But what is a continuous function?
$begingroup$
I have a very basic problem. I am confused about "continuous function" term.
What really is a continuous function? A function that is continuous for all of its domain or for all real numbers?
Let's say:
$ln|x|$ - the graph clearly says it's continuous for all real numbers except for $0$ which is not part of the domain. So is this function continuous or not? I could say same about $tan{x}$ or $frac{x+1}{x}$
And also what about:
$ln{x}$ - the graph clearly says it's continuous for all of its domain: $(0; infty)$ - so is this $f$ continuous or not?
Thanks for clarification.
real-analysis algebra-precalculus limits continuity
asked 6 hours ago
wenoweno
1739
$endgroup$
add a comment |
$begingroup$
I have a very basic problem. I am confused about "continuous function" term.
What really is a continuous function? A function that is continuous for all of its domain or for all real numbers?
Let's say:
$ln|x|$ - the graph clearly says it's continuous for all real numbers except for $0$ which is not part of the domain. So is this function continuous or not? I could say same about $tan{x}$ or $frac{x+1}{x}$
And also what about:
$ln{x}$ - the graph clearly says it's continuous for all of its domain: $(0; infty)$ - so is this $f$ continuous or not?
Thanks for clarification.
real-analysis algebra-precalculus limits continuity
asked 6 hours ago
wenoweno
1739
$endgroup$
$begingroup$
If a function is not defined at a point, then it certainly cannot be continuous at it. Is this your question?
$endgroup$
– LoveTooNap29
5 hours ago
$begingroup$
Not really. Question is: Is function $f(x) = ln{|x|}$ continuous? I assume it is?
$endgroup$
– weno
4 hours ago
1
$begingroup$
continuous on what set? As others have pointed out: continuity is a property of functions either at points or on sets but not a property of the function alone without regard to the domain. $log |x|$ is certainly continuous where it is defined. Is that clear?
$endgroup$
– LoveTooNap29
4 hours ago
2
$begingroup$
functions are continuous at specific points. For example: $f(x) = (x-2)^2; x < 0$ and $f(x) = 1$ if $0 le x < 1$ and $f(x) = sqrt x$ for $x ge 1$. Is continuous at $x = -7$ and $x = pi$ but is not continuous and $x = 0$. To say a function is "continuous" is short hand to mean in is continuous on every point in it's domain. It makes utterly no sense to talk of a function being continuous on a point not in its domain.
$endgroup$
– fleablood
3 hours ago
2
$begingroup$
That said some people will "abuse terminology" and say intuitive but wrong things such as "$f(x) = frac 1x$ is discontinuous at $x = 0$" this is... well, it's simply wrong.
$endgroup$
– fleablood
3 hours ago
add a comment |
$begingroup$
I have a very basic problem. I am confused about "continuous function" term.
What really is a continuous function? A function that is continuous for all of its domain or for all real numbers?
Let's say:
$ln|x|$ - the graph clearly says it's continuous for all real numbers except for $0$ which is not part of the domain. So is this function continuous or not? I could say same about $tan{x}$ or $frac{x+1}{x}$
And also what about:
$ln{x}$ - the graph clearly says it's continuous for all of its domain: $(0; infty)$ - so is this $f$ continuous or not?
Thanks for clarification.
real-analysis algebra-precalculus limits continuity
asked 6 hours ago
wenoweno
1739
$endgroup$
I have a very basic problem. I am confused about "continuous function" term.
What really is a continuous function? A function that is continuous for all of its domain or for all real numbers?
Let's say:
$ln|x|$ - the graph clearly says it's continuous for all real numbers except for $0$ which is not part of the domain. So is this function continuous or not? I could say same about $tan{x}$ or $frac{x+1}{x}$
And also what about:
$ln{x}$ - the graph clearly says it's continuous for all of its domain: $(0; infty)$ - so is this $f$ continuous or not?
Thanks for clarification.
real-analysis algebra-precalculus limits continuity
real-analysis algebra-precalculus limits continuity
asked 6 hours ago
wenoweno
1739
asked 6 hours ago
wenoweno
1739
asked 6 hours ago
wenoweno
1739
asked 6 hours ago
wenoweno
1739
asked 6 hours ago
wenoweno
1739
1739
$begingroup$
If a function is not defined at a point, then it certainly cannot be continuous at it. Is this your question?
