Limits of Rolle theoremRolle and Mean Value TheoremIs there a function from $mathbb R^2 to mathbb R^2$ that...
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Limits of Rolle theorem
Rolle and Mean Value TheoremIs there a function from $mathbb R^2 to mathbb R^2$ that is continuous, and *not* one to one, but still maps open sets to open sets?Piecewise continuous differentiableone-sided continuity and one-sided derivative?Does differentiablity of mod of a continuous function implies differentiablity of the function?Continuous inverse function theorem and open intervals?finding weird examples of continuous functions (or, maybe, they're tricks?)Pointwise limits of differentiable functions under constraintDifficulties in stating mean value theorem for functions which are not continuous on a closed interval.Continuous and differentiable function in Rolle's theorem
$begingroup$
I would like to see a function $f:[a,b]tomathbb{R}$ that is differentiable in $(a,b)$ but it is not continuous at least at one of the interval boundary points $a$ or $b$. Can you show me one?
This is a curiosity that would make me to see limits of Rolle theorem, because one of its hypothesis is that the function $f$ has to be continuous in the entire closed interval $[a,b]$, even if it could be differentiable only in the open $(a,b)$.
Thank you.
real-analysis continuity rolles-theorem
asked 1 hour ago
NamelessNameless
588
$endgroup$
add a comment |
$begingroup$
I would like to see a function $f:[a,b]tomathbb{R}$ that is differentiable in $(a,b)$ but it is not continuous at least at one of the interval boundary points $a$ or $b$. Can you show me one?
This is a curiosity that would make me to see limits of Rolle theorem, because one of its hypothesis is that the function $f$ has to be continuous in the entire closed interval $[a,b]$, even if it could be differentiable only in the open $(a,b)$.
Thank you.
real-analysis continuity rolles-theorem
asked 1 hour ago
NamelessNameless
588
$endgroup$
$begingroup$
What is the question?
$endgroup$
– Will M.
1 hour ago
$begingroup$
The first phrase. I edit to make the question clearer.
$endgroup$
– Nameless
1 hour ago
1
$begingroup$
I think the question is pretty clear: "why can we not relax the requirement of continuity on $[a,b]$ in Rolle's theorem?".
$endgroup$
– Patrick Stevens
1 hour ago
$begingroup$
@PatrickStevens I was truly unsure what the OP was intending to ask. The reply to my comment made even less sense. I am not sure how you have the ability to read peoples minds through what they write, but OK.
$endgroup$
– Will M.
32 mins ago
$begingroup$
I can not understand how my simple answer "the first sentence is my question" could not make sense, but OK.
$endgroup$
– Nameless
30 mins ago
add a comment |
$begingroup$
I would like to see a function $f:[a,b]tomathbb{R}$ that is differentiable in $(a,b)$ but it is not continuous at least at one of the interval boundary points $a$ or $b$. Can you show me one?
This is a curiosity that would make me to see limits of Rolle theorem, because one of its hypothesis is that the function $f$ has to be continuous in the entire closed interval $[a,b]$, even if it could be differentiable only in the open $(a,b)$.
Thank you.
real-analysis continuity rolles-theorem
asked 1 hour ago
NamelessNameless
588
$endgroup$
I would like to see a function $f:[a,b]tomathbb{R}$ that is differentiable in $(a,b)$ but it is not continuous at least at one of the interval boundary points $a$ or $b$. Can you show me one?
This is a curiosity that would make me to see limits of Rolle theorem, because one of its hypothesis is that the function $f$ has to be continuous in the entire closed interval $[a,b]$, even if it could be differentiable only in the open $(a,b)$.
Thank you.
real-analysis continuity rolles-theorem
real-analysis continuity rolles-theorem
asked 1 hour ago
NamelessNameless
588
asked 1 hour ago
NamelessNameless
588
edited 1 hour ago
Nameless
asked 1 hour ago
NamelessNameless
588
asked 1 hour ago
NamelessNameless
588
asked 1 hour ago
NamelessNameless
588
588
$begingroup$
What is the question?
$endgroup$
– Will M.
1 hour ago
$begingroup$
The first phrase. I edit to make the question clearer.
$endgroup$
– Nameless
1 hour ago
1
$begingroup$
I think the question is pretty clear: "why can we not relax the requirement of continuity on $[a,b]$ in Rolle's theorem?".
$endgroup$
– Patrick Stevens
1 hour ago
$begingroup$
@PatrickStevens I was truly unsure what the OP was intending to ask. The reply to my comment made even less sense. I am not sure how you have the ability to read peoples minds through what they write, but OK.
$endgroup$
– Will M.
32 mins ago
$begingroup$
I can not understand how my simple answer "the first sentence is my question" could not make sense, but OK.
$endgroup$
– Nameless
30 mins ago
add a comment |
$begingroup$
What is the question?
$endgroup$
– Will M.
1 hour ago
$begingroup$
The first phrase. I edit to make the question clearer.
