A proof for a continuous differentiable function [closed]
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I want to prove that for a continuous differentiable function, if there is not any minimum between two minima of the function then there is a maximum between them and this maximum is unique.
calculus
edited Aug 13 at 5:50
Henrik
5,81471930
asked Aug 13 at 5:25
M.Bashiri
1024
closed as off-topic by Henrik, Shailesh, Claude Leibovici, Adrian Keister, amWhy Aug 13 at 13:07
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." – Henrik, Shailesh, Claude Leibovici, Adrian Keister, amWhy
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I want to prove that for a continuous differentiable function, if there is not any minimum between two minima of the function then there is a maximum between them and this maximum is unique.
calculus
edited Aug 13 at 5:50
Henrik
5,81471930
asked Aug 13 at 5:25
M.Bashiri
1024
closed as off-topic by Henrik, Shailesh, Claude Leibovici, Adrian Keister, amWhy Aug 13 at 13:07
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." – Henrik, Shailesh, Claude Leibovici, Adrian Keister, amWhy
I want to prove that in a continuous function mins and maxes are sequentially
– M.Bashiri
Aug 13 at 5:31
It means that after a min there should be a maximum or min,max,min,max.....
– M.Bashiri
Aug 13 at 5:37
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to prove that for a continuous differentiable function, if there is not any minimum between two minima of the function then there is a maximum between them and this maximum is unique.
calculus
edited Aug 13 at 5:50
Henrik
5,81471930
asked Aug 13 at 5:25
M.Bashiri
1024
I want to prove that for a continuous differentiable function, if there is not any minimum between two minima of the function then there is a maximum between them and this maximum is unique.
calculus
edited Aug 13 at 5:50
Henrik
5,81471930
asked Aug 13 at 5:25
M.Bashiri
1024
edited Aug 13 at 5:50
Henrik
5,81471930
edited Aug 13 at 5:50
Henrik
5,81471930
edited Aug 13 at 5:50
Henrik
5,81471930
5,81471930
asked Aug 13 at 5:25
M.Bashiri
1024
asked Aug 13 at 5:25
M.Bashiri
1024
asked Aug 13 at 5:25
M.Bashiri
1024
1024
closed as off-topic by Henrik, Shailesh, Claude Leibovici, Adrian Keister, amWhy Aug 13 at 13:07
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." – Henrik, Shailesh, Claude Leibovici, Adrian Keister, amWhy
closed as off-topic by Henrik, Shailesh, Claude Leibovici, Adrian Keister, amWhy Aug 13 at 13:07
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." – Henrik, Shailesh, Claude Leibovici, Adrian Keister, amWhy
I want to prove that in a continuous function mins and maxes are sequentially
– M.Bashiri
Aug 13 at 5:31
It means that after a min there should be a maximum or min,max,min,max.....
– M.Bashiri
Aug 13 at 5:37
add a comment |Â
I want to prove that in a continuous function mins and maxes are sequentially
– M.Bashiri
Aug 13 at 5:31
It means that after a min there should be a maximum or min,max,min,max.....
– M.Bashiri
Aug 13 at 5:37
I want to prove that in a continuous function mins and maxes are sequentially
– M.Bashiri
Aug 13 at 5:31
I want to prove that in a continuous function mins and maxes are sequentially
– M.Bashiri
Aug 13 at 5:31
It means that after a min there should be a maximum or min,max,min,max.....
– M.Bashiri
Aug 13 at 5:37
It means that after a min there should be a maximum or min,max,min,max.....
– M.Bashiri
Aug 13 at 5:37
add a comment |Â
1 Answer
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Let $a,b$ the points where those minima are attained.
If $f(a)>f(b)$, then there is some point $xin(a,b)$ where $f(x)>f(a)$, otherwise $a$ wouldn't be a local minimum. Therefore, by the intermediate value theorem there is $b'in(x,b)$ such that $f(a)=f(b')$.
Similarly if $f(a)<f(b)$ there is $a'in(a,b)$ such that $f(a')=f(b)$.
Therefore, we can assume that $f(a)=f(b)$. Since $f$ can't be constant, there are points where the function is strictly larger than $f(a)$. Since $f$ is continuous is has a maximum in $[a,b]$, but since there are points in which the value is larger than $f(a)=f(b)$, the maximum must be in $(a,b)$.
If there were two maxima, then a similar argument as above, would show that in between them there are minima.
Beat me to it. In particular, this argument doesn't need to assume differentiability, much less continuous differentiability.
– Theo Bendit
Aug 13 at 5:46
thanks for your help
– M.Bashiri
Aug 13 at 5:50
Yes.its true. I did not try to prove in such a manner.I was using the derivative properties so I was forced to consider some constraints.
– M.Bashiri
Aug 13 at 6:22
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Let $a,b$ the points where those minima are attained.
If $f(a)>f(b)$, then there is some point $xin(a,b)$ where $f(x)>f(a)$, otherwise $a$ wouldn't be a local minimum. Therefore, by the intermediate value theorem there is $b'in(x,b)$ such that $f(a)=f(b')$.
Similarly if $f(a)<f(b)$ there is $a'in(a,b)$ such that $f(a')=f(b)$.
