Single point function is not a continuous function according to the intermediate value theorem [duplicate]
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Is a function defined at a single point continuous?
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According to the intermediate value theorem, we can infer that a single point function, for example $f(x) = sqrt x + sqrt(-x)$, which has a range of only $0$ and a domain of only $0$ is not a continuous function. Is my conclusion right or there is actually exception when it is continuous? or outside of the calculus realm of thinking single point function CAN BE continuous?
continuity
edited Nov 13 at 14:43
MysteryGuy
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asked Nov 13 at 14:19
Firman Nugraha
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marked as duplicate by StackTD, jgon, Lord Shark the Unknown, Kelvin Lois, the_candyman Nov 13 at 19:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
-2
down vote
favorite
This question already has an answer here:
Is a function defined at a single point continuous?
2 answers
According to the intermediate value theorem, we can infer that a single point function, for example $f(x) = sqrt x + sqrt(-x)$, which has a range of only $0$ and a domain of only $0$ is not a continuous function. Is my conclusion right or there is actually exception when it is continuous? or outside of the calculus realm of thinking single point function CAN BE continuous?
continuity
edited Nov 13 at 14:43
MysteryGuy
384216
asked Nov 13 at 14:19
Firman Nugraha
1
marked as duplicate by StackTD, jgon, Lord Shark the Unknown, Kelvin Lois, the_candyman Nov 13 at 19:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
What (precise) definition are you using? See also: math.stackexchange.com/questions/455296/….
– StackTD
Nov 13 at 14:26
2
The intermediate value theorem does not imply that $f(x) = sqrt x+sqrt{-x}$ is not continuous, since it states that $f$ takes all values between $f(0) = 0$ and... $f(0) = 0$.
– krirkrirk
Nov 13 at 14:28
1
No, we certainly cannot infer what you claim from the IVT. Why not? Nobody can say what the error in your reasoning is since you don't explain how you think IVT implies this...
– David C. Ullrich
Nov 13 at 15:48
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
This question already has an answer here:
Is a function defined at a single point continuous?
2 answers
According to the intermediate value theorem, we can infer that a single point function, for example $f(x) = sqrt x + sqrt(-x)$, which has a range of only $0$ and a domain of only $0$ is not a continuous function. Is my conclusion right or there is actually exception when it is continuous? or outside of the calculus realm of thinking single point function CAN BE continuous?
continuity
edited Nov 13 at 14:43
MysteryGuy
384216
asked Nov 13 at 14:19
Firman Nugraha
1
This question already has an answer here:
Is a function defined at a single point continuous?
2 answers
According to the intermediate value theorem, we can infer that a single point function, for example $f(x) = sqrt x + sqrt(-x)$, which has a range of only $0$ and a domain of only $0$ is not a continuous function. Is my conclusion right or there is actually exception when it is continuous? or outside of the calculus realm of thinking single point function CAN BE continuous?
This question already has an answer here:
Is a function defined at a single point continuous?
2 answers
continuity
continuity
edited Nov 13 at 14:43
MysteryGuy
384216
asked Nov 13 at 14:19
Firman Nugraha
1
edited Nov 13 at 14:43
MysteryGuy
384216
asked Nov 13 at 14:19
Firman Nugraha
1
edited Nov 13 at 14:43
MysteryGuy
384216
edited Nov 13 at 14:43
MysteryGuy
384216
edited Nov 13 at 14:43
MysteryGuy
384216
384216
asked Nov 13 at 14:19
Firman Nugraha
1
asked Nov 13 at 14:19
Firman Nugraha
1
asked Nov 13 at 14:19
Firman Nugraha
1
1
marked as duplicate by StackTD, jgon, Lord Shark the Unknown, Kelvin Lois, the_candyman Nov 13 at 19:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by StackTD, jgon, Lord Shark the Unknown, Kelvin Lois, the_candyman Nov 13 at 19:28
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
What (precise) definition are you using? See also: math.stackexchange.com/questions/455296/….
– StackTD
Nov 13 at 14:26
2
The intermediate value theorem does not imply that $f(x) = sqrt x+sqrt{-x}$ is not continuous, since it states that $f$ takes all values between $f(0) = 0$ and... $f(0) = 0$.
– krirkrirk
Nov 13 at 14:28
1
No, we certainly cannot infer what you claim from the IVT. Why not? Nobody can say what the error in your reasoning is since you don't explain how you think IVT implies this...
– David C. Ullrich
Nov 13 at 15:48
add a comment |
1
What (precise) definition are you using? See also: math.stackexchange.com/questions/455296/….
– StackTD
Nov 13 at 14:26
2
The intermediate value theorem does not imply that $f(x) = sqrt x+sqrt{-x}$ is not continuous, since it states that $f$ takes all values between $f(0) = 0$ and... $f(0) = 0$.
– krirkrirk
Nov 13 at 14:28
1
No, we certainly cannot infer what you claim from the IVT. Why not? Nobody can say what the error in your reasoning is since you don't explain how you think IVT implies this...
– David C. Ullrich
Nov 13 at 15:48
1
1
What (precise) definition are you using? See also: math.stackexchange.com/questions/455296/….
– StackTD
Nov 13 at 14:26
What (precise) definition are you using? See also: math.stackexchange.com/questions/455296/….
– StackTD
Nov 13 at 14:26
2
2
The intermediate value theorem does not imply that $f(x) = sqrt x+sqrt{-x}$ is not continuous, since it states that $f$ takes all values between $f(0) = 0$ and... $f(0) = 0$.