$endgroup$
– Nameless
1 hour ago
1
$begingroup$
I think the question is pretty clear: "why can we not relax the requirement of continuity on $[a,b]$ in Rolle's theorem?".
$endgroup$
– Patrick Stevens
1 hour ago
$begingroup$
@PatrickStevens I was truly unsure what the OP was intending to ask. The reply to my comment made even less sense. I am not sure how you have the ability to read peoples minds through what they write, but OK.
$endgroup$
– Will M.
32 mins ago
$begingroup$
I can not understand how my simple answer "the first sentence is my question" could not make sense, but OK.
$endgroup$
– Nameless
30 mins ago
$begingroup$
What is the question?
$endgroup$
– Will M.
1 hour ago
$begingroup$
What is the question?
$endgroup$
– Will M.
1 hour ago
$begingroup$
The first phrase. I edit to make the question clearer.
$endgroup$
– Nameless
1 hour ago
$begingroup$
The first phrase. I edit to make the question clearer.
$endgroup$
– Nameless
1 hour ago
1
1
$begingroup$
I think the question is pretty clear: "why can we not relax the requirement of continuity on $[a,b]$ in Rolle's theorem?".
$endgroup$
– Patrick Stevens
1 hour ago
$begingroup$
I think the question is pretty clear: "why can we not relax the requirement of continuity on $[a,b]$ in Rolle's theorem?".
$endgroup$
– Patrick Stevens
1 hour ago
$begingroup$
@PatrickStevens I was truly unsure what the OP was intending to ask. The reply to my comment made even less sense. I am not sure how you have the ability to read peoples minds through what they write, but OK.
$endgroup$
– Will M.
32 mins ago
$begingroup$
@PatrickStevens I was truly unsure what the OP was intending to ask. The reply to my comment made even less sense. I am not sure how you have the ability to read peoples minds through what they write, but OK.
$endgroup$
– Will M.
32 mins ago
$begingroup$
I can not understand how my simple answer "the first sentence is my question" could not make sense, but OK.
$endgroup$
– Nameless
30 mins ago
$begingroup$
I can not understand how my simple answer "the first sentence is my question" could not make sense, but OK.
$endgroup$
– Nameless
30 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider $f(x) = x$ on $(0, 1]$, and $f(0) = 1$.
answered 1 hour ago
Patrick StevensPatrick Stevens
28.8k52874
$endgroup$
1
$begingroup$
Who knows for what obscure reason I was imagining things incredibly more complex than that. Thank you very much. Sorry for the question that now seems very stupid to me.
$endgroup$
– Nameless
1 hour ago
1
$begingroup$
In first-year analysis, there are about five different counterexamples, all of them quite simple. Almost nothing you'll encounter will require really pathological counterexamples.
$endgroup$
– Patrick Stevens
1 hour ago
add a comment |
$begingroup$
Consider $f(x) = frac{1}{x}$ in $[0, 1]$ and define $f(0) = 0.5$.
Then by the extreme value theorem - which is needed to make Rolle's Theorem work - since f doesn't obtain a maximum, f is not continuous on [a,b].
answered 1 hour ago
Jossie CalderonJossie Calderon
250111
$endgroup$
$begingroup$
Thank you @Jossie Calderon.
$endgroup$
– Nameless
1 hour ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider $f(x) = x$ on $(0, 1]$, and $f(0) = 1$.
answered 1 hour ago
Patrick StevensPatrick Stevens
28.8k52874
$endgroup$
1
$begingroup$
Who knows for what obscure reason I was imagining things incredibly more complex than that. Thank you very much. Sorry for the question that now seems very stupid to me.
$endgroup$
– Nameless
1 hour ago
1
$begingroup$
In first-year analysis, there are about five different counterexamples, all of them quite simple. Almost nothing you'll encounter will require really pathological counterexamples.
$endgroup$
– Patrick Stevens
1 hour ago
add a comment |
$begingroup$
Consider $f(x) = x$ on $(0, 1]$, and $f(0) = 1$.
answered 1 hour ago
Patrick StevensPatrick Stevens
28.8k52874
$endgroup$
1
$begingroup$
Who knows for what obscure reason I was imagining things incredibly more complex than that. Thank you very much. Sorry for the question that now seems very stupid to me.
$endgroup$
– Nameless
1 hour ago
1
$begingroup$
In first-year analysis, there are about five different counterexamples, all of them quite simple. Almost nothing you'll encounter will require really pathological counterexamples.
$endgroup$
– Patrick Stevens
1 hour ago
add a comment |
$begingroup$
Consider $f(x) = x$ on $(0, 1]$, and $f(0) = 1$.
answered 1 hour ago
Patrick StevensPatrick Stevens
28.8k52874
$endgroup$
Consider $f(x) = x$ on $(0, 1]$, and $f(0) = 1$.
answered 1 hour ago
Patrick StevensPatrick Stevens
28.8k52874
answered 1 hour ago
Patrick StevensPatrick Stevens
28.8k52874
answered 1 hour ago
Patrick StevensPatrick Stevens
28.8k52874
answered 1 hour ago
Patrick StevensPatrick Stevens
28.8k52874
28.8k52874
1
$begingroup$
Who knows for what obscure reason I was imagining things incredibly more complex than that. Thank you very much. Sorry for the question that now seems very stupid to me.