$endgroup$
– LoveTooNap29
5 hours ago
$begingroup$
Not really. Question is: Is function $f(x) = ln{|x|}$ continuous? I assume it is?
$endgroup$
– weno
4 hours ago
1
$begingroup$
continuous on what set? As others have pointed out: continuity is a property of functions either at points or on sets but not a property of the function alone without regard to the domain. $log |x|$ is certainly continuous where it is defined. Is that clear?
$endgroup$
– LoveTooNap29
4 hours ago
2
$begingroup$
functions are continuous at specific points. For example: $f(x) = (x-2)^2; x < 0$ and $f(x) = 1$ if $0 le x < 1$ and $f(x) = sqrt x$ for $x ge 1$. Is continuous at $x = -7$ and $x = pi$ but is not continuous and $x = 0$. To say a function is "continuous" is short hand to mean in is continuous on every point in it's domain. It makes utterly no sense to talk of a function being continuous on a point not in its domain.
$endgroup$
– fleablood
3 hours ago
2
$begingroup$
That said some people will "abuse terminology" and say intuitive but wrong things such as "$f(x) = frac 1x$ is discontinuous at $x = 0$" this is... well, it's simply wrong.
$endgroup$
– fleablood
3 hours ago
add a comment |
$begingroup$
If a function is not defined at a point, then it certainly cannot be continuous at it. Is this your question?
$endgroup$
– LoveTooNap29
5 hours ago
$begingroup$
Not really. Question is: Is function $f(x) = ln{|x|}$ continuous? I assume it is?
$endgroup$
– weno
4 hours ago
1
$begingroup$
continuous on what set? As others have pointed out: continuity is a property of functions either at points or on sets but not a property of the function alone without regard to the domain. $log |x|$ is certainly continuous where it is defined. Is that clear?
$endgroup$
– LoveTooNap29
4 hours ago
2
$begingroup$
functions are continuous at specific points. For example: $f(x) = (x-2)^2; x < 0$ and $f(x) = 1$ if $0 le x < 1$ and $f(x) = sqrt x$ for $x ge 1$. Is continuous at $x = -7$ and $x = pi$ but is not continuous and $x = 0$. To say a function is "continuous" is short hand to mean in is continuous on every point in it's domain. It makes utterly no sense to talk of a function being continuous on a point not in its domain.
$endgroup$
– fleablood
3 hours ago
2
$begingroup$
That said some people will "abuse terminology" and say intuitive but wrong things such as "$f(x) = frac 1x$ is discontinuous at $x = 0$" this is... well, it's simply wrong.
$endgroup$
– fleablood
3 hours ago
$begingroup$
If a function is not defined at a point, then it certainly cannot be continuous at it. Is this your question?
$endgroup$
– LoveTooNap29
5 hours ago
$begingroup$
If a function is not defined at a point, then it certainly cannot be continuous at it. Is this your question?
$endgroup$
– LoveTooNap29
5 hours ago
$begingroup$
Not really. Question is: Is function $f(x) = ln{|x|}$ continuous? I assume it is?
$endgroup$
– weno
4 hours ago
$begingroup$
Not really. Question is: Is function $f(x) = ln{|x|}$ continuous? I assume it is?
$endgroup$
– weno
4 hours ago
1
1
$begingroup$
continuous on what set? As others have pointed out: continuity is a property of functions either at points or on sets but not a property of the function alone without regard to the domain. $log |x|$ is certainly continuous where it is defined. Is that clear?
$endgroup$
– LoveTooNap29
4 hours ago
$begingroup$
continuous on what set? As others have pointed out: continuity is a property of functions either at points or on sets but not a property of the function alone without regard to the domain. $log |x|$ is certainly continuous where it is defined. Is that clear?
$endgroup$
– LoveTooNap29
4 hours ago
2
2
$begingroup$
functions are continuous at specific points. For example: $f(x) = (x-2)^2; x < 0$ and $f(x) = 1$ if $0 le x < 1$ and $f(x) = sqrt x$ for $x ge 1$. Is continuous at $x = -7$ and $x = pi$ but is not continuous and $x = 0$. To say a function is "continuous" is short hand to mean in is continuous on every point in it's domain. It makes utterly no sense to talk of a function being continuous on a point not in its domain.