Therefore, we can assume that $f(a)=f(b)$. Since $f$ can't be constant, there are points where the function is strictly larger than $f(a)$. Since $f$ is continuous is has a maximum in $[a,b]$, but since there are points in which the value is larger than $f(a)=f(b)$, the maximum must be in $(a,b)$.
If there were two maxima, then a similar argument as above, would show that in between them there are minima.
Beat me to it. In particular, this argument doesn't need to assume differentiability, much less continuous differentiability.
– Theo Bendit
Aug 13 at 5:46
thanks for your help
– M.Bashiri
Aug 13 at 5:50
Yes.its true. I did not try to prove in such a manner.I was using the derivative properties so I was forced to consider some constraints.
– M.Bashiri
Aug 13 at 6:22
add a comment |Â
up vote
2
down vote
Let $a,b$ the points where those minima are attained.
If $f(a)>f(b)$, then there is some point $xin(a,b)$ where $f(x)>f(a)$, otherwise $a$ wouldn't be a local minimum. Therefore, by the intermediate value theorem there is $b'in(x,b)$ such that $f(a)=f(b')$.
Similarly if $f(a)<f(b)$ there is $a'in(a,b)$ such that $f(a')=f(b)$.
Therefore, we can assume that $f(a)=f(b)$. Since $f$ can't be constant, there are points where the function is strictly larger than $f(a)$. Since $f$ is continuous is has a maximum in $[a,b]$, but since there are points in which the value is larger than $f(a)=f(b)$, the maximum must be in $(a,b)$.
If there were two maxima, then a similar argument as above, would show that in between them there are minima.
Beat me to it. In particular, this argument doesn't need to assume differentiability, much less continuous differentiability.
– Theo Bendit
Aug 13 at 5:46
thanks for your help
– M.Bashiri
Aug 13 at 5:50
Yes.its true. I did not try to prove in such a manner.I was using the derivative properties so I was forced to consider some constraints.
– M.Bashiri
Aug 13 at 6:22
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $a,b$ the points where those minima are attained.
If $f(a)>f(b)$, then there is some point $xin(a,b)$ where $f(x)>f(a)$, otherwise $a$ wouldn't be a local minimum. Therefore, by the intermediate value theorem there is $b'in(x,b)$ such that $f(a)=f(b')$.
Similarly if $f(a)<f(b)$ there is $a'in(a,b)$ such that $f(a')=f(b)$.
Therefore, we can assume that $f(a)=f(b)$. Since $f$ can't be constant, there are points where the function is strictly larger than $f(a)$. Since $f$ is continuous is has a maximum in $[a,b]$, but since there are points in which the value is larger than $f(a)=f(b)$, the maximum must be in $(a,b)$.
If there were two maxima, then a similar argument as above, would show that in between them there are minima.
Let $a,b$ the points where those minima are attained.
If $f(a)>f(b)$, then there is some point $xin(a,b)$ where $f(x)>f(a)$, otherwise $a$ wouldn't be a local minimum. Therefore, by the intermediate value theorem there is $b'in(x,b)$ such that $f(a)=f(b')$.
Similarly if $f(a)<f(b)$ there is $a'in(a,b)$ such that $f(a')=f(b)$.
Therefore, we can assume that $f(a)=f(b)$. Since $f$ can't be constant, there are points where the function is strictly larger than $f(a)$. Since $f$ is continuous is has a maximum in $[a,b]$, but since there are points in which the value is larger than $f(a)=f(b)$, the maximum must be in $(a,b)$.
If there were two maxima, then a similar argument as above, would show that in between them there are minima.
edited Aug 13 at 5:47
answered Aug 13 at 5:45
user583185
Beat me to it. In particular, this argument doesn't need to assume differentiability, much less continuous differentiability.
– Theo Bendit
Aug 13 at 5:46
thanks for your help
– M.Bashiri
Aug 13 at 5:50
Yes.its true. I did not try to prove in such a manner.I was using the derivative properties so I was forced to consider some constraints.
– M.Bashiri
Aug 13 at 6:22
add a comment |Â
Beat me to it. In particular, this argument doesn't need to assume differentiability, much less continuous differentiability.
– Theo Bendit
Aug 13 at 5:46
thanks for your help
– M.Bashiri
Aug 13 at 5:50
Yes.its true. I did not try to prove in such a manner.I was using the derivative properties so I was forced to consider some constraints.
– M.Bashiri
Aug 13 at 6:22
Beat me to it. In particular, this argument doesn't need to assume differentiability, much less continuous differentiability.
– Theo Bendit
Aug 13 at 5:46
Beat me to it. In particular, this argument doesn't need to assume differentiability, much less continuous differentiability.
– Theo Bendit
Aug 13 at 5:46
thanks for your help
– M.Bashiri
Aug 13 at 5:50
thanks for your help
– M.Bashiri
Aug 13 at 5:50
Yes.its true. I did not try to prove in such a manner.I was using the derivative properties so I was forced to consider some constraints.
– M.Bashiri
Aug 13 at 6:22
Yes.its true. I did not try to prove in such a manner.I was using the derivative properties so I was forced to consider some constraints.
– M.Bashiri
Aug 13 at 6:22
add a comment |Â
I want to prove that in a continuous function mins and maxes are sequentially
– M.Bashiri
Aug 13 at 5:31
It means that after a min there should be a maximum or min,max,min,max.....
– M.Bashiri
Aug 13 at 5:37