Mixing
Intermediate Value Theorem in $ mathbb{R}^2$ and $ mathbb{R}^3$
$begingroup$
Let $f:mathbb{R}^n rightarrow mathbb{R} $ be a continuous function and let $a,b in mathbb{R}^n$. Let $g:mathbb{R} rightarrow mathbb{R} $ be defined as:
$ g(t) = f(ta + (1-t)b) $.
i) Prove that $g$ is continuous.
My answer: I have done this using the fact that g is a composition of continuous functions and the fact that f is continuous
ii) Prove that the following Intermediate Value Theorem is true for $f$:
If $d$ is a number between $f(a)$ and $f(b)$ then there exists a point $c in mathbb{R}^n $ such that $f(c)=d$.
My attempt at a proof:
First we make the observation that $g(0)=f(b)$ and that $g(1)=f(a)$.
Since $g$ is a real and continuous function $implies$ there exists a $c' in[0,1]$ such that $g(c')=d$ since d is a number between $f(a)$ and $f(b)$. Thus $f(c'a+(1-c')b)=d$ and were are done.
iii) State and prove a similar Intermediate Value Theorem for continuous functions $h:R rightarrow mathbb{R}$ and $h':B rightarrow mathbb{R}$ where $R$ is an open or closes rectangle in $mathbb{R}^2$ and B is an open or closed ball in $mathbb{R}^3$
My thoughts: Since we have proven it for $mathbb{R}^n$ is the arguement not just to say that both sets are subsets of $mathbb{R}^n$?
I would like to know if the answer to iii) is that simple and also if my proof in ii) is correct? Thanks in advance!
real-analysis functions continuity
asked Jan 27 at 11:14
CruZCruZ
638
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}^n rightarrow mathbb{R} $ be a continuous function and let $a,b in mathbb{R}^n$. Let $g:mathbb{R} rightarrow mathbb{R} $ be defined as:
$ g(t) = f(ta + (1-t)b) $.
i) Prove that $g$ is continuous.
My answer: I have done this using the fact that g is a composition of continuous functions and the fact that f is continuous
ii) Prove that the following Intermediate Value Theorem is true for $f$:
If $d$ is a number between $f(a)$ and $f(b)$ then there exists a point $c in mathbb{R}^n $ such that $f(c)=d$.
My attempt at a proof:
First we make the observation that $g(0)=f(b)$ and that $g(1)=f(a)$.
Since $g$ is a real and continuous function $implies$ there exists a $c' in[0,1]$ such that $g(c')=d$ since d is a number between $f(a)$ and $f(b)$. Thus $f(c'a+(1-c')b)=d$ and were are done.
iii) State and prove a similar Intermediate Value Theorem for continuous functions $h:R rightarrow mathbb{R}$ and $h':B rightarrow mathbb{R}$ where $R$ is an open or closes rectangle in $mathbb{R}^2$ and B is an open or closed ball in $mathbb{R}^3$
My thoughts: Since we have proven it for $mathbb{R}^n$ is the arguement not just to say that both sets are subsets of $mathbb{R}^n$?
I would like to know if the answer to iii) is that simple and also if my proof in ii) is correct? Thanks in advance!
real-analysis functions continuity
asked Jan 27 at 11:14
CruZCruZ
638
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}^n rightarrow mathbb{R} $ be a continuous function and let $a,b in mathbb{R}^n$. Let $g:mathbb{R} rightarrow mathbb{R} $ be defined as:
$ g(t) = f(ta + (1-t)b) $.
i) Prove that $g$ is continuous.
My answer: I have done this using the fact that g is a composition of continuous functions and the fact that f is continuous
ii) Prove that the following Intermediate Value Theorem is true for $f$:
If $d$ is a number between $f(a)$ and $f(b)$ then there exists a point $c in mathbb{R}^n $ such that $f(c)=d$.
My attempt at a proof:
First we make the observation that $g(0)=f(b)$ and that $g(1)=f(a)$.
Since $g$ is a real and continuous function $implies$ there exists a $c' in[0,1]$ such that $g(c')=d$ since d is a number between $f(a)$ and $f(b)$. Thus $f(c'a+(1-c')b)=d$ and were are done.
iii) State and prove a similar Intermediate Value Theorem for continuous functions $h:R rightarrow mathbb{R}$ and $h':B rightarrow mathbb{R}$ where $R$ is an open or closes rectangle in $mathbb{R}^2$ and B is an open or closed ball in $mathbb{R}^3$
My thoughts: Since we have proven it for $mathbb{R}^n$ is the arguement not just to say that both sets are subsets of $mathbb{R}^n$?
I would like to know if the answer to iii) is that simple and also if my proof in ii) is correct? Thanks in advance!
real-analysis functions continuity
asked Jan 27 at 11:14
CruZCruZ
638
$endgroup$
Let $f:mathbb{R}^n rightarrow mathbb{R} $ be a continuous function and let $a,b in mathbb{R}^n$. Let $g:mathbb{R} rightarrow mathbb{R} $ be defined as:
$ g(t) = f(ta + (1-t)b) $.
i) Prove that $g$ is continuous.