$endgroup$
– fleablood
3 hours ago
$begingroup$
functions are continuous at specific points. For example: $f(x) = (x-2)^2; x < 0$ and $f(x) = 1$ if $0 le x < 1$ and $f(x) = sqrt x$ for $x ge 1$. Is continuous at $x = -7$ and $x = pi$ but is not continuous and $x = 0$. To say a function is "continuous" is short hand to mean in is continuous on every point in it's domain. It makes utterly no sense to talk of a function being continuous on a point not in its domain.
$endgroup$
– fleablood
3 hours ago
2
2
$begingroup$
That said some people will "abuse terminology" and say intuitive but wrong things such as "$f(x) = frac 1x$ is discontinuous at $x = 0$" this is... well, it's simply wrong.
$endgroup$
– fleablood
3 hours ago
$begingroup$
That said some people will "abuse terminology" and say intuitive but wrong things such as "$f(x) = frac 1x$ is discontinuous at $x = 0$" this is... well, it's simply wrong.
$endgroup$
– fleablood
3 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Mathematicians (but not all calculus books) mean "continuous at every point of its domain" when they say a function is "continuous." The functions $f(x) = 1/x$ and $f(x)=ln x$ are continuous functions.
answered 4 hours ago
Ted ShifrinTed Shifrin
63.4k44489
$endgroup$
add a comment |
$begingroup$
"Continuous" is not, in and of itself, a property of a function. You have to talk about being continuous at a given point, or on a collection of points as you have above.
It is generally safe to assume that if somebody leaves off the set, they intend to say that the function is continuous on its domain (as both of your examples are); but, I tend to believe that explicit is better than implicit.
answered 6 hours ago
Nick PetersonNick Peterson
26.5k23961
$endgroup$
add a comment |
$begingroup$
The exact answer depends on your chosen definition of "function" (there is more than one). For most uses, a function is regarded as being continuous on an interval $(a,b)$ if for every number $c$ in $(a,b)$, $f(x)=lim_{xto c} f(x)$.
In your example $f(x)=ln{x}$ is continuous on the interval $(0,infty)$ and either undefined or complex/multivalued everywhere else, depending on whether you consider the codomain (range) of $f$ to include the complex numbers or not.
In other words, no function is ever just 'continuous' - it is continuous within an interval (which may or may not be its domain).
answered 6 hours ago
R. BurtonR. Burton
50819
$endgroup$
$begingroup$
No. Not for most uses, a function is regarded as being continuous on an interval.
$endgroup$
– Math_QED
5 hours ago
$begingroup$
There is a case to be made for 'pac-man' continuity via quotient space. Also, not every function is a function of a single, real variable, so the notion of "interval" may not always apply.
$endgroup$
– R. Burton
5 hours ago
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Mathematicians (but not all calculus books) mean "continuous at every point of its domain" when they say a function is "continuous." The functions $f(x) = 1/x$ and $f(x)=ln x$ are continuous functions.
answered 4 hours ago
Ted ShifrinTed Shifrin
63.4k44489
$endgroup$
add a comment |
$begingroup$
Mathematicians (but not all calculus books) mean "continuous at every point of its domain" when they say a function is "continuous." The functions $f(x) = 1/x$ and $f(x)=ln x$ are continuous functions.
answered 4 hours ago
Ted ShifrinTed Shifrin
63.4k44489
$endgroup$
add a comment |
$begingroup$
Mathematicians (but not all calculus books) mean "continuous at every point of its domain" when they say a function is "continuous." The functions $f(x) = 1/x$ and $f(x)=ln x$ are continuous functions.
answered 4 hours ago
Ted ShifrinTed Shifrin
63.4k44489
$endgroup$
Mathematicians (but not all calculus books) mean "continuous at every point of its domain" when they say a function is "continuous." The functions $f(x) = 1/x$ and $f(x)=ln x$ are continuous functions.
answered 4 hours ago
Ted ShifrinTed Shifrin
63.4k44489
answered 4 hours ago
Ted ShifrinTed Shifrin
63.4k44489
answered 4 hours ago
Ted ShifrinTed Shifrin
63.4k44489
answered 4 hours ago
Ted ShifrinTed Shifrin
63.4k44489
63.4k44489
add a comment |
add a comment |
$begingroup$
"Continuous" is not, in and of itself, a property of a function. You have to talk about being continuous at a given point, or on a collection of points as you have above.
