Inverse Function like theorem for monotoncity
$begingroup$
Inverse function theorem says that locally a function is $1-1$ if its derivate is non-zero. Do we have a similar conclusion for monotonous function i.e. statement of the following kind
If $f(x)$ is continuous and has a non-zero derivative at $x_0$, then their exists an open ball around $x_0$ such that $f(x)$ is monotonous?
calculus
asked Dec 5 '18 at 19:58
henceprovedhenceproved
1628
$endgroup$
add a comment |
$begingroup$
Inverse function theorem says that locally a function is $1-1$ if its derivate is non-zero. Do we have a similar conclusion for monotonous function i.e. statement of the following kind
If $f(x)$ is continuous and has a non-zero derivative at $x_0$, then their exists an open ball around $x_0$ such that $f(x)$ is monotonous?
calculus
asked Dec 5 '18 at 19:58
henceprovedhenceproved
1628
$endgroup$
$begingroup$
"Open ball" doesn't have much sense over an one variable function. Probably you mean an "interval".
$endgroup$
– Rebellos
Dec 5 '18 at 20:02
$begingroup$
Consider $fcolon(-1,1)tomathbb R$ defined by $f(x)=tan(x)$ for rational $x$ and $f(x)=x$ otherwise. Now $f$ is continuous in $x_0=0$ with derivative $f'(0)=1$, but there doesn't exist a neighbourhood of $x_0$ where $f$ is monotonous.
$endgroup$
– Michael Hoppe
Dec 5 '18 at 20:06
$begingroup$
@MichaelHoppe I don't think $f(x)$ is continuous on an interval? It is only continuous at a point.
$endgroup$
– henceproved
Dec 5 '18 at 20:10
$begingroup$
Well, you asked for a function, such "$f(x)$ is continuous and has a non-zero derivative at $x_0$". I gave you one.
$endgroup$
– Michael Hoppe
Dec 5 '18 at 20:13
$begingroup$
For the "lemma" you want to derive, you may need a more special "smoothness" and characteristics in general of any functions regarded. Simple continuity won't do the trick.
$endgroup$
– Rebellos
Dec 5 '18 at 20:16
add a comment |
$begingroup$
Inverse function theorem says that locally a function is $1-1$ if its derivate is non-zero. Do we have a similar conclusion for monotonous function i.e. statement of the following kind
If $f(x)$ is continuous and has a non-zero derivative at $x_0$, then their exists an open ball around $x_0$ such that $f(x)$ is monotonous?
calculus
asked Dec 5 '18 at 19:58
henceprovedhenceproved
1628
$endgroup$
Inverse function theorem says that locally a function is $1-1$ if its derivate is non-zero. Do we have a similar conclusion for monotonous function i.e. statement of the following kind
If $f(x)$ is continuous and has a non-zero derivative at $x_0$, then their exists an open ball around $x_0$ such that $f(x)$ is monotonous?
calculus
calculus
asked Dec 5 '18 at 19:58
henceprovedhenceproved
1628
asked Dec 5 '18 at 19:58
henceprovedhenceproved
1628
asked Dec 5 '18 at 19:58
henceprovedhenceproved
1628
asked Dec 5 '18 at 19:58
henceprovedhenceproved
1628
asked Dec 5 '18 at 19:58
henceprovedhenceproved
1628
1628
$begingroup$
"Open ball" doesn't have much sense over an one variable function. Probably you mean an "interval".
$endgroup$
– Rebellos
Dec 5 '18 at 20:02
$begingroup$
Consider $fcolon(-1,1)tomathbb R$ defined by $f(x)=tan(x)$ for rational $x$ and $f(x)=x$ otherwise. Now $f$ is continuous in $x_0=0$ with derivative $f'(0)=1$, but there doesn't exist a neighbourhood of $x_0$ where $f$ is monotonous.
$endgroup$
– Michael Hoppe
Dec 5 '18 at 20:06
$begingroup$
@MichaelHoppe I don't think $f(x)$ is continuous on an interval? It is only continuous at a point.
$endgroup$
– henceproved
Dec 5 '18 at 20:10
$begingroup$
Well, you asked for a function, such "$f(x)$ is continuous and has a non-zero derivative at $x_0$". I gave you one.
$endgroup$
– Michael Hoppe
Dec 5 '18 at 20:13
$begingroup$
For the "lemma" you want to derive, you may need a more special "smoothness" and characteristics in general of any functions regarded. Simple continuity won't do the trick.
$endgroup$
– Rebellos
Dec 5 '18 at 20:16
add a comment |
$begingroup$
"Open ball" doesn't have much sense over an one variable function. Probably you mean an "interval".