My answer: I have done this using the fact that g is a composition of continuous functions and the fact that f is continuous
ii) Prove that the following Intermediate Value Theorem is true for $f$:
If $d$ is a number between $f(a)$ and $f(b)$ then there exists a point $c in mathbb{R}^n $ such that $f(c)=d$.
My attempt at a proof:
First we make the observation that $g(0)=f(b)$ and that $g(1)=f(a)$.
Since $g$ is a real and continuous function $implies$ there exists a $c' in[0,1]$ such that $g(c')=d$ since d is a number between $f(a)$ and $f(b)$. Thus $f(c'a+(1-c')b)=d$ and were are done.
iii) State and prove a similar Intermediate Value Theorem for continuous functions $h:R rightarrow mathbb{R}$ and $h':B rightarrow mathbb{R}$ where $R$ is an open or closes rectangle in $mathbb{R}^2$ and B is an open or closed ball in $mathbb{R}^3$
My thoughts: Since we have proven it for $mathbb{R}^n$ is the arguement not just to say that both sets are subsets of $mathbb{R}^n$?
I would like to know if the answer to iii) is that simple and also if my proof in ii) is correct? Thanks in advance!
real-analysis functions continuity
real-analysis functions continuity
asked Jan 27 at 11:14
CruZCruZ
638
asked Jan 27 at 11:14
CruZCruZ
638
asked Jan 27 at 11:14
CruZCruZ
638
asked Jan 27 at 11:14
CruZCruZ
638
asked Jan 27 at 11:14
CruZCruZ
638
638
add a comment |
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I would like to know if the answer to iii) is that simple and also if my proof in ii) is correct? Thanks in advance!
No, you have to do a little bit more, because for given $a,bin R$ and $d$ between $h(a)$ and $h(b)$, you have to check that $c$ is still in $R$ and not in $mathbb R^2setminus R$. But using the same ansatz as above, you have just to do a slight change to get a $cin R$ such that $h(c)=d$.
Your proof i) and ii) is correct, but at ii) you should define $c:=c'a+(1-c')b$ at the end to get the $c$.
answered Jan 27 at 11:23
Mundron SchmidtMundron Schmidt
7,5042729
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$begingroup$
I would like to know if the answer to iii) is that simple and also if my proof in ii) is correct? Thanks in advance!
No, you have to do a little bit more, because for given $a,bin R$ and $d$ between $h(a)$ and $h(b)$, you have to check that $c$ is still in $R$ and not in $mathbb R^2setminus R$. But using the same ansatz as above, you have just to do a slight change to get a $cin R$ such that $h(c)=d$.
Your proof i) and ii) is correct, but at ii) you should define $c:=c'a+(1-c')b$ at the end to get the $c$.
answered Jan 27 at 11:23
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
add a comment |
$begingroup$
I would like to know if the answer to iii) is that simple and also if my proof in ii) is correct? Thanks in advance!
No, you have to do a little bit more, because for given $a,bin R$ and $d$ between $h(a)$ and $h(b)$, you have to check that $c$ is still in $R$ and not in $mathbb R^2setminus R$. But using the same ansatz as above, you have just to do a slight change to get a $cin R$ such that $h(c)=d$.
Your proof i) and ii) is correct, but at ii) you should define $c:=c'a+(1-c')b$ at the end to get the $c$.
answered Jan 27 at 11:23
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
add a comment |
$begingroup$
I would like to know if the answer to iii) is that simple and also if my proof in ii) is correct? Thanks in advance!
No, you have to do a little bit more, because for given $a,bin R$ and $d$ between $h(a)$ and $h(b)$, you have to check that $c$ is still in $R$ and not in $mathbb R^2setminus R$. But using the same ansatz as above, you have just to do a slight change to get a $cin R$ such that $h(c)=d$.
Your proof i) and ii) is correct, but at ii) you should define $c:=c'a+(1-c')b$ at the end to get the $c$.
answered Jan 27 at 11:23
Mundron SchmidtMundron Schmidt
7,5042729
$endgroup$
I would like to know if the answer to iii) is that simple and also if my proof in ii) is correct? Thanks in advance!
No, you have to do a little bit more, because for given $a,bin R$ and $d$ between $h(a)$ and $h(b)$, you have to check that $c$ is still in $R$ and not in $mathbb R^2setminus R$. But using the same ansatz as above, you have just to do a slight change to get a $cin R$ such that $h(c)=d$.
Your proof i) and ii) is correct, but at ii) you should define $c:=c'a+(1-c')b$ at the end to get the $c$.
answered Jan 27 at 11:23
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 27 at 11:23
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 27 at 11:23
Mundron SchmidtMundron Schmidt
7,5042729
answered Jan 27 at 11:23
Mundron SchmidtMundron Schmidt
7,5042729
7,5042729
add a comment |
add a comment |
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