Determining a functions differentiability
$begingroup$
If a function has a point of discontinuity, such that the slope of tangents at points before and after that point are equal, will the function be differentiable?
functions continuity graphing-functions
asked Dec 26 '18 at 19:33
user574937user574937
384
$endgroup$
add a comment |
$begingroup$
If a function has a point of discontinuity, such that the slope of tangents at points before and after that point are equal, will the function be differentiable?
functions continuity graphing-functions
asked Dec 26 '18 at 19:33
user574937user574937
384
$endgroup$
1
$begingroup$
differntiability (at a point) implies continuity (at that point)
$endgroup$
– Hagen von Eitzen
Dec 26 '18 at 19:37
add a comment |
$begingroup$
If a function has a point of discontinuity, such that the slope of tangents at points before and after that point are equal, will the function be differentiable?
functions continuity graphing-functions
asked Dec 26 '18 at 19:33
user574937user574937
384
$endgroup$
If a function has a point of discontinuity, such that the slope of tangents at points before and after that point are equal, will the function be differentiable?
functions continuity graphing-functions
functions continuity graphing-functions
asked Dec 26 '18 at 19:33
user574937user574937
384
asked Dec 26 '18 at 19:33
user574937user574937
384
asked Dec 26 '18 at 19:33
user574937user574937
384
asked Dec 26 '18 at 19:33
user574937user574937
384
asked Dec 26 '18 at 19:33
user574937user574937
384
384
1
$begingroup$
differntiability (at a point) implies continuity (at that point)
$endgroup$
– Hagen von Eitzen
Dec 26 '18 at 19:37
add a comment |
1
$begingroup$
differntiability (at a point) implies continuity (at that point)
$endgroup$
– Hagen von Eitzen
Dec 26 '18 at 19:37
1
1
$begingroup$
differntiability (at a point) implies continuity (at that point)
$endgroup$
– Hagen von Eitzen
Dec 26 '18 at 19:37
$begingroup$
differntiability (at a point) implies continuity (at that point)
$endgroup$
– Hagen von Eitzen
Dec 26 '18 at 19:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that
$$lim_{hto 0} frac {f(x+h)-f(x)}{h} =f'(x) $$
$$implies lim _{hto 0} {f(x+h)-f(x)} =lim _{hto 0}h f'(x)=0$$
That is, if the function is differentiable at $x$ then it is also continuous at $x$.
answered Dec 26 '18 at 19:52
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
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%2f3053242%2fdetermining-a-functions-differentiability%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that
$$lim_{hto 0} frac {f(x+h)-f(x)}{h} =f'(x) $$
$$implies lim _{hto 0} {f(x+h)-f(x)} =lim _{hto 0}h f'(x)=0$$
That is, if the function is differentiable at $x$ then it is also continuous at $x$.
answered Dec 26 '18 at 19:52
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
Note that
$$lim_{hto 0} frac {f(x+h)-f(x)}{h} =f'(x) $$
$$implies lim _{hto 0} {f(x+h)-f(x)} =lim _{hto 0}h f'(x)=0$$
That is, if the function is differentiable at $x$ then it is also continuous at $x$.
answered Dec 26 '18 at 19:52
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
Note that
$$lim_{hto 0} frac {f(x+h)-f(x)}{h} =f'(x) $$
$$implies lim _{hto 0} {f(x+h)-f(x)} =lim _{hto 0}h f'(x)=0$$
That is, if the function is differentiable at $x$ then it is also continuous at $x$.
answered Dec 26 '18 at 19:52
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
Note that
$$lim_{hto 0} frac {f(x+h)-f(x)}{h} =f'(x) $$
$$implies lim _{hto 0} {f(x+h)-f(x)} =lim _{hto 0}h f'(x)=0$$
That is, if the function is differentiable at $x$ then it is also continuous at $x$.
answered Dec 26 '18 at 19:52
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 26 '18 at 19:52
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 26 '18 at 19:52
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 26 '18 at 19:52
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
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%2f3053242%2fdetermining-a-functions-differentiability%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
$begingroup$
differntiability (at a point) implies continuity (at that point)
$endgroup$
– Hagen von Eitzen
Dec 26 '18 at 19:37