$endgroup$
– Rebellos
Dec 5 '18 at 20:02
$begingroup$
Consider $fcolon(-1,1)tomathbb R$ defined by $f(x)=tan(x)$ for rational $x$ and $f(x)=x$ otherwise. Now $f$ is continuous in $x_0=0$ with derivative $f'(0)=1$, but there doesn't exist a neighbourhood of $x_0$ where $f$ is monotonous.
$endgroup$
– Michael Hoppe
Dec 5 '18 at 20:06
$begingroup$
@MichaelHoppe I don't think $f(x)$ is continuous on an interval? It is only continuous at a point.
$endgroup$
– henceproved
Dec 5 '18 at 20:10
$begingroup$
Well, you asked for a function, such "$f(x)$ is continuous and has a non-zero derivative at $x_0$". I gave you one.
$endgroup$
– Michael Hoppe
Dec 5 '18 at 20:13
$begingroup$
For the "lemma" you want to derive, you may need a more special "smoothness" and characteristics in general of any functions regarded. Simple continuity won't do the trick.
$endgroup$
– Rebellos
Dec 5 '18 at 20:16
$begingroup$
"Open ball" doesn't have much sense over an one variable function. Probably you mean an "interval".
$endgroup$
– Rebellos
Dec 5 '18 at 20:02
$begingroup$
"Open ball" doesn't have much sense over an one variable function. Probably you mean an "interval".
$endgroup$
– Rebellos
Dec 5 '18 at 20:02
$begingroup$
Consider $fcolon(-1,1)tomathbb R$ defined by $f(x)=tan(x)$ for rational $x$ and $f(x)=x$ otherwise. Now $f$ is continuous in $x_0=0$ with derivative $f'(0)=1$, but there doesn't exist a neighbourhood of $x_0$ where $f$ is monotonous.
$endgroup$
– Michael Hoppe
Dec 5 '18 at 20:06
$begingroup$
Consider $fcolon(-1,1)tomathbb R$ defined by $f(x)=tan(x)$ for rational $x$ and $f(x)=x$ otherwise. Now $f$ is continuous in $x_0=0$ with derivative $f'(0)=1$, but there doesn't exist a neighbourhood of $x_0$ where $f$ is monotonous.
$endgroup$
– Michael Hoppe
Dec 5 '18 at 20:06
$begingroup$
@MichaelHoppe I don't think $f(x)$ is continuous on an interval? It is only continuous at a point.
$endgroup$
– henceproved
Dec 5 '18 at 20:10
$begingroup$
@MichaelHoppe I don't think $f(x)$ is continuous on an interval? It is only continuous at a point.
$endgroup$
– henceproved
Dec 5 '18 at 20:10
$begingroup$
Well, you asked for a function, such "$f(x)$ is continuous and has a non-zero derivative at $x_0$". I gave you one.
$endgroup$
– Michael Hoppe
Dec 5 '18 at 20:13
$begingroup$
Well, you asked for a function, such "$f(x)$ is continuous and has a non-zero derivative at $x_0$". I gave you one.
$endgroup$
– Michael Hoppe
Dec 5 '18 at 20:13
$begingroup$
For the "lemma" you want to derive, you may need a more special "smoothness" and characteristics in general of any functions regarded. Simple continuity won't do the trick.
$endgroup$
– Rebellos
Dec 5 '18 at 20:16
$begingroup$
For the "lemma" you want to derive, you may need a more special "smoothness" and characteristics in general of any functions regarded. Simple continuity won't do the trick.
$endgroup$
– Rebellos
Dec 5 '18 at 20:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Differentiability is not enough but continuous differentiability is. $f(x) = x^{2}sin(1/x) + x$ is differentiable at $0$ with derivative $1$, but is not monotonic on any interval containing $0$.
answered Dec 6 '18 at 11:03
preferred_anonpreferred_anon
13k11742
$endgroup$
$begingroup$
Is the theorem well know, if so then can you point out to the source or else can you show me how to prove it?
$endgroup$
– henceproved
Dec 6 '18 at 12:14
$begingroup$
It's the Inverse Function Theorem in 1 dimension.
$endgroup$
– preferred_anon
Dec 6 '18 at 12:16
add a comment |
$begingroup$
Consider $fcolon(-1,1)tomathbb R$ defined by $f(x)=tan(x)$ for rational $x$ and $f(x)=x$ otherwise. Now $f$ is continuous in $x_0=0$ with derivative $f'(0)=1$, but there doesn't exist a neighbourhood of $x_0$ where $f$ is monotonous.
