Limits of Rolle theorem












3












$begingroup$


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 derivatives continuity examples-counterexamples rolles-theorem







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is the question?
    $endgroup$
    – Will M.
    5 hours ago










  • $begingroup$
    The first phrase. I edit to make the question clearer.
    $endgroup$
    – Nameless
    5 hours 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
    5 hours 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.
    4 hours ago










  • $begingroup$
    I can not understand how my simple answer "the first sentence is my question" could not make sense, but OK.
    $endgroup$
    – Nameless
    4 hours ago
















3












$begingroup$


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 derivatives continuity examples-counterexamples rolles-theorem







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is the question?
    $endgroup$
    – Will M.
    5 hours ago










  • $begingroup$
    The first phrase. I edit to make the question clearer.
    $endgroup$
    – Nameless
    5 hours 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
    5 hours 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.
    4 hours ago










  • $begingroup$
    I can not understand how my simple answer "the first sentence is my question" could not make sense, but OK.
    $endgroup$
    – Nameless
    4 hours ago














3












3








3





$begingroup$


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 derivatives continuity examples-counterexamples rolles-theorem







share|cite|improve this question











$endgroup$




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 derivatives continuity examples-counterexamples rolles-theorem




real-analysis derivatives continuity examples-counterexamples rolles-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Martin Sleziak

44.7k10119272




44.7k10119272










asked 5 hours ago









NamelessNameless

638




638








  • 1




    $begingroup$
    What is the question?
    $endgroup$
    – Will M.
    5 hours ago










  • $begingroup$
    The first phrase. I edit to make the question clearer.
    $endgroup$
    – Nameless
    5 hours 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
    5 hours 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.
    4 hours ago










  • $begingroup$
    I can not understand how my simple answer "the first sentence is my question" could not make sense, but OK.
    $endgroup$
    – Nameless
    4 hours ago














  • 1




    $begingroup$
    What is the question?
    $endgroup$
    – Will M.
    5 hours ago










  • $begingroup$
    The first phrase. I edit to make the question clearer.
    $endgroup$
    – Nameless
    5 hours 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
    5 hours 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.
    4 hours ago










  • $begingroup$
    I can not understand how my simple answer "the first sentence is my question" could not make sense, but OK.
    $endgroup$
    – Nameless
    4 hours ago








1




1




$begingroup$
What is the question?
$endgroup$
– Will M.
5 hours ago




$begingroup$
What is the question?
$endgroup$
– Will M.
5 hours ago












$begingroup$
The first phrase. I edit to make the question clearer.
$endgroup$
– Nameless
5 hours ago




$begingroup$
The first phrase. I edit to make the question clearer.
$endgroup$
– Nameless
5 hours 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
5 hours 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
5 hours 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.
4 hours 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.
4 hours ago












$begingroup$
I can not understand how my simple answer "the first sentence is my question" could not make sense, but OK.
$endgroup$
– Nameless
4 hours ago




$begingroup$
I can not understand how my simple answer "the first sentence is my question" could not make sense, but OK.
$endgroup$
– Nameless
4 hours ago










2 Answers
2






active

oldest

votes


















6












$begingroup$

Consider $f(x) = x$ on $(0, 1]$, and $f(0) = 1$.






share|cite|improve this answer









$endgroup$









  • 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
    5 hours ago






  • 3




    $begingroup$
    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
    5 hours ago



















2












$begingroup$

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 $[a,b]$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you @Jossie Calderon.
    $endgroup$
    – Nameless
    5 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Consider $f(x) = x$ on $(0, 1]$, and $f(0) = 1$.






share|cite|improve this answer









$endgroup$









  • 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
    5 hours ago






  • 3




    $begingroup$
    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
    5 hours ago
















6












$begingroup$

Consider $f(x) = x$ on $(0, 1]$, and $f(0) = 1$.






share|cite|improve this answer









$endgroup$









  • 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
    5 hours ago






  • 3




    $begingroup$
    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
    5 hours ago














6












6








6





$begingroup$

Consider $f(x) = x$ on $(0, 1]$, and $f(0) = 1$.






share|cite|improve this answer









$endgroup$



Consider $f(x) = x$ on $(0, 1]$, and $f(0) = 1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 5 hours 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
    5 hours ago






  • 3




    $begingroup$
    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
    5 hours ago














  • 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
    5 hours ago






  • 3




    $begingroup$
    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
    5 hours 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
5 hours 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
5 hours ago




3




3




$begingroup$
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
5 hours 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
5 hours ago











2












$begingroup$

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 $[a,b]$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you @Jossie Calderon.
    $endgroup$
    – Nameless
    5 hours ago
















2












$begingroup$

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 $[a,b]$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you @Jossie Calderon.
    $endgroup$
    – Nameless
    5 hours ago














2












2








2





$begingroup$

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 $[a,b]$.






share|cite|improve this answer











$endgroup$



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 $[a,b]$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago









Martin Sleziak

44.7k10119272




44.7k10119272










answered 5 hours ago









Jossie CalderonJossie Calderon

248111




248111












  • $begingroup$
    Thank you @Jossie Calderon.
    $endgroup$
    – Nameless
    5 hours ago


















  • $begingroup$
    Thank you @Jossie Calderon.
    $endgroup$
    – Nameless
    5 hours ago
















$begingroup$
Thank you @Jossie Calderon.
$endgroup$
– Nameless
5 hours ago




$begingroup$
Thank you @Jossie Calderon.
$endgroup$
– Nameless
5 hours ago


















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