Mixing
Closed or Open Intervals in Extreme Value Theorem, Rolle's Theorem, and Mean Value Theorem
Clash Royale CLAN TAG#URR8PPP
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I am a little confused by the openness of intervals in Extreme Value Theorem (EVT), Rolle's Theorem (RT), and Mean Value Theorem (MVT):
EVT: Given a function $f:[a,b]rightarrow R$ is continuous, $f$ attains its maximum and minimum at some $cin [a,b]$. (closed intervals)
RT: Given $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, if $f(a)=f(b)$, then there exists $cin (a,b)$ such that $f'(c)=0$. (I believe if $f$ is differentiable on $[a,b]$, then, there exists $cin [a,b]$ such that $f'(c)=0$. Why do we need an open interval in this statement?)
MVT: Given $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $cin (a,b)$ such that $f'(c)=fracf(b)-f(a)b-a$. (Obviously, MVT goes with RT. Again, I believe we can just replace "differentiable on $(a,b)$ by "differentiable on $[a,b]$", and then get "there exists $cin [a,b]$ such that $dots$." Do we use an open interval for some specific reason?)
Thanks in advance for any explanation.
real-analysis derivatives continuity
asked Jul 27 at 1:27
James Wang
876
add a comment |Â
up vote
2
down vote
favorite
I am a little confused by the openness of intervals in Extreme Value Theorem (EVT), Rolle's Theorem (RT), and Mean Value Theorem (MVT):
EVT: Given a function $f:[a,b]rightarrow R$ is continuous, $f$ attains its maximum and minimum at some $cin [a,b]$. (closed intervals)
RT: Given $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, if $f(a)=f(b)$, then there exists $cin (a,b)$ such that $f'(c)=0$. (I believe if $f$ is differentiable on $[a,b]$, then, there exists $cin [a,b]$ such that $f'(c)=0$. Why do we need an open interval in this statement?)
MVT: Given $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $cin (a,b)$ such that $f'(c)=fracf(b)-f(a)b-a$. (Obviously, MVT goes with RT. Again, I believe we can just replace "differentiable on $(a,b)$ by "differentiable on $[a,b]$", and then get "there exists $cin [a,b]$ such that $dots$." Do we use an open interval for some specific reason?)
Thanks in advance for any explanation.
real-analysis derivatives continuity
asked Jul 27 at 1:27
James Wang
876
4
RT: Your second version, asks for more (differentiability at more points), and gives less (the point $c$ is less clear where it is since now it could also be at $a$ or at $b$). The original version is stronger. The second version is true, though.
– user578878
Jul 27 at 1:32
3
For RT, to see why open intervals are "right" for differentiability, think about the upper half of a circle.
– Randall
Jul 27 at 1:36
4
The extreme value theorem requires a closed interval. The max / min may be at an endpoint. Over an open interval there may not be a max or a min. Rolles theorem / MVT still hold over closed intervals, but they telll you that there will be special points in the interior of the interval, i.e. not at the end points.
– Doug M
Jul 27 at 1:50
2
AH, FINALLY SOMEONE ASKS THIS!
– Nick
Jul 27 at 1:57
2
I put questions like this on topology exams all the time. It's good for students to think about why these theorems are written this way.
– Randall
Jul 27 at 1:57
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am a little confused by the openness of intervals in Extreme Value Theorem (EVT), Rolle's Theorem (RT), and Mean Value Theorem (MVT):
EVT: Given a function $f:[a,b]rightarrow R$ is continuous, $f$ attains its maximum and minimum at some $cin [a,b]$. (closed intervals)
RT: Given $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, if $f(a)=f(b)$, then there exists $cin (a,b)$ such that $f'(c)=0$. (I believe if $f$ is differentiable on $[a,b]$, then, there exists $cin [a,b]$ such that $f'(c)=0$. Why do we need an open interval in this statement?)
MVT: Given $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $cin (a,b)$ such that $f'(c)=fracf(b)-f(a)b-a$. (Obviously, MVT goes with RT. Again, I believe we can just replace "differentiable on $(a,b)$ by "differentiable on $[a,b]$", and then get "there exists $cin [a,b]$ such that $dots$." Do we use an open interval for some specific reason?)
Thanks in advance for any explanation.
real-analysis derivatives continuity
asked Jul 27 at 1:27
James Wang
876
I am a little confused by the openness of intervals in Extreme Value Theorem (EVT), Rolle's Theorem (RT), and Mean Value Theorem (MVT):
EVT: Given a function $f:[a,b]rightarrow R$ is continuous, $f$ attains its maximum and minimum at some $cin [a,b]$. (closed intervals)
RT: Given $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, if $f(a)=f(b)$, then there exists $cin (a,b)$ such that $f'(c)=0$. (I believe if $f$ is differentiable on $[a,b]$, then, there exists $cin [a,b]$ such that $f'(c)=0$. Why do we need an open interval in this statement?)