All we can say here that there's a "punctual" monotony: We only can claim that for $x<0<y$ we have $f(x)<f(y)$.
answered Dec 5 '18 at 20:11
Michael HoppeMichael Hoppe
11.1k31837
$endgroup$
$begingroup$
The $f(x)$ that you construct is not a continuous function on any interval. My question loosely is if $f(x)$ is continuous in an interval, then does their exists an subinterval where $f(x)$ is monotonous.
$endgroup$
– henceproved
Dec 6 '18 at 0:11
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027573%2finverse-function-like-theorem-for-monotoncity%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Differentiability is not enough but continuous differentiability is. $f(x) = x^{2}sin(1/x) + x$ is differentiable at $0$ with derivative $1$, but is not monotonic on any interval containing $0$.
answered Dec 6 '18 at 11:03
preferred_anonpreferred_anon
13k11742
$endgroup$
$begingroup$
Is the theorem well know, if so then can you point out to the source or else can you show me how to prove it?
$endgroup$
– henceproved
Dec 6 '18 at 12:14
$begingroup$
It's the Inverse Function Theorem in 1 dimension.
$endgroup$
– preferred_anon
Dec 6 '18 at 12:16
add a comment |
$begingroup$
Differentiability is not enough but continuous differentiability is. $f(x) = x^{2}sin(1/x) + x$ is differentiable at $0$ with derivative $1$, but is not monotonic on any interval containing $0$.
answered Dec 6 '18 at 11:03
preferred_anonpreferred_anon
13k11742
$endgroup$
$begingroup$
Is the theorem well know, if so then can you point out to the source or else can you show me how to prove it?
$endgroup$
– henceproved
Dec 6 '18 at 12:14
$begingroup$
It's the Inverse Function Theorem in 1 dimension.
$endgroup$
– preferred_anon
Dec 6 '18 at 12:16
add a comment |
$begingroup$
Differentiability is not enough but continuous differentiability is. $f(x) = x^{2}sin(1/x) + x$ is differentiable at $0$ with derivative $1$, but is not monotonic on any interval containing $0$.
answered Dec 6 '18 at 11:03
preferred_anonpreferred_anon
13k11742
$endgroup$
Differentiability is not enough but continuous differentiability is. $f(x) = x^{2}sin(1/x) + x$ is differentiable at $0$ with derivative $1$, but is not monotonic on any interval containing $0$.
answered Dec 6 '18 at 11:03
preferred_anonpreferred_anon
13k11742
answered Dec 6 '18 at 11:03
preferred_anonpreferred_anon
13k11742
answered Dec 6 '18 at 11:03
preferred_anonpreferred_anon
13k11742
answered Dec 6 '18 at 11:03
preferred_anonpreferred_anon
13k11742
13k11742
$begingroup$
Is the theorem well know, if so then can you point out to the source or else can you show me how to prove it?
$endgroup$
– henceproved
Dec 6 '18 at 12:14
$begingroup$
It's the Inverse Function Theorem in 1 dimension.
$endgroup$
– preferred_anon
Dec 6 '18 at 12:16
add a comment |
$begingroup$
Is the theorem well know, if so then can you point out to the source or else can you show me how to prove it?
$endgroup$
– henceproved
Dec 6 '18 at 12:14
$begingroup$
It's the Inverse Function Theorem in 1 dimension.
$endgroup$
– preferred_anon
Dec 6 '18 at 12:16
$begingroup$
Is the theorem well know, if so then can you point out to the source or else can you show me how to prove it?
$endgroup$
– henceproved
Dec 6 '18 at 12:14
$begingroup$
Is the theorem well know, if so then can you point out to the source or else can you show me how to prove it?
$endgroup$
– henceproved
Dec 6 '18 at 12:14
$begingroup$
It's the Inverse Function Theorem in 1 dimension.
$endgroup$
– preferred_anon
Dec 6 '18 at 12:16
$begingroup$
It's the Inverse Function Theorem in 1 dimension.
$endgroup$
– preferred_anon
Dec 6 '18 at 12:16
add a comment |
$begingroup$
Consider $fcolon(-1,1)tomathbb R$ defined by $f(x)=tan(x)$ for rational $x$ and $f(x)=x$ otherwise. Now $f$ is continuous in $x_0=0$ with derivative $f'(0)=1$, but there doesn't exist a neighbourhood of $x_0$ where $f$ is monotonous.
All we can say here that there's a "punctual" monotony: We only can claim that for $x<0<y$ we have $f(x)<f(y)$.
answered Dec 5 '18 at 20:11
Michael HoppeMichael Hoppe
11.1k31837
$endgroup$
$begingroup$
The $f(x)$ that you construct is not a continuous function on any interval. My question loosely is if $f(x)$ is continuous in an interval, then does their exists an subinterval where $f(x)$ is monotonous.