It is generally safe to assume that if somebody leaves off the set, they intend to say that the function is continuous on its domain (as both of your examples are); but, I tend to believe that explicit is better than implicit.
answered 6 hours ago
Nick PetersonNick Peterson
26.5k23961
$endgroup$
add a comment |
$begingroup$
"Continuous" is not, in and of itself, a property of a function. You have to talk about being continuous at a given point, or on a collection of points as you have above.
It is generally safe to assume that if somebody leaves off the set, they intend to say that the function is continuous on its domain (as both of your examples are); but, I tend to believe that explicit is better than implicit.
answered 6 hours ago
Nick PetersonNick Peterson
26.5k23961
$endgroup$
add a comment |
$begingroup$
"Continuous" is not, in and of itself, a property of a function. You have to talk about being continuous at a given point, or on a collection of points as you have above.
It is generally safe to assume that if somebody leaves off the set, they intend to say that the function is continuous on its domain (as both of your examples are); but, I tend to believe that explicit is better than implicit.
answered 6 hours ago
Nick PetersonNick Peterson
26.5k23961
$endgroup$
"Continuous" is not, in and of itself, a property of a function. You have to talk about being continuous at a given point, or on a collection of points as you have above.
It is generally safe to assume that if somebody leaves off the set, they intend to say that the function is continuous on its domain (as both of your examples are); but, I tend to believe that explicit is better than implicit.
answered 6 hours ago
Nick PetersonNick Peterson
26.5k23961
answered 6 hours ago
Nick PetersonNick Peterson
26.5k23961
answered 6 hours ago
Nick PetersonNick Peterson
26.5k23961
answered 6 hours ago
Nick PetersonNick Peterson
26.5k23961
26.5k23961
add a comment |
add a comment |
$begingroup$
The exact answer depends on your chosen definition of "function" (there is more than one). For most uses, a function is regarded as being continuous on an interval $(a,b)$ if for every number $c$ in $(a,b)$, $f(x)=lim_{xto c} f(x)$.
In your example $f(x)=ln{x}$ is continuous on the interval $(0,infty)$ and either undefined or complex/multivalued everywhere else, depending on whether you consider the codomain (range) of $f$ to include the complex numbers or not.
In other words, no function is ever just 'continuous' - it is continuous within an interval (which may or may not be its domain).
answered 6 hours ago
R. BurtonR. Burton
50819
$endgroup$
$begingroup$
No. Not for most uses, a function is regarded as being continuous on an interval.
$endgroup$
– Math_QED
5 hours ago
$begingroup$
There is a case to be made for 'pac-man' continuity via quotient space. Also, not every function is a function of a single, real variable, so the notion of "interval" may not always apply.
$endgroup$
– R. Burton
5 hours ago
add a comment |
$begingroup$
The exact answer depends on your chosen definition of "function" (there is more than one). For most uses, a function is regarded as being continuous on an interval $(a,b)$ if for every number $c$ in $(a,b)$, $f(x)=lim_{xto c} f(x)$.
In your example $f(x)=ln{x}$ is continuous on the interval $(0,infty)$ and either undefined or complex/multivalued everywhere else, depending on whether you consider the codomain (range) of $f$ to include the complex numbers or not.
In other words, no function is ever just 'continuous' - it is continuous within an interval (which may or may not be its domain).
answered 6 hours ago
R. BurtonR. Burton
50819
$endgroup$
$begingroup$
No. Not for most uses, a function is regarded as being continuous on an interval.
$endgroup$
– Math_QED
5 hours ago
$begingroup$
There is a case to be made for 'pac-man' continuity via quotient space. Also, not every function is a function of a single, real variable, so the notion of "interval" may not always apply.