– krirkrirk
Nov 13 at 14:28
The intermediate value theorem does not imply that $f(x) = sqrt x+sqrt{-x}$ is not continuous, since it states that $f$ takes all values between $f(0) = 0$ and... $f(0) = 0$.
– krirkrirk
Nov 13 at 14:28
1
1
No, we certainly cannot infer what you claim from the IVT. Why not? Nobody can say what the error in your reasoning is since you don't explain how you think IVT implies this...
– David C. Ullrich
Nov 13 at 15:48
No, we certainly cannot infer what you claim from the IVT. Why not? Nobody can say what the error in your reasoning is since you don't explain how you think IVT implies this...
– David C. Ullrich
Nov 13 at 15:48
add a comment |
1 Answer
1
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Although there are similar duplicates here a quick reasoning:
- You consider a real-valued function with domain $D = {0}$, so you can use the limit definition of continuity:
$$f mbox{ is continuous at } x_0 = 0 $$
$$Leftrightarrow lim_{ntoinfty}f(x_n) = f(0) mbox{ for all sequences } {x_n}_{n in D}subset D mbox{ with } lim_{ntoinfty}x_n = 0$$ - The only possible sequence converging to $0$ in the domain ${0}$ is the constant sequence $x_n = 0$ for all $n in mathbb{N}$.
$$lim_{ntoinfty}f(x_n) = f(0) = 0$$
So, $f$ is contiuous.
answered Nov 13 at 14:36
trancelocation
8,1291519
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Although there are similar duplicates here a quick reasoning:
- You consider a real-valued function with domain $D = {0}$, so you can use the limit definition of continuity:
$$f mbox{ is continuous at } x_0 = 0 $$
$$Leftrightarrow lim_{ntoinfty}f(x_n) = f(0) mbox{ for all sequences } {x_n}_{n in D}subset D mbox{ with } lim_{ntoinfty}x_n = 0$$ - The only possible sequence converging to $0$ in the domain ${0}$ is the constant sequence $x_n = 0$ for all $n in mathbb{N}$.
$$lim_{ntoinfty}f(x_n) = f(0) = 0$$
So, $f$ is contiuous.
answered Nov 13 at 14:36
trancelocation
8,1291519
add a comment |
up vote
0
down vote
Although there are similar duplicates here a quick reasoning:
- You consider a real-valued function with domain $D = {0}$, so you can use the limit definition of continuity:
$$f mbox{ is continuous at } x_0 = 0 $$
$$Leftrightarrow lim_{ntoinfty}f(x_n) = f(0) mbox{ for all sequences } {x_n}_{n in D}subset D mbox{ with } lim_{ntoinfty}x_n = 0$$ - The only possible sequence converging to $0$ in the domain ${0}$ is the constant sequence $x_n = 0$ for all $n in mathbb{N}$.
$$lim_{ntoinfty}f(x_n) = f(0) = 0$$
So, $f$ is contiuous.
answered Nov 13 at 14:36
trancelocation
8,1291519
add a comment |
up vote
0
down vote
up vote
0
down vote
Although there are similar duplicates here a quick reasoning:
- You consider a real-valued function with domain $D = {0}$, so you can use the limit definition of continuity:
$$f mbox{ is continuous at } x_0 = 0 $$
$$Leftrightarrow lim_{ntoinfty}f(x_n) = f(0) mbox{ for all sequences } {x_n}_{n in D}subset D mbox{ with } lim_{ntoinfty}x_n = 0$$ - The only possible sequence converging to $0$ in the domain ${0}$ is the constant sequence $x_n = 0$ for all $n in mathbb{N}$.
$$lim_{ntoinfty}f(x_n) = f(0) = 0$$
So, $f$ is contiuous.
answered Nov 13 at 14:36
trancelocation
8,1291519
Although there are similar duplicates here a quick reasoning:
- You consider a real-valued function with domain $D = {0}$, so you can use the limit definition of continuity:
$$f mbox{ is continuous at } x_0 = 0 $$
$$Leftrightarrow lim_{ntoinfty}f(x_n) = f(0) mbox{ for all sequences } {x_n}_{n in D}subset D mbox{ with } lim_{ntoinfty}x_n = 0$$ - The only possible sequence converging to $0$ in the domain ${0}$ is the constant sequence $x_n = 0$ for all $n in mathbb{N}$.
$$lim_{ntoinfty}f(x_n) = f(0) = 0$$
So, $f$ is contiuous.
answered Nov 13 at 14:36
trancelocation
8,1291519
answered Nov 13 at 14:36
trancelocation
8,1291519
answered Nov 13 at 14:36
trancelocation
8,1291519
answered Nov 13 at 14:36
trancelocation
8,1291519
8,1291519
add a comment |
add a comment |
1
What (precise) definition are you using? See also: math.stackexchange.com/questions/455296/….
– StackTD
Nov 13 at 14:26
2
The intermediate value theorem does not imply that $f(x) = sqrt x+sqrt{-x}$ is not continuous, since it states that $f$ takes all values between $f(0) = 0$ and... $f(0) = 0$.
– krirkrirk
Nov 13 at 14:28
1
No, we certainly cannot infer what you claim from the IVT. Why not? Nobody can say what the error in your reasoning is since you don't explain how you think IVT implies this...
– David C. Ullrich
Nov 13 at 15:48