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Why do mean value theorems have open interval for differentiablity while closed for continuity?

Why can we only talk about derivatives on an open interval?Does $f~'$ have to be continuous for application of Mean Value Theorem?Mean Value Theorem Proofing For all $ x>0$Mean-value theorems (other than Cauchy's, Lagrange's or Rolle's)Mean value theorem for vector valued function (not integral form)Why do we call this theorem in vector valued functions “mean value theorem”?Why does the Mean Value Theorem require a closed interval for continuity and an open interval for differentiability?Usage of mean value theorem ; bounded derivative and open intervalPresenting a lecture on Mean value theorems.How to use the Mean-Value Theorem on a function continuous over an open interval.

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2

$begingroup$

For mean value theorems like Lagrange and rolles, we have the condition as

For applying mean value theorem to any function $f(x)$ for the domain $[a,b]$ , it should be

(1) continuous in $[a,b]$

(2) differentiable in $(a,b)$

So why is it that for the criteria of differentiablity, we have the open interval ?? Is it possible for a function differentiable in $(a,b) $ and continuous in $[a,b]$ to be non- differentable at the end points.

Also why is the first statement needed ?? Doesn't the second statement of differentiablity also mean that the function is continuous ??
I'm not very experienced in calculus and still in highschool, so it might be something too obvious I'm missing , please help :)

calculus differential

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edited 23 mins ago

DivMit

asked 8 hours ago

DivMitDivMit

57811 silver badge1010 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it is possible for a function differentiable in $(a,b)$ and continuous in $[a,b]$ to be non-differentiable at the end points. With that encouragement, would you care to seek examples on your own? And in the last paragraph, note that differentiability on the interior interval $(a,b)$ implies continuity there, but fails to imply (one-sided) continuity at the endpoints. Again, perhaps you will want to think about how weakening the assumptions (removing continuity at the endpoints) opens the door to functions that do not satisfy the MVT.
  $endgroup$
  – hardmath
  8 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

For mean value theorems like Lagrange and rolles, we have the condition as

For applying mean value theorem to any function $f(x)$ for the domain $[a,b]$ , it should be

(1) continuous in $[a,b]$

(2) differentiable in $(a,b)$

So why is it that for the criteria of differentiablity, we have the open interval ?? Is it possible for a function differentiable in $(a,b) $ and continuous in $[a,b]$ to be non- differentable at the end points.

Also why is the first statement needed ?? Doesn't the second statement of differentiablity also mean that the function is continuous ??
I'm not very experienced in calculus and still in highschool, so it might be something too obvious I'm missing , please help :)

calculus differential

share|cite|improve this question

edited 23 mins ago

DivMit

asked 8 hours ago

DivMitDivMit

57811 silver badge1010 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it is possible for a function differentiable in $(a,b)$ and continuous in $[a,b]$ to be non-differentiable at the end points. With that encouragement, would you care to seek examples on your own? And in the last paragraph, note that differentiability on the interior interval $(a,b)$ implies continuity there, but fails to imply (one-sided) continuity at the endpoints. Again, perhaps you will want to think about how weakening the assumptions (removing continuity at the endpoints) opens the door to functions that do not satisfy the MVT.
  $endgroup$
  – hardmath
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

For mean value theorems like Lagrange and rolles, we have the condition as

For applying mean value theorem to any function $f(x)$ for the domain $[a,b]$ , it should be

(1) continuous in $[a,b]$

(2) differentiable in $(a,b)$

So why is it that for the criteria of differentiablity, we have the open interval ?? Is it possible for a function differentiable in $(a,b) $ and continuous in $[a,b]$ to be non- differentable at the end points.

Also why is the first statement needed ?? Doesn't the second statement of differentiablity also mean that the function is continuous ??
I'm not very experienced in calculus and still in highschool, so it might be something too obvious I'm missing , please help :)

calculus differential

share|cite|improve this question

edited 23 mins ago

DivMit

asked 8 hours ago

DivMitDivMit

57811 silver badge1010 bronze badges

$endgroup$

share|cite|improve this question

edited 23 mins ago

DivMit

asked 8 hours ago

DivMitDivMit

57811 silver badge1010 bronze badges

share|cite|improve this question

edited 23 mins ago

DivMit

asked 8 hours ago

DivMitDivMit

57811 silver badge1010 bronze badges

asked 8 hours ago

DivMitDivMit

57811 silver badge1010 bronze badges

asked 8 hours ago

DivMitDivMit

57811 silver badge1010 bronze badges

2 Answers
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6

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For example, the function $sqrtx(1-x)$ is continuous in $[0,1]$ but only differentiable on $(0,1)$.

