Did I make a mistake when finding the intervals this function is continous on?
$begingroup$
I was given a function f given by $f(x) =
begin{cases}
frac{x^2 - a^2}{x - a ;} textit{if} ; x neq a, \
2a ; textit{if} ; x=a \
end{cases}$,
and told to find the intervals over which $f$ is continous, in terms of $a$. I just figured it is continous everywhere except $a$, so the intervals would be $]-infty, a[$ and $]a, infty[$. However, my textbook gives the answer, without working, as $]-infty, -a[$ and $]-a, infty[$. I'm like 99% sure this is a mistake by the textbook author, not me, but continuty and the like is not my strong side, so I just want to make sure I didn't make a mistake here. Thanks in advance!
continuity piecewise-continuity
asked Jan 1 at 17:55
Christoffer Corfield AakreChristoffer Corfield Aakre
284
$endgroup$
|
show 5 more comments
$begingroup$
I was given a function f given by $f(x) =
begin{cases}
frac{x^2 - a^2}{x - a ;} textit{if} ; x neq a, \
2a ; textit{if} ; x=a \
end{cases}$,
and told to find the intervals over which $f$ is continous, in terms of $a$. I just figured it is continous everywhere except $a$, so the intervals would be $]-infty, a[$ and $]a, infty[$. However, my textbook gives the answer, without working, as $]-infty, -a[$ and $]-a, infty[$. I'm like 99% sure this is a mistake by the textbook author, not me, but continuty and the like is not my strong side, so I just want to make sure I didn't make a mistake here. Thanks in advance!
continuity piecewise-continuity
asked Jan 1 at 17:55
Christoffer Corfield AakreChristoffer Corfield Aakre
284
$endgroup$
1
$begingroup$
Yes, and I am giving the answer in terms of $a$.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 17:59
$begingroup$
You are right, the function is continuos everywhere except at $x = a$ ... Yet another textbook error :)
$endgroup$
– caverac
Jan 1 at 18:04
$begingroup$
Why is this not continuous at $a$?
$endgroup$
– Mark Bennet
Jan 1 at 18:04
$begingroup$
@MarkBennet it is not continous because $limlimits_{x to a}f(x)$ is undefined.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:06
$begingroup$
@caverac, thanks :) I can't accept your answer since you only posted a comment.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:07
|
show 5 more comments
$begingroup$
I was given a function f given by $f(x) =
begin{cases}
frac{x^2 - a^2}{x - a ;} textit{if} ; x neq a, \
2a ; textit{if} ; x=a \
end{cases}$,
and told to find the intervals over which $f$ is continous, in terms of $a$. I just figured it is continous everywhere except $a$, so the intervals would be $]-infty, a[$ and $]a, infty[$. However, my textbook gives the answer, without working, as $]-infty, -a[$ and $]-a, infty[$. I'm like 99% sure this is a mistake by the textbook author, not me, but continuty and the like is not my strong side, so I just want to make sure I didn't make a mistake here. Thanks in advance!
continuity piecewise-continuity
asked Jan 1 at 17:55
Christoffer Corfield AakreChristoffer Corfield Aakre
284
$endgroup$
I was given a function f given by $f(x) =
begin{cases}
frac{x^2 - a^2}{x - a ;} textit{if} ; x neq a, \
2a ; textit{if} ; x=a \
end{cases}$,
and told to find the intervals over which $f$ is continous, in terms of $a$. I just figured it is continous everywhere except $a$, so the intervals would be $]-infty, a[$ and $]a, infty[$. However, my textbook gives the answer, without working, as $]-infty, -a[$ and $]-a, infty[$. I'm like 99% sure this is a mistake by the textbook author, not me, but continuty and the like is not my strong side, so I just want to make sure I didn't make a mistake here. Thanks in advance!
continuity piecewise-continuity
continuity piecewise-continuity
asked Jan 1 at 17:55
Christoffer Corfield AakreChristoffer Corfield Aakre
284
asked Jan 1 at 17:55
Christoffer Corfield AakreChristoffer Corfield Aakre
284
asked Jan 1 at 17:55
Christoffer Corfield AakreChristoffer Corfield Aakre
284
asked Jan 1 at 17:55
Christoffer Corfield AakreChristoffer Corfield Aakre
284
asked Jan 1 at 17:55
Christoffer Corfield AakreChristoffer Corfield Aakre
284
284
1
$begingroup$
Yes, and I am giving the answer in terms of $a$.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 17:59
$begingroup$
You are right, the function is continuos everywhere except at $x = a$ ... Yet another textbook error :)
$endgroup$
– caverac
Jan 1 at 18:04
$begingroup$
Why is this not continuous at $a$?
