Why do mean value theorems have open interval for differentiablity while closed for continuity?Why can we only talk about derivatives on an open interval?Does $f~'$ have to be continuous for application of Mean Value Theorem?Mean Value Theorem Proofing For all $ x>0$Mean-value theorems (other than Cauchy's, Lagrange's or Rolle's)Mean value theorem for vector valued function (not integral form)Why do we call this theorem in vector valued functions “mean value theorem”?Why does the Mean Value Theorem require a closed interval for continuity and an open interval for differentiability?Usage of mean value theorem ; bounded derivative and open intervalPresenting a lecture on Mean value theorems.How to use the Mean-Value Theorem on a function continuous over an open interval.
Possible isometry groups of open manifolds
What exactly is the Tension force?
Why didn't Al Powell investigate the lights at the top of the building?
What is this old "lemon-squeezer" shaped pan
Won 50K! Now what should I do with it
Is this more than a packing puzzle?
Commutator subgroup of Heisenberg group.
I quit, and boss offered me 3 month "grace period" where I could still come back
Adding a vertical line at the right end of the horizontal line in frac
Alternatives to using writing paper for writing practice
How to fit a linear model in the Bayesian way in Mathematica?
Connect neutrals together in 3-gang box (load side) with 3x 3-way switches?
Crab Nebula short story from 1960s or '70s
Construct a pentagon avoiding compass use
Why do they not say "The Baby"
Why linear regression uses "vertical" distance to the best-fit-line, instead of actual distance?
Could the crash sites of the Apollo 11 and 16 LMs be seen by the LRO?
Would letting a multiclass character rebuild their character to be single-classed be game-breaking?
Remove intersect line for one circle using venndiagram2sets
Is killing off one of my queer characters homophobic?
Why do mean value theorems have open interval for differentiablity while closed for continuity?
How did Southern slaveholders in the United States relate to the Caribbean and Latin America?
Getting fresh water in the middle of hypersaline lake in the Bronze Age
Asking for higher salary after I increased my initial figure
Why do mean value theorems have open interval for differentiablity while closed for continuity?
Why can we only talk about derivatives on an open interval?Does $f~'$ have to be continuous for application of Mean Value Theorem?Mean Value Theorem Proofing For all $ x>0$Mean-value theorems (other than Cauchy's, Lagrange's or Rolle's)Mean value theorem for vector valued function (not integral form)Why do we call this theorem in vector valued functions “mean value theorem”?Why does the Mean Value Theorem require a closed interval for continuity and an open interval for differentiability?Usage of mean value theorem ; bounded derivative and open intervalPresenting a lecture on Mean value theorems.How to use the Mean-Value Theorem on a function continuous over an open interval.
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
For mean value theorems like Lagrange and rolles, we have the condition as
For applying mean value theorem to any function $f(x)$ for the domain $[a,b]$ , it should be
(1) continuous in $[a,b]$
(2) differentiable in $(a,b)$
So why is it that for the criteria of differentiablity, we have the open interval ?? Is it possible for a function differentiable in $(a,b) $ and continuous in $[a,b]$ to be non- differentable at the end points.
Also why is the first statement needed ?? Doesn't the second statement of differentiablity also mean that the function is continuous ??
I'm not very experienced in calculus and still in highschool, so it might be something too obvious I'm missing , please help :)
calculus differential
asked 8 hours ago
DivMitDivMit
5781 silver badge10 bronze badges
$endgroup$
add a comment |
$begingroup$
For mean value theorems like Lagrange and rolles, we have the condition as
For applying mean value theorem to any function $f(x)$ for the domain $[a,b]$ , it should be
(1) continuous in $[a,b]$
(2) differentiable in $(a,b)$
So why is it that for the criteria of differentiablity, we have the open interval ?? Is it possible for a function differentiable in $(a,b) $ and continuous in $[a,b]$ to be non- differentable at the end points.
Also why is the first statement needed ?? Doesn't the second statement of differentiablity also mean that the function is continuous ??
I'm not very experienced in calculus and still in highschool, so it might be something too obvious I'm missing , please help :)
calculus differential
asked 8 hours ago
DivMitDivMit
5781 silver badge10 bronze badges
$endgroup$
1
$begingroup$
Yes it is possible for a function differentiable in $(a,b)$ and continuous in $[a,b]$ to be non-differentiable at the end points. With that encouragement, would you care to seek examples on your own? And in the last paragraph, note that differentiability on the interior interval $(a,b)$ implies continuity there, but fails to imply (one-sided) continuity at the endpoints. Again, perhaps you will want to think about how weakening the assumptions (removing continuity at the endpoints) opens the door to functions that do not satisfy the MVT.
$endgroup$
– hardmath
8 hours ago
add a comment |
$begingroup$
For mean value theorems like Lagrange and rolles, we have the condition as
For applying mean value theorem to any function $f(x)$ for the domain $[a,b]$ , it should be
(1) continuous in $[a,b]$
(2) differentiable in $(a,b)$
So why is it that for the criteria of differentiablity, we have the open interval ?? Is it possible for a function differentiable in $(a,b) $ and continuous in $[a,b]$ to be non- differentable at the end points.
