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Order between one to one functions and their inverses

3

1

$begingroup$

Let $f,g :R to R $ one to one functions such that
$f(x)< g(x), forall x in R $

Is it true that $f^-1(x)>g^-1(x), forall x in R$??

I'd say yes, thinking at their graphs: if the graph of f is below the graph of g, when we take the symmetry with respect to the line $y=x$ the graph of $f^-1(x)$ will be above the graph of $g^-1$. But i am interested in a rigorously formulated proof . If it is true...

functions

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asked 4 hours ago

amarius8312amarius8312

29018

$endgroup$

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3

1

$begingroup$

Let $f,g :R to R $ one to one functions such that
$f(x)< g(x), forall x in R $

Is it true that $f^-1(x)>g^-1(x), forall x in R$??

I'd say yes, thinking at their graphs: if the graph of f is below the graph of g, when we take the symmetry with respect to the line $y=x$ the graph of $f^-1(x)$ will be above the graph of $g^-1$. But i am interested in a rigorously formulated proof . If it is true...

functions

share|cite|improve this question

asked 4 hours ago

amarius8312amarius8312

29018

$endgroup$

add a comment | 

$begingroup$

Let $f,g :R to R $ one to one functions such that
$f(x)< g(x), forall x in R $

Is it true that $f^-1(x)>g^-1(x), forall x in R$??

I'd say yes, thinking at their graphs: if the graph of f is below the graph of g, when we take the symmetry with respect to the line $y=x$ the graph of $f^-1(x)$ will be above the graph of $g^-1$. But i am interested in a rigorously formulated proof . If it is true...

functions

share|cite|improve this question

asked 4 hours ago

amarius8312amarius8312

29018

$endgroup$

share|cite|improve this question

asked 4 hours ago

amarius8312amarius8312

29018

share|cite|improve this question

asked 4 hours ago

amarius8312amarius8312

29018

asked 4 hours ago

amarius8312amarius8312

29018

asked 4 hours ago

amarius8312amarius8312

29018

2 Answers
2

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6

$begingroup$

No, if the graph of $g(x)$ is above the graph of $f(x)$, then the graph of $g^-1(x)$ is to the right of the graph of $f^-1(x)$. If these functions are increasing, then being to the right of another function is the same as being below it. However, if they are decreasing, then being to the right of another function is the same as being above it.

For a concrete counterexample, take $f(x)=-x$ and $g(x)=-x+1$. Then $f^-1(x)=-x$ and $g^-1(x)=-x+1$, so $g^-1(x)>f^-1(x)$ for all $x$.

share|cite|improve this answer

answered 4 hours ago

kccukccu

11.6k11231

$endgroup$

add a comment | 

0

$begingroup$

Right!
I'd say like this:
Let $f^-1(x)=a, g^-1(x)=b$
So we have $x=f(a)=g(b)$

Then from the hypothesis $g(b)>f(b)$ so $f(a)>f(b)$

And if we assume that f is strictly increasing it follows that $a>b$
So it's sufficient to add the condition that $f$ is increasing,
$g$ doesn't need to be.

share|cite|improve this answer

answered 3 hours ago

amarius8312amarius8312

29018

$endgroup$

add a comment | 

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6

$begingroup$

No, if the graph of $g(x)$ is above the graph of $f(x)$, then the graph of $g^-1(x)$ is to the right of the graph of $f^-1(x)$. If these functions are increasing, then being to the right of another function is the same as being below it. However, if they are decreasing, then being to the right of another function is the same as being above it.

For a concrete counterexample, take $f(x)=-x$ and $g(x)=-x+1$. Then $f^-1(x)=-x$ and $g^-1(x)=-x+1$, so $g^-1(x)>f^-1(x)$ for all $x$.

share|cite|improve this answer

answered 4 hours ago

kccukccu

11.6k11231

$endgroup$

add a comment | 

6

$begingroup$

No, if the graph of $g(x)$ is above the graph of $f(x)$, then the graph of $g^-1(x)$ is to the right of the graph of $f^-1(x)$. If these functions are increasing, then being to the right of another function is the same as being below it. However, if they are decreasing, then being to the right of another function is the same as being above it.

For a concrete counterexample, take $f(x)=-x$ and $g(x)=-x+1$. Then $f^-1(x)=-x$ and $g^-1(x)=-x+1$, so $g^-1(x)>f^-1(x)$ for all $x$.

share|cite|improve this answer

answered 4 hours ago

kccukccu

11.6k11231

$endgroup$

add a comment | 

$begingroup$

No, if the graph of $g(x)$ is above the graph of $f(x)$, then the graph of $g^-1(x)$ is to the right of the graph of $f^-1(x)$. If these functions are increasing, then being to the right of another function is the same as being below it. However, if they are decreasing, then being to the right of another function is the same as being above it.

For a concrete counterexample, take $f(x)=-x$ and $g(x)=-x+1$. Then $f^-1(x)=-x$ and $g^-1(x)=-x+1$, so $g^-1(x)>f^-1(x)$ for all $x$.

share|cite|improve this answer

answered 4 hours ago

kccukccu

11.6k11231

$endgroup$

share|cite|improve this answer

answered 4 hours ago

kccukccu

11.6k11231

answered 4 hours ago

kccukccu

11.6k11231

answered 4 hours ago

kccukccu

11.6k11231

0

$begingroup$

Right!
I'd say like this:
Let $f^-1(x)=a, g^-1(x)=b$
So we have $x=f(a)=g(b)$

Then from the hypothesis $g(b)>f(b)$ so $f(a)>f(b)$

And if we assume that f is strictly increasing it follows that $a>b$
So it's sufficient to add the condition that $f$ is increasing,
$g$ doesn't need to be.

share|cite|improve this answer

answered 3 hours ago

amarius8312amarius8312

29018

$endgroup$

add a comment | 

0

$begingroup$

Right!
I'd say like this:
Let $f^-1(x)=a, g^-1(x)=b$
So we have $x=f(a)=g(b)$

Then from the hypothesis $g(b)>f(b)$ so $f(a)>f(b)$

And if we assume that f is strictly increasing it follows that $a>b$
So it's sufficient to add the condition that $f$ is increasing,
$g$ doesn't need to be.

share|cite|improve this answer

answered 3 hours ago

amarius8312amarius8312

29018

$endgroup$

add a comment | 

$begingroup$

Right!
I'd say like this:
Let $f^-1(x)=a, g^-1(x)=b$
So we have $x=f(a)=g(b)$

Then from the hypothesis $g(b)>f(b)$ so $f(a)>f(b)$

And if we assume that f is strictly increasing it follows that $a>b$
So it's sufficient to add the condition that $f$ is increasing,
$g$ doesn't need to be.

share|cite|improve this answer

answered 3 hours ago

amarius8312amarius8312

29018

$endgroup$

share|cite|improve this answer

answered 3 hours ago

amarius8312amarius8312

29018

answered 3 hours ago

amarius8312amarius8312

29018

answered 3 hours ago

amarius8312amarius8312

29018