$endgroup$
– henceproved
Dec 6 '18 at 0:11
add a comment |
$begingroup$
Consider $fcolon(-1,1)tomathbb R$ defined by $f(x)=tan(x)$ for rational $x$ and $f(x)=x$ otherwise. Now $f$ is continuous in $x_0=0$ with derivative $f'(0)=1$, but there doesn't exist a neighbourhood of $x_0$ where $f$ is monotonous.
All we can say here that there's a "punctual" monotony: We only can claim that for $x<0<y$ we have $f(x)<f(y)$.
answered Dec 5 '18 at 20:11
Michael HoppeMichael Hoppe
11.1k31837
$endgroup$
$begingroup$
The $f(x)$ that you construct is not a continuous function on any interval. My question loosely is if $f(x)$ is continuous in an interval, then does their exists an subinterval where $f(x)$ is monotonous.
$endgroup$
– henceproved
Dec 6 '18 at 0:11
add a comment |
$begingroup$
Consider $fcolon(-1,1)tomathbb R$ defined by $f(x)=tan(x)$ for rational $x$ and $f(x)=x$ otherwise. Now $f$ is continuous in $x_0=0$ with derivative $f'(0)=1$, but there doesn't exist a neighbourhood of $x_0$ where $f$ is monotonous.
All we can say here that there's a "punctual" monotony: We only can claim that for $x<0<y$ we have $f(x)<f(y)$.
answered Dec 5 '18 at 20:11
Michael HoppeMichael Hoppe
11.1k31837
$endgroup$
Consider $fcolon(-1,1)tomathbb R$ defined by $f(x)=tan(x)$ for rational $x$ and $f(x)=x$ otherwise. Now $f$ is continuous in $x_0=0$ with derivative $f'(0)=1$, but there doesn't exist a neighbourhood of $x_0$ where $f$ is monotonous.
All we can say here that there's a "punctual" monotony: We only can claim that for $x<0<y$ we have $f(x)<f(y)$.
answered Dec 5 '18 at 20:11
Michael HoppeMichael Hoppe
11.1k31837
answered Dec 5 '18 at 20:11
Michael HoppeMichael Hoppe
11.1k31837
answered Dec 5 '18 at 20:11
Michael HoppeMichael Hoppe
11.1k31837
answered Dec 5 '18 at 20:11
Michael HoppeMichael Hoppe
11.1k31837
11.1k31837
$begingroup$
The $f(x)$ that you construct is not a continuous function on any interval. My question loosely is if $f(x)$ is continuous in an interval, then does their exists an subinterval where $f(x)$ is monotonous.
$endgroup$
– henceproved
Dec 6 '18 at 0:11
add a comment |
$begingroup$
The $f(x)$ that you construct is not a continuous function on any interval. My question loosely is if $f(x)$ is continuous in an interval, then does their exists an subinterval where $f(x)$ is monotonous.
$endgroup$
– henceproved
Dec 6 '18 at 0:11
$begingroup$
The $f(x)$ that you construct is not a continuous function on any interval. My question loosely is if $f(x)$ is continuous in an interval, then does their exists an subinterval where $f(x)$ is monotonous.
$endgroup$
– henceproved
Dec 6 '18 at 0:11
$begingroup$
The $f(x)$ that you construct is not a continuous function on any interval. My question loosely is if $f(x)$ is continuous in an interval, then does their exists an subinterval where $f(x)$ is monotonous.
$endgroup$
– henceproved
Dec 6 '18 at 0:11
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027573%2finverse-function-like-theorem-for-monotoncity%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
"Open ball" doesn't have much sense over an one variable function. Probably you mean an "interval".
$endgroup$
– Rebellos
Dec 5 '18 at 20:02
$begingroup$
Consider $fcolon(-1,1)tomathbb R$ defined by $f(x)=tan(x)$ for rational $x$ and $f(x)=x$ otherwise. Now $f$ is continuous in $x_0=0$ with derivative $f'(0)=1$, but there doesn't exist a neighbourhood of $x_0$ where $f$ is monotonous.
$endgroup$
– Michael Hoppe
Dec 5 '18 at 20:06
$begingroup$
@MichaelHoppe I don't think $f(x)$ is continuous on an interval? It is only continuous at a point.
$endgroup$
– henceproved
Dec 5 '18 at 20:10
$begingroup$
Well, you asked for a function, such "$f(x)$ is continuous and has a non-zero derivative at $x_0$". I gave you one.
$endgroup$
– Michael Hoppe
Dec 5 '18 at 20:13
$begingroup$
For the "lemma" you want to derive, you may need a more special "smoothness" and characteristics in general of any functions regarded. Simple continuity won't do the trick.
$endgroup$
– Rebellos
Dec 5 '18 at 20:16