Did I make a mistake when finding the intervals this function is continous on?












3












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















3












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














3












3








3





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















4












$begingroup$

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 $x$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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
});


}
});














draft saved

draft discarded


















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









4












$begingroup$

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 $x$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















4












$begingroup$

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 $x$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














4












4








4





$begingroup$

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 $x$






share|cite|improve this answer









$endgroup$



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 $x$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 1 at 18:10









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.6k42866




76.6k42866












  • $begingroup$
    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








  • 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


















draft saved

draft discarded




















































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.




draft saved


draft discarded














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





















































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







-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more