$endgroup$
– R. Burton
5 hours ago
add a comment |
$begingroup$
The exact answer depends on your chosen definition of "function" (there is more than one). For most uses, a function is regarded as being continuous on an interval $(a,b)$ if for every number $c$ in $(a,b)$, $f(x)=lim_{xto c} f(x)$.
In your example $f(x)=ln{x}$ is continuous on the interval $(0,infty)$ and either undefined or complex/multivalued everywhere else, depending on whether you consider the codomain (range) of $f$ to include the complex numbers or not.
In other words, no function is ever just 'continuous' - it is continuous within an interval (which may or may not be its domain).
answered 6 hours ago
R. BurtonR. Burton
50819
$endgroup$
The exact answer depends on your chosen definition of "function" (there is more than one). For most uses, a function is regarded as being continuous on an interval $(a,b)$ if for every number $c$ in $(a,b)$, $f(x)=lim_{xto c} f(x)$.
In your example $f(x)=ln{x}$ is continuous on the interval $(0,infty)$ and either undefined or complex/multivalued everywhere else, depending on whether you consider the codomain (range) of $f$ to include the complex numbers or not.
In other words, no function is ever just 'continuous' - it is continuous within an interval (which may or may not be its domain).
answered 6 hours ago
R. BurtonR. Burton
50819
answered 6 hours ago
R. BurtonR. Burton
50819
answered 6 hours ago
R. BurtonR. Burton
50819
answered 6 hours ago
R. BurtonR. Burton
50819
50819
$begingroup$
No. Not for most uses, a function is regarded as being continuous on an interval.
$endgroup$
– Math_QED
5 hours ago
$begingroup$
There is a case to be made for 'pac-man' continuity via quotient space. Also, not every function is a function of a single, real variable, so the notion of "interval" may not always apply.
$endgroup$
– R. Burton
5 hours ago
add a comment |
$begingroup$
No. Not for most uses, a function is regarded as being continuous on an interval.
$endgroup$
– Math_QED
5 hours ago
$begingroup$
There is a case to be made for 'pac-man' continuity via quotient space. Also, not every function is a function of a single, real variable, so the notion of "interval" may not always apply.
$endgroup$
– R. Burton
5 hours ago
$begingroup$
No. Not for most uses, a function is regarded as being continuous on an interval.
$endgroup$
– Math_QED
5 hours ago
$begingroup$
No. Not for most uses, a function is regarded as being continuous on an interval.
$endgroup$
– Math_QED
5 hours ago
$begingroup$
There is a case to be made for 'pac-man' continuity via quotient space. Also, not every function is a function of a single, real variable, so the notion of "interval" may not always apply.
$endgroup$
– R. Burton
5 hours ago
$begingroup$
There is a case to be made for 'pac-man' continuity via quotient space. Also, not every function is a function of a single, real variable, so the notion of "interval" may not always apply.
$endgroup$
– R. Burton
5 hours ago
add a comment |
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$begingroup$
If a function is not defined at a point, then it certainly cannot be continuous at it. Is this your question?
$endgroup$
– LoveTooNap29
5 hours ago
$begingroup$
Not really. Question is: Is function $f(x) = ln{|x|}$ continuous? I assume it is?
$endgroup$
– weno
4 hours ago
1
$begingroup$
continuous on what set? As others have pointed out: continuity is a property of functions either at points or on sets but not a property of the function alone without regard to the domain. $log |x|$ is certainly continuous where it is defined. Is that clear?
$endgroup$
– LoveTooNap29
4 hours ago
2
$begingroup$
functions are continuous at specific points. For example: $f(x) = (x-2)^2; x < 0$ and $f(x) = 1$ if $0 le x < 1$ and $f(x) = sqrt x$ for $x ge 1$. Is continuous at $x = -7$ and $x = pi$ but is not continuous and $x = 0$. To say a function is "continuous" is short hand to mean in is continuous on every point in it's domain. It makes utterly no sense to talk of a function being continuous on a point not in its domain.
$endgroup$
– fleablood
3 hours ago
2
$begingroup$
That said some people will "abuse terminology" and say intuitive but wrong things such as "$f(x) = frac 1x$ is discontinuous at $x = 0$" this is... well, it's simply wrong.
$endgroup$
– fleablood
3 hours ago