MVT: Given $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $cin (a,b)$ such that $f'(c)=fracf(b)-f(a)b-a$. (Obviously, MVT goes with RT. Again, I believe we can just replace "differentiable on $(a,b)$ by "differentiable on $[a,b]$", and then get "there exists $cin [a,b]$ such that $dots$." Do we use an open interval for some specific reason?)
Thanks in advance for any explanation.
real-analysis derivatives continuity
asked Jul 27 at 1:27
James Wang
876
asked Jul 27 at 1:27
James Wang
876
asked Jul 27 at 1:27
James Wang
876
asked Jul 27 at 1:27
James Wang
876
876
4
RT: Your second version, asks for more (differentiability at more points), and gives less (the point $c$ is less clear where it is since now it could also be at $a$ or at $b$). The original version is stronger. The second version is true, though.
– user578878
Jul 27 at 1:32
3
For RT, to see why open intervals are "right" for differentiability, think about the upper half of a circle.
– Randall
Jul 27 at 1:36
4
The extreme value theorem requires a closed interval. The max / min may be at an endpoint. Over an open interval there may not be a max or a min. Rolles theorem / MVT still hold over closed intervals, but they telll you that there will be special points in the interior of the interval, i.e. not at the end points.
– Doug M
Jul 27 at 1:50
2
AH, FINALLY SOMEONE ASKS THIS!
– Nick
Jul 27 at 1:57
2
I put questions like this on topology exams all the time. It's good for students to think about why these theorems are written this way.
– Randall
Jul 27 at 1:57
add a comment |Â
4
RT: Your second version, asks for more (differentiability at more points), and gives less (the point $c$ is less clear where it is since now it could also be at $a$ or at $b$). The original version is stronger. The second version is true, though.
– user578878
Jul 27 at 1:32
3
For RT, to see why open intervals are "right" for differentiability, think about the upper half of a circle.
– Randall
Jul 27 at 1:36
4
The extreme value theorem requires a closed interval. The max / min may be at an endpoint. Over an open interval there may not be a max or a min. Rolles theorem / MVT still hold over closed intervals, but they telll you that there will be special points in the interior of the interval, i.e. not at the end points.
– Doug M
Jul 27 at 1:50
2
AH, FINALLY SOMEONE ASKS THIS!
– Nick
Jul 27 at 1:57
2
I put questions like this on topology exams all the time. It's good for students to think about why these theorems are written this way.
– Randall
Jul 27 at 1:57
4
4
RT: Your second version, asks for more (differentiability at more points), and gives less (the point $c$ is less clear where it is since now it could also be at $a$ or at $b$). The original version is stronger. The second version is true, though.
– user578878
Jul 27 at 1:32
RT: Your second version, asks for more (differentiability at more points), and gives less (the point $c$ is less clear where it is since now it could also be at $a$ or at $b$). The original version is stronger. The second version is true, though.
– user578878
Jul 27 at 1:32
3
3
For RT, to see why open intervals are "right" for differentiability, think about the upper half of a circle.
– Randall
Jul 27 at 1:36
For RT, to see why open intervals are "right" for differentiability, think about the upper half of a circle.
– Randall
Jul 27 at 1:36
4
4
The extreme value theorem requires a closed interval. The max / min may be at an endpoint. Over an open interval there may not be a max or a min. Rolles theorem / MVT still hold over closed intervals, but they telll you that there will be special points in the interior of the interval, i.e. not at the end points.
– Doug M
Jul 27 at 1:50
The extreme value theorem requires a closed interval. The max / min may be at an endpoint. Over an open interval there may not be a max or a min. Rolles theorem / MVT still hold over closed intervals, but they telll you that there will be special points in the interior of the interval, i.e. not at the end points.
– Doug M
Jul 27 at 1:50
2
2
AH, FINALLY SOMEONE ASKS THIS!
– Nick
Jul 27 at 1:57
AH, FINALLY SOMEONE ASKS THIS!
– Nick
Jul 27 at 1:57
2
2
I put questions like this on topology exams all the time. It's good for students to think about why these theorems are written this way.
– Randall
Jul 27 at 1:57
I put questions like this on topology exams all the time. It's good for students to think about why these theorems are written this way.