$endgroup$
– Mark Bennet
Jan 1 at 18:04
$begingroup$
@MarkBennet it is not continous because $limlimits_{x to a}f(x)$ is undefined.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:06
$begingroup$
@caverac, thanks :) I can't accept your answer since you only posted a comment.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:07
|
show 5 more comments
1
$begingroup$
Yes, and I am giving the answer in terms of $a$.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 17:59
$begingroup$
You are right, the function is continuos everywhere except at $x = a$ ... Yet another textbook error :)
$endgroup$
– caverac
Jan 1 at 18:04
$begingroup$
Why is this not continuous at $a$?
$endgroup$
– Mark Bennet
Jan 1 at 18:04
$begingroup$
@MarkBennet it is not continous because $limlimits_{x to a}f(x)$ is undefined.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:06
$begingroup$
@caverac, thanks :) I can't accept your answer since you only posted a comment.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:07
1
1
$begingroup$
Yes, and I am giving the answer in terms of $a$.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 17:59
$begingroup$
Yes, and I am giving the answer in terms of $a$.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 17:59
$begingroup$
You are right, the function is continuos everywhere except at $x = a$ ... Yet another textbook error :)
$endgroup$
– caverac
Jan 1 at 18:04
$begingroup$
You are right, the function is continuos everywhere except at $x = a$ ... Yet another textbook error :)
$endgroup$
– caverac
Jan 1 at 18:04
$begingroup$
Why is this not continuous at $a$?
$endgroup$
– Mark Bennet
Jan 1 at 18:04
$begingroup$
Why is this not continuous at $a$?
$endgroup$
– Mark Bennet
Jan 1 at 18:04
$begingroup$
@MarkBennet it is not continous because $limlimits_{x to a}f(x)$ is undefined.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:06
$begingroup$
@MarkBennet it is not continous because $limlimits_{x to a}f(x)$ is undefined.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:06
$begingroup$
@caverac, thanks :) I can't accept your answer since you only posted a comment.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:07
$begingroup$
@caverac, thanks :) I can't accept your answer since you only posted a comment.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:07
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
It is $$frac{x^2-a^2}{x-a}=x+a$$ if $xneq$ $a$ and
$$lim_{xto a}frac{x^2-a^2}{x-a}=2a$$ thus our function is continous for all real $x$
answered Jan 1 at 18:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.2k42865
$endgroup$
$begingroup$
I think this is the right answer. I just made a mistake earlier when evaluating $limlimits_{x to a}f(x)$, but now I agree it is contionus for all $xinmathbb{R}$
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:14
1
$begingroup$
I think the given solution is wrong(in your textbook)
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 18:18
$begingroup$
Yeah I figured. Thanks a lot.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:21
$begingroup$
To the proposer: $f(x)=x+a$ for all $x$. Clearly this is continuous for all $x.$
$endgroup$
– DanielWainfleet
Jan 1 at 18:43
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%2f3058699%2fdid-i-make-a-mistake-when-finding-the-intervals-this-function-is-continous-on%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
$begingroup$
It is $$frac{x^2-a^2}{x-a}=x+a$$ if $xneq$ $a$ and
$$lim_{xto a}frac{x^2-a^2}{x-a}=2a$$ thus our function is continous for all real $x$
answered Jan 1 at 18:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.2k42865
$endgroup$
$begingroup$
I think this is the right answer. I just made a mistake earlier when evaluating $limlimits_{x to a}f(x)$, but now I agree it is contionus for all $xinmathbb{R}$
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:14
1
$begingroup$
I think the given solution is wrong(in your textbook)
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 18:18
$begingroup$
Yeah I figured. Thanks a lot.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:21
$begingroup$
To the proposer: $f(x)=x+a$ for all $x$. Clearly this is continuous for all $x.$
$endgroup$
– DanielWainfleet
Jan 1 at 18:43
add a comment |
$begingroup$
It is $$frac{x^2-a^2}{x-a}=x+a$$ if $xneq$ $a$ and
$$lim_{xto a}frac{x^2-a^2}{x-a}=2a$$ thus our function is continous for all real $x$
answered Jan 1 at 18:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.2k42865
$endgroup$
$begingroup$