On the other hand, $1/x$ is differentiable in $(0,infty)$ but is not continuous at $0$
(even if you define it at $0$).

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

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add a comment | 

4

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Here are a couple of examples to think about:

1. $f(x) = sqrtx$ on the interval $[0,1]$. This is continuous on $[0,1]$ and differentiable on $(0,1)$, but not differentiable at 0 in any reasonable sense. Yet the mean value theorem applies to it.




2. $f(x) = begincases 0, & x = 0 \ 1, & x > 0. endcases$ This is differentiable on $(0,1)$ but the mean value theorem does not apply to it.

share|cite|improve this answer

answered 8 hours ago

Nate EldredgeNate Eldredge

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6

$begingroup$

For example, the function $sqrtx(1-x)$ is continuous in $[0,1]$ but only differentiable on $(0,1)$.

On the other hand, $1/x$ is differentiable in $(0,infty)$ but is not continuous at $0$
(even if you define it at $0$).

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

342k2323 gold badges234234 silver badges495495 bronze badges

$endgroup$

add a comment | 

6

$begingroup$

For example, the function $sqrtx(1-x)$ is continuous in $[0,1]$ but only differentiable on $(0,1)$.

On the other hand, $1/x$ is differentiable in $(0,infty)$ but is not continuous at $0$
(even if you define it at $0$).

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

342k2323 gold badges234234 silver badges495495 bronze badges

$endgroup$

add a comment | 

$begingroup$

For example, the function $sqrtx(1-x)$ is continuous in $[0,1]$ but only differentiable on $(0,1)$.

On the other hand, $1/x$ is differentiable in $(0,infty)$ but is not continuous at $0$
(even if you define it at $0$).

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

342k2323 gold badges234234 silver badges495495 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Robert IsraelRobert Israel

342k2323 gold badges234234 silver badges495495 bronze badges

answered 8 hours ago

Robert IsraelRobert Israel

342k2323 gold badges234234 silver badges495495 bronze badges

answered 8 hours ago

Robert IsraelRobert Israel

342k2323 gold badges234234 silver badges495495 bronze badges

4

$begingroup$

Here are a couple of examples to think about:

1. $f(x) = sqrtx$ on the interval $[0,1]$. This is continuous on $[0,1]$ and differentiable on $(0,1)$, but not differentiable at 0 in any reasonable sense. Yet the mean value theorem applies to it.




2. $f(x) = begincases 0, & x = 0 \ 1, & x > 0. endcases$ This is differentiable on $(0,1)$ but the mean value theorem does not apply to it.

share|cite|improve this answer

answered 8 hours ago

Nate EldredgeNate Eldredge

67.1k99 gold badges8585 silver badges180180 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

Here are a couple of examples to think about:

1. $f(x) = sqrtx$ on the interval $[0,1]$. This is continuous on $[0,1]$ and differentiable on $(0,1)$, but not differentiable at 0 in any reasonable sense. Yet the mean value theorem applies to it.




2. $f(x) = begincases 0, & x = 0 \ 1, & x > 0. endcases$ This is differentiable on $(0,1)$ but the mean value theorem does not apply to it.

share|cite|improve this answer

answered 8 hours ago

Nate EldredgeNate Eldredge

67.1k99 gold badges8585 silver badges180180 bronze badges

$endgroup$

add a comment | 

$begingroup$

Here are a couple of examples to think about:

1. $f(x) = sqrtx$ on the interval $[0,1]$. This is continuous on $[0,1]$ and differentiable on $(0,1)$, but not differentiable at 0 in any reasonable sense. Yet the mean value theorem applies to it.




2. $f(x) = begincases 0, & x = 0 \ 1, & x > 0. endcases$ This is differentiable on $(0,1)$ but the mean value theorem does not apply to it.

share|cite|improve this answer

answered 8 hours ago

Nate EldredgeNate Eldredge

67.1k99 gold badges8585 silver badges180180 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Nate EldredgeNate Eldredge

67.1k99 gold badges8585 silver badges180180 bronze badges

answered 8 hours ago

Nate EldredgeNate Eldredge

67.1k99 gold badges8585 silver badges180180 bronze badges

answered 8 hours ago

Nate EldredgeNate Eldredge

67.1k99 gold badges8585 silver badges180180 bronze badges