prove if f is differentiable and uniformly continuous on (a b) then f ' is unbounded on (a b). [closed]
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prove if f is differentiable and uniformly continuous on (a b) then f ' is unbounded on (a b). Hint : use contradiction
The function can be sqrt of x
so f'(x)=1/2 sqrt x
real-analysis calculus
asked Dec 4 at 0:43
Archi Aguero
11
closed as off-topic by Rory Daulton, Leucippus, Umberto P., user10354138, Shailesh Dec 4 at 2:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Rory Daulton, Leucippus, Umberto P., user10354138, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-3
down vote
favorite
prove if f is differentiable and uniformly continuous on (a b) then f ' is unbounded on (a b). Hint : use contradiction
The function can be sqrt of x
so f'(x)=1/2 sqrt x
real-analysis calculus
asked Dec 4 at 0:43
Archi Aguero
11
closed as off-topic by Rory Daulton, Leucippus, Umberto P., user10354138, Shailesh Dec 4 at 2:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Rory Daulton, Leucippus, Umberto P., user10354138, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
This is an inappropriate question without background. You need to explain what you've tried.
– Ben W
Dec 4 at 0:43
Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: many of us want to see that you have put significant work into the problem.
– Rory Daulton
Dec 4 at 0:51
1
Huh? $x mapsto sqrt{x}$ on $(0,1)$?
– MathematicsStudent1122
Dec 4 at 0:59
@MathematicsStudent1122 . Right. The proposition is false. We can even have a strictly monotonic everywhere-differentiable $f:Bbb R to Bbb R$ such that $f'$ is unbounded on $[0,1]$
– DanielWainfleet
Dec 4 at 2:37
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
prove if f is differentiable and uniformly continuous on (a b) then f ' is unbounded on (a b). Hint : use contradiction
The function can be sqrt of x
so f'(x)=1/2 sqrt x
real-analysis calculus
asked Dec 4 at 0:43
Archi Aguero
11
prove if f is differentiable and uniformly continuous on (a b) then f ' is unbounded on (a b). Hint : use contradiction
The function can be sqrt of x
so f'(x)=1/2 sqrt x
real-analysis calculus
real-analysis calculus
asked Dec 4 at 0:43
Archi Aguero
11
asked Dec 4 at 0:43
Archi Aguero
11
edited Dec 6 at 18:27
asked Dec 4 at 0:43
Archi Aguero
11
asked Dec 4 at 0:43
Archi Aguero
11
asked Dec 4 at 0:43
Archi Aguero
11
11
closed as off-topic by Rory Daulton, Leucippus, Umberto P., user10354138, Shailesh Dec 4 at 2:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Rory Daulton, Leucippus, Umberto P., user10354138, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Rory Daulton, Leucippus, Umberto P., user10354138, Shailesh Dec 4 at 2:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Rory Daulton, Leucippus, Umberto P., user10354138, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
This is an inappropriate question without background. You need to explain what you've tried.
– Ben W
Dec 4 at 0:43
Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: many of us want to see that you have put significant work into the problem.
– Rory Daulton
Dec 4 at 0:51
1
Huh? $x mapsto sqrt{x}$ on $(0,1)$?
– MathematicsStudent1122
Dec 4 at 0:59
@MathematicsStudent1122 . Right. The proposition is false. We can even have a strictly monotonic everywhere-differentiable $f:Bbb R to Bbb R$ such that $f'$ is unbounded on $[0,1]$
– DanielWainfleet
Dec 4 at 2:37
add a comment |
This is an inappropriate question without background. You need to explain what you've tried.
– Ben W
Dec 4 at 0:43
Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: many of us want to see that you have put significant work into the problem.
– Rory Daulton
Dec 4 at 0:51
1
Huh? $x mapsto sqrt{x}$ on $(0,1)$?
– MathematicsStudent1122
Dec 4 at 0:59
@MathematicsStudent1122 . Right. The proposition is false. We can even have a strictly monotonic everywhere-differentiable $f:Bbb R to Bbb R$ such that $f'$ is unbounded on $[0,1]$
– DanielWainfleet
Dec 4 at 2:37
This is an inappropriate question without background. You need to explain what you've tried.
– Ben W
Dec 4 at 0:43
This is an inappropriate question without background. You need to explain what you've tried.
– Ben W
Dec 4 at 0:43
Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: many of us want to see that you have put significant work into the problem.
– Rory Daulton
Dec 4 at 0:51
Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: many of us want to see that you have put significant work into the problem.
– Rory Daulton
Dec 4 at 0:51
1
1
Huh? $x mapsto sqrt{x}$ on $(0,1)$?
– MathematicsStudent1122
Dec 4 at 0:59
Huh? $x mapsto sqrt{x}$ on $(0,1)$?
– MathematicsStudent1122
Dec 4 at 0:59
@MathematicsStudent1122 . Right. The proposition is false. We can even have a strictly monotonic everywhere-differentiable $f:Bbb R to Bbb R$ such that $f'$ is unbounded on $[0,1]$
– DanielWainfleet
Dec 4 at 2:37
@MathematicsStudent1122 . Right. The proposition is false. We can even have a strictly monotonic everywhere-differentiable $f:Bbb R to Bbb R$ such that $f'$ is unbounded on $[0,1]$
– DanielWainfleet
Dec 4 at 2:37
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This is an inappropriate question without background. You need to explain what you've tried.
– Ben W
Dec 4 at 0:43
Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: many of us want to see that you have put significant work into the problem.
– Rory Daulton
Dec 4 at 0:51
1
Huh? $x mapsto sqrt{x}$ on $(0,1)$?
– MathematicsStudent1122
Dec 4 at 0:59
@MathematicsStudent1122 . Right. The proposition is false. We can even have a strictly monotonic everywhere-differentiable $f:Bbb R to Bbb R$ such that $f'$ is unbounded on $[0,1]$
– DanielWainfleet
Dec 4 at 2:37