a point $c$ such that $g'(c) = 0$. But $g'(c) = f'(c)$, so we are done.
So, it doesn't really matter which one we use, as both versions are seen to be equivalent to each other.
answered 2 days ago
Brahadeesh
3,25731144
If by 'right statement' OP means 'most general', then no we don't need f(a),f(b) to equal zero. Better to state in the more general. And Rolle's Theorem not always about roots (f(a)=f(b)=0). The version I was taught was simply a formalization that between two points with equal y-value, there is a peak with zero derivative.
– smci
17 hours ago
This could be an answer on its own.
– Brahadeesh
13 hours ago
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
You are right, taking $f(a) = f(b)$ is sufficient.
But, one can prove the theorem in this general scenario using the theorem for the case $f(a) = 0 = f(b)$, as follows:
Assume Rolle's theorem as stated in the question details is true. Let $f$ be a function satisfying the same hypotheses, except that $f(a) = f(b) = k$, where $k$ is not necessarily equal to zero. Then, the function $g(x) = f(x) - k$ satisfies the hypotheses of Rolle's theorem, and so there is a point $c$ such that $g'(c) = 0$. But $g'(c) = f'(c)$, so we are done.
So, it doesn't really matter which one we use, as both versions are seen to be equivalent to each other.
answered 2 days ago
Brahadeesh
3,25731144
You are right, taking $f(a) = f(b)$ is sufficient.
But, one can prove the theorem in this general scenario using the theorem for the case $f(a) = 0 = f(b)$, as follows:
Assume Rolle's theorem as stated in the question details is true. Let $f$ be a function satisfying the same hypotheses, except that $f(a) = f(b) = k$, where $k$ is not necessarily equal to zero. Then, the function $g(x) = f(x) - k$ satisfies the hypotheses of Rolle's theorem, and so there is a point $c$ such that $g'(c) = 0$. But $g'(c) = f'(c)$, so we are done.
So, it doesn't really matter which one we use, as both versions are seen to be equivalent to each other.
answered 2 days ago
Brahadeesh
3,25731144
edited yesterday
answered 2 days ago
Brahadeesh
3,25731144
answered 2 days ago
Brahadeesh
3,25731144
answered 2 days ago
Brahadeesh
3,25731144
3,25731144
If by 'right statement' OP means 'most general', then no we don't need f(a),f(b) to equal zero. Better to state in the more general. And Rolle's Theorem not always about roots (f(a)=f(b)=0). The version I was taught was simply a formalization that between two points with equal y-value, there is a peak with zero derivative.
– smci
17 hours ago
This could be an answer on its own.
– Brahadeesh
13 hours ago
add a comment |Â
If by 'right statement' OP means 'most general', then no we don't need f(a),f(b) to equal zero. Better to state in the more general. And Rolle's Theorem not always about roots (f(a)=f(b)=0). The version I was taught was simply a formalization that between two points with equal y-value, there is a peak with zero derivative.
– smci
17 hours ago
This could be an answer on its own.
– Brahadeesh
13 hours ago
If by 'right statement' OP means 'most general', then no we don't need f(a),f(b) to equal zero. Better to state in the more general. And Rolle's Theorem not always about roots (f(a)=f(b)=0). The version I was taught was simply a formalization that between two points with equal y-value, there is a peak with zero derivative.
– smci
17 hours ago
If by 'right statement' OP means 'most general', then no we don't need f(a),f(b) to equal zero. Better to state in the more general. And Rolle's Theorem not always about roots (f(a)=f(b)=0). The version I was taught was simply a formalization that between two points with equal y-value, there is a peak with zero derivative.
– smci
17 hours ago
This could be an answer on its own.
– Brahadeesh
13 hours ago
This could be an answer on its own.
– Brahadeesh
13 hours ago
add a comment |Â
up vote
8
down vote
Yes, you are right: $f(a)=f(b)$ is enough. However, it is usual to add the condition $=0$ after $f(a)=f(b)$ so that the theorem becomes: between two roots of $f$, there's a root of $f'$, which is the original theorem due Michel Rolle (who stated it for polynomials only).
Anyway, this is a moot point, since both statements are just a particular case of the Mean Value Theorem.
answered 2 days ago
José Carlos Santos
111k1695172
Thank you. It was really helpful.
– U Know me
2 days ago
I'm glad I could help.
– José Carlos Santos
2 days ago
3
OTOH, the usual way to prove the Mean Value Theorem is by applying the theorem of Rolle. Depending on how much one 'groks' working with derivatives, the step from the general ($f(a) = f(b)$ but not necessarily $=0$) version of Rolle to the Mean Value Theorem is not much bigger than from the "$f(a) = f(b)=0$" version of Rolle to the general one.
– Ingix
2 days ago
2
+1 for the history of these theorems.
– Paramanand Singh
yesterday
add a comment |Â
up vote
8
down vote
Yes, you are right: $f(a)=f(b)$ is enough. However, it is usual to add the condition $=0$ after $f(a)=f(b)$ so that the theorem becomes: between two roots of $f$, there's a root of $f'$, which is the original theorem due Michel Rolle (who stated it for polynomials only).
Anyway, this is a moot point, since both statements are just a particular case of the Mean Value Theorem.
answered 2 days ago
José Carlos Santos
111k1695172
Thank you. It was really helpful.
– U Know me
2 days ago
I'm glad I could help.
– José Carlos Santos
2 days ago
3
OTOH, the usual way to prove the Mean Value Theorem is by applying the theorem of Rolle. Depending on how much one 'groks' working with derivatives, the step from the general ($f(a) = f(b)$ but not necessarily $=0$) version of Rolle to the Mean Value Theorem is not much bigger than from the "$f(a) = f(b)=0$" version of Rolle to the general one.