– Randall
Jul 27 at 1:57
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
For the EVT, the statement is clearly potentially false if you delete endpoints. E.g., the function $f(x) = frac1x$ on $(0,1)$ has neither a max nor a min. If you try to get cute and give it domain $(0,1]$ then it has a min but still no max. If you want to be guaranteed both, you need a closed and bounded interval.
For RT, consider the upper half of a circle. Let's take $f(x) =sqrt1-x^2$ for definiteness. For the "right" RT, note that this is continuous on $[-1,1]$ and differentiable on $(-1,1)$. Connecting the dots at the ends, RT says there should be a horizontal tangent somewhere. Well, sure, it's at the north pole $(0,1)$. However, you cannot demand a version of RT that requires differentiability on $[-1,1]$ and retain this example: the function is not differentiable on the closed interval $[-1,1]$ as the derivative blows up at the endpoints (vertical tangents). This is what user nextpuzzle was getting at when they said the original version is stronger.
As you've noted, RT is roughly equivalent to the MVT, so the same reason suffices.
answered Jul 27 at 1:41
Randall
7,2201825
"you cannot ask this to work with differentiability on $[-1,1]$" can be understood as saying that the theorem wouldn't be true. You want to say that asking for differentiability on $[-1,1]$ would give a theorem that, while true, couldn't be applied to that particular $f$.
– user578878
Jul 27 at 1:52
@nextpuzzle Fair point. By "this" I meant this particular function is lost as an example.
– Randall
Jul 27 at 1:53
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
For the EVT, the statement is clearly potentially false if you delete endpoints. E.g., the function $f(x) = frac1x$ on $(0,1)$ has neither a max nor a min. If you try to get cute and give it domain $(0,1]$ then it has a min but still no max. If you want to be guaranteed both, you need a closed and bounded interval.
For RT, consider the upper half of a circle. Let's take $f(x) =sqrt1-x^2$ for definiteness. For the "right" RT, note that this is continuous on $[-1,1]$ and differentiable on $(-1,1)$. Connecting the dots at the ends, RT says there should be a horizontal tangent somewhere. Well, sure, it's at the north pole $(0,1)$. However, you cannot demand a version of RT that requires differentiability on $[-1,1]$ and retain this example: the function is not differentiable on the closed interval $[-1,1]$ as the derivative blows up at the endpoints (vertical tangents). This is what user nextpuzzle was getting at when they said the original version is stronger.
As you've noted, RT is roughly equivalent to the MVT, so the same reason suffices.
answered Jul 27 at 1:41
Randall
7,2201825
"you cannot ask this to work with differentiability on $[-1,1]$" can be understood as saying that the theorem wouldn't be true. You want to say that asking for differentiability on $[-1,1]$ would give a theorem that, while true, couldn't be applied to that particular $f$.
– user578878
Jul 27 at 1:52
@nextpuzzle Fair point. By "this" I meant this particular function is lost as an example.
– Randall
Jul 27 at 1:53
add a comment |Â
up vote
4
down vote
accepted
For the EVT, the statement is clearly potentially false if you delete endpoints. E.g., the function $f(x) = frac1x$ on $(0,1)$ has neither a max nor a min. If you try to get cute and give it domain $(0,1]$ then it has a min but still no max. If you want to be guaranteed both, you need a closed and bounded interval.
For RT, consider the upper half of a circle. Let's take $f(x) =sqrt1-x^2$ for definiteness. For the "right" RT, note that this is continuous on $[-1,1]$ and differentiable on $(-1,1)$. Connecting the dots at the ends, RT says there should be a horizontal tangent somewhere. Well, sure, it's at the north pole $(0,1)$. However, you cannot demand a version of RT that requires differentiability on $[-1,1]$ and retain this example: the function is not differentiable on the closed interval $[-1,1]$ as the derivative blows up at the endpoints (vertical tangents). This is what user nextpuzzle was getting at when they said the original version is stronger.
As you've noted, RT is roughly equivalent to the MVT, so the same reason suffices.
answered Jul 27 at 1:41
Randall
7,2201825
"you cannot ask this to work with differentiability on $[-1,1]$" can be understood as saying that the theorem wouldn't be true. You want to say that asking for differentiability on $[-1,1]$ would give a theorem that, while true, couldn't be applied to that particular $f$.
– user578878
Jul 27 at 1:52
@nextpuzzle Fair point. By "this" I meant this particular function is lost as an example.
– Randall
Jul 27 at 1:53
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
For the EVT, the statement is clearly potentially false if you delete endpoints. E.g., the function $f(x) = frac1x$ on $(0,1)$ has neither a max nor a min. If you try to get cute and give it domain $(0,1]$ then it has a min but still no max. If you want to be guaranteed both, you need a closed and bounded interval.