Also why is the first statement needed ?? Doesn't the second statement of differentiablity also mean that the function is continuous ??
I'm not very experienced in calculus and still in highschool, so it might be something too obvious I'm missing , please help :)
calculus differential
asked 8 hours ago
DivMitDivMit
5781 silver badge10 bronze badges
$endgroup$
For mean value theorems like Lagrange and rolles, we have the condition as
For applying mean value theorem to any function $f(x)$ for the domain $[a,b]$ , it should be
(1) continuous in $[a,b]$
(2) differentiable in $(a,b)$
So why is it that for the criteria of differentiablity, we have the open interval ?? Is it possible for a function differentiable in $(a,b) $ and continuous in $[a,b]$ to be non- differentable at the end points.
Also why is the first statement needed ?? Doesn't the second statement of differentiablity also mean that the function is continuous ??
I'm not very experienced in calculus and still in highschool, so it might be something too obvious I'm missing , please help :)
calculus differential
calculus differential
asked 8 hours ago
DivMitDivMit
5781 silver badge10 bronze badges
asked 8 hours ago
DivMitDivMit
5781 silver badge10 bronze badges
edited 18 mins ago
DivMit
asked 8 hours ago
DivMitDivMit
5781 silver badge10 bronze badges
asked 8 hours ago
DivMitDivMit
5781 silver badge10 bronze badges
asked 8 hours ago
DivMitDivMit
5781 silver badge10 bronze badges
5781 silver badge10 bronze badges
1
$begingroup$
Yes it is possible for a function differentiable in $(a,b)$ and continuous in $[a,b]$ to be non-differentiable at the end points. With that encouragement, would you care to seek examples on your own? And in the last paragraph, note that differentiability on the interior interval $(a,b)$ implies continuity there, but fails to imply (one-sided) continuity at the endpoints. Again, perhaps you will want to think about how weakening the assumptions (removing continuity at the endpoints) opens the door to functions that do not satisfy the MVT.
$endgroup$
– hardmath
8 hours ago
add a comment |
1
$begingroup$
Yes it is possible for a function differentiable in $(a,b)$ and continuous in $[a,b]$ to be non-differentiable at the end points. With that encouragement, would you care to seek examples on your own? And in the last paragraph, note that differentiability on the interior interval $(a,b)$ implies continuity there, but fails to imply (one-sided) continuity at the endpoints. Again, perhaps you will want to think about how weakening the assumptions (removing continuity at the endpoints) opens the door to functions that do not satisfy the MVT.
$endgroup$
– hardmath
8 hours ago
1
1
$begingroup$
Yes it is possible for a function differentiable in $(a,b)$ and continuous in $[a,b]$ to be non-differentiable at the end points. With that encouragement, would you care to seek examples on your own? And in the last paragraph, note that differentiability on the interior interval $(a,b)$ implies continuity there, but fails to imply (one-sided) continuity at the endpoints. Again, perhaps you will want to think about how weakening the assumptions (removing continuity at the endpoints) opens the door to functions that do not satisfy the MVT.
$endgroup$
– hardmath
8 hours ago
$begingroup$
Yes it is possible for a function differentiable in $(a,b)$ and continuous in $[a,b]$ to be non-differentiable at the end points. With that encouragement, would you care to seek examples on your own? And in the last paragraph, note that differentiability on the interior interval $(a,b)$ implies continuity there, but fails to imply (one-sided) continuity at the endpoints. Again, perhaps you will want to think about how weakening the assumptions (removing continuity at the endpoints) opens the door to functions that do not satisfy the MVT.
$endgroup$
– hardmath
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For example, the function $sqrtx(1-x)$ is continuous in $[0,1]$ but only differentiable on $(0,1)$.
On the other hand, $1/x$ is differentiable in $(0,infty)$ but is not continuous at $0$
(even if you define it at $0$).
answered 8 hours ago
Robert IsraelRobert Israel
342k23 gold badges234 silver badges495 bronze badges
$endgroup$
add a comment |
$begingroup$
Here are a couple of examples to think about:
$f(x) = sqrtx$ on the interval $[0,1]$. This is continuous on $[0,1]$ and differentiable on $(0,1)$, but not differentiable at 0 in any reasonable sense. Yet the mean value theorem applies to it.
$f(x) = begincases 0, & x = 0 \ 1, & x > 0. endcases$ This is differentiable on $(0,1)$ but the mean value theorem does not apply to it.
answered 8 hours ago
Nate EldredgeNate Eldredge
67.1k9 gold badges85 silver badges180 bronze badges
$endgroup$
add a comment |
Your Answer
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%2f3293097%2fwhy-do-mean-value-theorems-have-open-interval-for-differentiablity-while-closed%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
$begingroup$
For example, the function $sqrtx(1-x)$ is continuous in $[0,1]$ but only differentiable on $(0,1)$.