– Ingix
2 days ago
2
+1 for the history of these theorems.
– Paramanand Singh
yesterday
add a comment |Â
up vote
8
down vote
up vote
8
down vote
Yes, you are right: $f(a)=f(b)$ is enough. However, it is usual to add the condition $=0$ after $f(a)=f(b)$ so that the theorem becomes: between two roots of $f$, there's a root of $f'$, which is the original theorem due Michel Rolle (who stated it for polynomials only).
Anyway, this is a moot point, since both statements are just a particular case of the Mean Value Theorem.
answered 2 days ago
José Carlos Santos
111k1695172
Yes, you are right: $f(a)=f(b)$ is enough. However, it is usual to add the condition $=0$ after $f(a)=f(b)$ so that the theorem becomes: between two roots of $f$, there's a root of $f'$, which is the original theorem due Michel Rolle (who stated it for polynomials only).
Anyway, this is a moot point, since both statements are just a particular case of the Mean Value Theorem.
answered 2 days ago
José Carlos Santos
111k1695172
answered 2 days ago
José Carlos Santos
111k1695172
answered 2 days ago
José Carlos Santos
111k1695172
answered 2 days ago
José Carlos Santos
111k1695172
111k1695172
Thank you. It was really helpful.
– U Know me
2 days ago
I'm glad I could help.
– José Carlos Santos
2 days ago
3
OTOH, the usual way to prove the Mean Value Theorem is by applying the theorem of Rolle. Depending on how much one 'groks' working with derivatives, the step from the general ($f(a) = f(b)$ but not necessarily $=0$) version of Rolle to the Mean Value Theorem is not much bigger than from the "$f(a) = f(b)=0$" version of Rolle to the general one.
– Ingix
2 days ago
2
+1 for the history of these theorems.
– Paramanand Singh
yesterday
add a comment |Â
Thank you. It was really helpful.
– U Know me
2 days ago
I'm glad I could help.
– José Carlos Santos
2 days ago
3
OTOH, the usual way to prove the Mean Value Theorem is by applying the theorem of Rolle. Depending on how much one 'groks' working with derivatives, the step from the general ($f(a) = f(b)$ but not necessarily $=0$) version of Rolle to the Mean Value Theorem is not much bigger than from the "$f(a) = f(b)=0$" version of Rolle to the general one.
– Ingix
2 days ago
2
+1 for the history of these theorems.
– Paramanand Singh
yesterday
Thank you. It was really helpful.
– U Know me
2 days ago
Thank you. It was really helpful.
– U Know me
2 days ago
I'm glad I could help.
– José Carlos Santos
2 days ago
I'm glad I could help.
– José Carlos Santos
2 days ago
3
3
OTOH, the usual way to prove the Mean Value Theorem is by applying the theorem of Rolle. Depending on how much one 'groks' working with derivatives, the step from the general ($f(a) = f(b)$ but not necessarily $=0$) version of Rolle to the Mean Value Theorem is not much bigger than from the "$f(a) = f(b)=0$" version of Rolle to the general one.
– Ingix
2 days ago
OTOH, the usual way to prove the Mean Value Theorem is by applying the theorem of Rolle. Depending on how much one 'groks' working with derivatives, the step from the general ($f(a) = f(b)$ but not necessarily $=0$) version of Rolle to the Mean Value Theorem is not much bigger than from the "$f(a) = f(b)=0$" version of Rolle to the general one.
– Ingix
2 days ago
2
2
+1 for the history of these theorems.
– Paramanand Singh
yesterday
+1 for the history of these theorems.
– Paramanand Singh
yesterday
add a comment |Â
up vote
4
down vote
The two versions are clearly equivalent. Suppose just $f(a)=f(b)$. Let $g(t)=f(t)-f(a)$. Then $g(a)=g(b)=0$, so the formally weaker version shows $g'(c)=0$, hence $f'(c)=0$ for some $c$.
answered 2 days ago
David C. Ullrich
53.7k33481
add a comment |Â
up vote
4
down vote
The two versions are clearly equivalent. Suppose just $f(a)=f(b)$. Let $g(t)=f(t)-f(a)$. Then $g(a)=g(b)=0$, so the formally weaker version shows $g'(c)=0$, hence $f'(c)=0$ for some $c$.
answered 2 days ago
David C. Ullrich
53.7k33481
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The two versions are clearly equivalent. Suppose just $f(a)=f(b)$. Let $g(t)=f(t)-f(a)$. Then $g(a)=g(b)=0$, so the formally weaker version shows $g'(c)=0$, hence $f'(c)=0$ for some $c$.
answered 2 days ago
David C. Ullrich
53.7k33481
The two versions are clearly equivalent. Suppose just $f(a)=f(b)$. Let $g(t)=f(t)-f(a)$. Then $g(a)=g(b)=0$, so the formally weaker version shows $g'(c)=0$, hence $f'(c)=0$ for some $c$.
answered 2 days ago
David C. Ullrich
53.7k33481
answered 2 days ago
David C. Ullrich
53.7k33481
answered 2 days ago
David C. Ullrich
53.7k33481
answered 2 days ago
David C. Ullrich
53.7k33481
53.7k33481
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2872218%2frolles-theorem-whats-the-right-statement-of-the-theorem%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
2 days ago
If by 'right statement' you mean 'most general', then no we don't need f(a),f(b) to equal zero.
– smci
18 hours ago