I think this is the right answer. I just made a mistake earlier when evaluating $limlimits_{x to a}f(x)$, but now I agree it is contionus for all $xinmathbb{R}$
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:14
1
$begingroup$
I think the given solution is wrong(in your textbook)
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 18:18
$begingroup$
Yeah I figured. Thanks a lot.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:21
$begingroup$
To the proposer: $f(x)=x+a$ for all $x$. Clearly this is continuous for all $x.$
$endgroup$
– DanielWainfleet
Jan 1 at 18:43
add a comment |
$begingroup$
It is $$frac{x^2-a^2}{x-a}=x+a$$ if $xneq$ $a$ and
$$lim_{xto a}frac{x^2-a^2}{x-a}=2a$$ thus our function is continous for all real $x$
answered Jan 1 at 18:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.2k42865
$endgroup$
It is $$frac{x^2-a^2}{x-a}=x+a$$ if $xneq$ $a$ and
$$lim_{xto a}frac{x^2-a^2}{x-a}=2a$$ thus our function is continous for all real $x$
answered Jan 1 at 18:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.2k42865
answered Jan 1 at 18:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.2k42865
answered Jan 1 at 18:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.2k42865
answered Jan 1 at 18:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.2k42865
74.2k42865
$begingroup$
I think this is the right answer. I just made a mistake earlier when evaluating $limlimits_{x to a}f(x)$, but now I agree it is contionus for all $xinmathbb{R}$
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:14
1
$begingroup$
I think the given solution is wrong(in your textbook)
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 18:18
$begingroup$
Yeah I figured. Thanks a lot.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:21
$begingroup$
To the proposer: $f(x)=x+a$ for all $x$. Clearly this is continuous for all $x.$
$endgroup$
– DanielWainfleet
Jan 1 at 18:43
add a comment |
$begingroup$
I think this is the right answer. I just made a mistake earlier when evaluating $limlimits_{x to a}f(x)$, but now I agree it is contionus for all $xinmathbb{R}$
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:14
1
$begingroup$
I think the given solution is wrong(in your textbook)
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 18:18
$begingroup$
Yeah I figured. Thanks a lot.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:21
$begingroup$
To the proposer: $f(x)=x+a$ for all $x$. Clearly this is continuous for all $x.$
$endgroup$
– DanielWainfleet
Jan 1 at 18:43
$begingroup$
I think this is the right answer. I just made a mistake earlier when evaluating $limlimits_{x to a}f(x)$, but now I agree it is contionus for all $xinmathbb{R}$
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:14
$begingroup$
I think this is the right answer. I just made a mistake earlier when evaluating $limlimits_{x to a}f(x)$, but now I agree it is contionus for all $xinmathbb{R}$
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:14
1
1
$begingroup$
I think the given solution is wrong(in your textbook)
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 18:18
$begingroup$
I think the given solution is wrong(in your textbook)
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 18:18
$begingroup$
Yeah I figured. Thanks a lot.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:21
$begingroup$
Yeah I figured. Thanks a lot.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:21
$begingroup$
To the proposer: $f(x)=x+a$ for all $x$. Clearly this is continuous for all $x.$
$endgroup$
– DanielWainfleet
Jan 1 at 18:43
$begingroup$
To the proposer: $f(x)=x+a$ for all $x$. Clearly this is continuous for all $x.$
$endgroup$
– DanielWainfleet
Jan 1 at 18:43
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%2f3058699%2fdid-i-make-a-mistake-when-finding-the-intervals-this-function-is-continous-on%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, and I am giving the answer in terms of $a$.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 17:59
$begingroup$
You are right, the function is continuos everywhere except at $x = a$ ... Yet another textbook error :)
$endgroup$
– caverac
Jan 1 at 18:04
$begingroup$
Why is this not continuous at $a$?
$endgroup$
– Mark Bennet
Jan 1 at 18:04
$begingroup$
@MarkBennet it is not continous because $limlimits_{x to a}f(x)$ is undefined.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:06
$begingroup$
@caverac, thanks :) I can't accept your answer since you only posted a comment.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:07