$endgroup$
– Nameless
1 hour ago
1
$begingroup$
In first-year analysis, there are about five different counterexamples, all of them quite simple. Almost nothing you'll encounter will require really pathological counterexamples.
$endgroup$
– Patrick Stevens
1 hour ago
add a comment |
1
$begingroup$
Who knows for what obscure reason I was imagining things incredibly more complex than that. Thank you very much. Sorry for the question that now seems very stupid to me.
$endgroup$
– Nameless
1 hour ago
1
$begingroup$
In first-year analysis, there are about five different counterexamples, all of them quite simple. Almost nothing you'll encounter will require really pathological counterexamples.
$endgroup$
– Patrick Stevens
1 hour ago
1
1
$begingroup$
Who knows for what obscure reason I was imagining things incredibly more complex than that. Thank you very much. Sorry for the question that now seems very stupid to me.
$endgroup$
– Nameless
1 hour ago
$begingroup$
Who knows for what obscure reason I was imagining things incredibly more complex than that. Thank you very much. Sorry for the question that now seems very stupid to me.
$endgroup$
– Nameless
1 hour ago
1
1
$begingroup$
In first-year analysis, there are about five different counterexamples, all of them quite simple. Almost nothing you'll encounter will require really pathological counterexamples.
$endgroup$
– Patrick Stevens
1 hour ago
$begingroup$
In first-year analysis, there are about five different counterexamples, all of them quite simple. Almost nothing you'll encounter will require really pathological counterexamples.
$endgroup$
– Patrick Stevens
1 hour ago
add a comment |
$begingroup$
Consider $f(x) = frac{1}{x}$ in $[0, 1]$ and define $f(0) = 0.5$.
Then by the extreme value theorem - which is needed to make Rolle's Theorem work - since f doesn't obtain a maximum, f is not continuous on [a,b].
answered 1 hour ago
Jossie CalderonJossie Calderon
250111
$endgroup$
$begingroup$
Thank you @Jossie Calderon.
$endgroup$
– Nameless
1 hour ago
add a comment |
$begingroup$
Consider $f(x) = frac{1}{x}$ in $[0, 1]$ and define $f(0) = 0.5$.
Then by the extreme value theorem - which is needed to make Rolle's Theorem work - since f doesn't obtain a maximum, f is not continuous on [a,b].
answered 1 hour ago
Jossie CalderonJossie Calderon
250111
$endgroup$
$begingroup$
Thank you @Jossie Calderon.
$endgroup$
– Nameless
1 hour ago
add a comment |
$begingroup$
Consider $f(x) = frac{1}{x}$ in $[0, 1]$ and define $f(0) = 0.5$.
Then by the extreme value theorem - which is needed to make Rolle's Theorem work - since f doesn't obtain a maximum, f is not continuous on [a,b].
answered 1 hour ago
Jossie CalderonJossie Calderon
250111
$endgroup$
Consider $f(x) = frac{1}{x}$ in $[0, 1]$ and define $f(0) = 0.5$.
Then by the extreme value theorem - which is needed to make Rolle's Theorem work - since f doesn't obtain a maximum, f is not continuous on [a,b].
answered 1 hour ago
Jossie CalderonJossie Calderon
250111
edited 1 hour ago
answered 1 hour ago
Jossie CalderonJossie Calderon
250111
answered 1 hour ago
Jossie CalderonJossie Calderon
250111
answered 1 hour ago
Jossie CalderonJossie Calderon
250111
250111
$begingroup$
Thank you @Jossie Calderon.
$endgroup$
– Nameless
1 hour ago
add a comment |
$begingroup$
Thank you @Jossie Calderon.
$endgroup$
– Nameless
1 hour ago
$begingroup$
Thank you @Jossie Calderon.
$endgroup$
– Nameless
1 hour ago
$begingroup$
Thank you @Jossie Calderon.
$endgroup$
– Nameless
1 hour ago
add a comment |
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$begingroup$
What is the question?
$endgroup$
– Will M.
1 hour ago
$begingroup$
The first phrase. I edit to make the question clearer.
$endgroup$
– Nameless
1 hour ago
1
$begingroup$
I think the question is pretty clear: "why can we not relax the requirement of continuity on $[a,b]$ in Rolle's theorem?".
$endgroup$
– Patrick Stevens
1 hour ago
$begingroup$
@PatrickStevens I was truly unsure what the OP was intending to ask. The reply to my comment made even less sense. I am not sure how you have the ability to read peoples minds through what they write, but OK.
$endgroup$
– Will M.
32 mins ago
$begingroup$
I can not understand how my simple answer "the first sentence is my question" could not make sense, but OK.
$endgroup$
– Nameless
30 mins ago