For RT, consider the upper half of a circle. Let's take $f(x) =sqrt1-x^2$ for definiteness. For the "right" RT, note that this is continuous on $[-1,1]$ and differentiable on $(-1,1)$. Connecting the dots at the ends, RT says there should be a horizontal tangent somewhere. Well, sure, it's at the north pole $(0,1)$. However, you cannot demand a version of RT that requires differentiability on $[-1,1]$ and retain this example: the function is not differentiable on the closed interval $[-1,1]$ as the derivative blows up at the endpoints (vertical tangents). This is what user nextpuzzle was getting at when they said the original version is stronger.
As you've noted, RT is roughly equivalent to the MVT, so the same reason suffices.
answered Jul 27 at 1:41
Randall
7,2201825
For the EVT, the statement is clearly potentially false if you delete endpoints. E.g., the function $f(x) = frac1x$ on $(0,1)$ has neither a max nor a min. If you try to get cute and give it domain $(0,1]$ then it has a min but still no max. If you want to be guaranteed both, you need a closed and bounded interval.
For RT, consider the upper half of a circle. Let's take $f(x) =sqrt1-x^2$ for definiteness. For the "right" RT, note that this is continuous on $[-1,1]$ and differentiable on $(-1,1)$. Connecting the dots at the ends, RT says there should be a horizontal tangent somewhere. Well, sure, it's at the north pole $(0,1)$. However, you cannot demand a version of RT that requires differentiability on $[-1,1]$ and retain this example: the function is not differentiable on the closed interval $[-1,1]$ as the derivative blows up at the endpoints (vertical tangents). This is what user nextpuzzle was getting at when they said the original version is stronger.
As you've noted, RT is roughly equivalent to the MVT, so the same reason suffices.
answered Jul 27 at 1:41
Randall
7,2201825
edited Jul 27 at 1:56
answered Jul 27 at 1:41
Randall
7,2201825
answered Jul 27 at 1:41
Randall
7,2201825
answered Jul 27 at 1:41
Randall
7,2201825
7,2201825
"you cannot ask this to work with differentiability on $[-1,1]$" can be understood as saying that the theorem wouldn't be true. You want to say that asking for differentiability on $[-1,1]$ would give a theorem that, while true, couldn't be applied to that particular $f$.
– user578878
Jul 27 at 1:52
@nextpuzzle Fair point. By "this" I meant this particular function is lost as an example.
– Randall
Jul 27 at 1:53
add a comment |Â
"you cannot ask this to work with differentiability on $[-1,1]$" can be understood as saying that the theorem wouldn't be true. You want to say that asking for differentiability on $[-1,1]$ would give a theorem that, while true, couldn't be applied to that particular $f$.
– user578878
Jul 27 at 1:52
@nextpuzzle Fair point. By "this" I meant this particular function is lost as an example.
– Randall
Jul 27 at 1:53
"you cannot ask this to work with differentiability on $[-1,1]$" can be understood as saying that the theorem wouldn't be true. You want to say that asking for differentiability on $[-1,1]$ would give a theorem that, while true, couldn't be applied to that particular $f$.
– user578878
Jul 27 at 1:52
"you cannot ask this to work with differentiability on $[-1,1]$" can be understood as saying that the theorem wouldn't be true. You want to say that asking for differentiability on $[-1,1]$ would give a theorem that, while true, couldn't be applied to that particular $f$.
– user578878
Jul 27 at 1:52
@nextpuzzle Fair point. By "this" I meant this particular function is lost as an example.
– Randall
Jul 27 at 1:53
@nextpuzzle Fair point. By "this" I meant this particular function is lost as an example.
– Randall
Jul 27 at 1:53
add a comment |Â
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4
RT: Your second version, asks for more (differentiability at more points), and gives less (the point $c$ is less clear where it is since now it could also be at $a$ or at $b$). The original version is stronger. The second version is true, though.
– user578878
Jul 27 at 1:32
3
For RT, to see why open intervals are "right" for differentiability, think about the upper half of a circle.
– Randall
Jul 27 at 1:36
4
The extreme value theorem requires a closed interval. The max / min may be at an endpoint. Over an open interval there may not be a max or a min. Rolles theorem / MVT still hold over closed intervals, but they telll you that there will be special points in the interior of the interval, i.e. not at the end points.
– Doug M
Jul 27 at 1:50
2
AH, FINALLY SOMEONE ASKS THIS!
– Nick
Jul 27 at 1:57
2
I put questions like this on topology exams all the time. It's good for students to think about why these theorems are written this way.
– Randall
Jul 27 at 1:57