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
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You are right, the function is continuos everywhere except at $x = a$ ... Yet another textbook error :)
$endgroup$
– caverac
Jan 1 at 18:04
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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
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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
76.6k42866
$endgroup$
$begingroup$
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 |
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1 Answer
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$begingroup$
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
76.6k42866
$endgroup$
$begingroup$
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
76.6k42866
$endgroup$
$begingroup$
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
76.6k42866
$endgroup$
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
76.6k42866
answered Jan 1 at 18:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.6k42866
answered Jan 1 at 18:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.6k42866
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
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 |
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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