On the other hand, $1/x$ is differentiable in $(0,infty)$ but is not continuous at $0$
(even if you define it at $0$).
answered 8 hours ago
Robert IsraelRobert Israel
342k23 gold badges234 silver badges495 bronze badges
$endgroup$
add a comment |
$begingroup$
For example, the function $sqrtx(1-x)$ is continuous in $[0,1]$ but only differentiable on $(0,1)$.
On the other hand, $1/x$ is differentiable in $(0,infty)$ but is not continuous at $0$
(even if you define it at $0$).
answered 8 hours ago
Robert IsraelRobert Israel
342k23 gold badges234 silver badges495 bronze badges
$endgroup$
add a comment |
$begingroup$
For example, the function $sqrtx(1-x)$ is continuous in $[0,1]$ but only differentiable on $(0,1)$.
On the other hand, $1/x$ is differentiable in $(0,infty)$ but is not continuous at $0$
(even if you define it at $0$).
answered 8 hours ago
Robert IsraelRobert Israel
342k23 gold badges234 silver badges495 bronze badges
$endgroup$
For example, the function $sqrtx(1-x)$ is continuous in $[0,1]$ but only differentiable on $(0,1)$.
On the other hand, $1/x$ is differentiable in $(0,infty)$ but is not continuous at $0$
(even if you define it at $0$).
answered 8 hours ago
Robert IsraelRobert Israel
342k23 gold badges234 silver badges495 bronze badges
answered 8 hours ago
Robert IsraelRobert Israel
342k23 gold badges234 silver badges495 bronze badges
answered 8 hours ago
Robert IsraelRobert Israel
342k23 gold badges234 silver badges495 bronze badges
answered 8 hours ago
Robert IsraelRobert Israel
342k23 gold badges234 silver badges495 bronze badges
342k23 gold badges234 silver badges495 bronze badges
add a comment |
add a comment |
$begingroup$
Here are a couple of examples to think about:
$f(x) = sqrtx$ on the interval $[0,1]$. This is continuous on $[0,1]$ and differentiable on $(0,1)$, but not differentiable at 0 in any reasonable sense. Yet the mean value theorem applies to it.
$f(x) = begincases 0, & x = 0 \ 1, & x > 0. endcases$ This is differentiable on $(0,1)$ but the mean value theorem does not apply to it.
answered 8 hours ago
Nate EldredgeNate Eldredge
67.1k9 gold badges85 silver badges180 bronze badges
$endgroup$
add a comment |
$begingroup$
Here are a couple of examples to think about:
$f(x) = sqrtx$ on the interval $[0,1]$. This is continuous on $[0,1]$ and differentiable on $(0,1)$, but not differentiable at 0 in any reasonable sense. Yet the mean value theorem applies to it.
$f(x) = begincases 0, & x = 0 \ 1, & x > 0. endcases$ This is differentiable on $(0,1)$ but the mean value theorem does not apply to it.
answered 8 hours ago
Nate EldredgeNate Eldredge
67.1k9 gold badges85 silver badges180 bronze badges
$endgroup$
add a comment |
$begingroup$
Here are a couple of examples to think about:
$f(x) = sqrtx$ on the interval $[0,1]$. This is continuous on $[0,1]$ and differentiable on $(0,1)$, but not differentiable at 0 in any reasonable sense. Yet the mean value theorem applies to it.
$f(x) = begincases 0, & x = 0 \ 1, & x > 0. endcases$ This is differentiable on $(0,1)$ but the mean value theorem does not apply to it.
answered 8 hours ago
Nate EldredgeNate Eldredge
67.1k9 gold badges85 silver badges180 bronze badges
$endgroup$
Here are a couple of examples to think about:
$f(x) = sqrtx$ on the interval $[0,1]$. This is continuous on $[0,1]$ and differentiable on $(0,1)$, but not differentiable at 0 in any reasonable sense. Yet the mean value theorem applies to it.
$f(x) = begincases 0, & x = 0 \ 1, & x > 0. endcases$ This is differentiable on $(0,1)$ but the mean value theorem does not apply to it.
answered 8 hours ago
Nate EldredgeNate Eldredge
67.1k9 gold badges85 silver badges180 bronze badges
answered 8 hours ago
Nate EldredgeNate Eldredge
67.1k9 gold badges85 silver badges180 bronze badges
answered 8 hours ago
Nate EldredgeNate Eldredge
67.1k9 gold badges85 silver badges180 bronze badges
answered 8 hours ago
Nate EldredgeNate Eldredge
67.1k9 gold badges85 silver badges180 bronze badges
67.1k9 gold badges85 silver badges180 bronze badges
add a comment |
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%2f3293097%2fwhy-do-mean-value-theorems-have-open-interval-for-differentiablity-while-closed%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
1
$begingroup$
Yes it is possible for a function differentiable in $(a,b)$ and continuous in $[a,b]$ to be non-differentiable at the end points. With that encouragement, would you care to seek examples on your own? And in the last paragraph, note that differentiability on the interior interval $(a,b)$ implies continuity there, but fails to imply (one-sided) continuity at the endpoints. Again, perhaps you will want to think about how weakening the assumptions (removing continuity at the endpoints) opens the door to functions that do not satisfy the MVT.
$endgroup$
– hardmath
8 hours ago