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How to prove that $f(x) = frac-2x+1(2x-1)^2-1$ is one-to-one on $(0,1)$?
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How do I prove that the function
$$f(x) = frac-2x+1(2x-1)^2-1$$
is one-to-one on the interval $(0,1)$? I have simplified the function $f(x)=f(y)$ and have a multivariable quadratic equation equal to zero. How do I conclude that the function is one to one?
real-analysis rational-functions
edited Sep 3 at 7:56
Asaf Karagila♦
294k31410737
asked Sep 2 at 17:47
Bret Hisey
1006
add a comment |Â
up vote
3
down vote
favorite
How do I prove that the function
$$f(x) = frac-2x+1(2x-1)^2-1$$
is one-to-one on the interval $(0,1)$? I have simplified the function $f(x)=f(y)$ and have a multivariable quadratic equation equal to zero. How do I conclude that the function is one to one?
real-analysis rational-functions
edited Sep 3 at 7:56
Asaf Karagila♦
294k31410737
asked Sep 2 at 17:47
Bret Hisey
1006
I apologize for leaving out a piece of information, I only need to know if the function is one to one from the interval (a,b).
– Bret Hisey
Sep 2 at 18:47
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
How do I prove that the function
$$f(x) = frac-2x+1(2x-1)^2-1$$
is one-to-one on the interval $(0,1)$? I have simplified the function $f(x)=f(y)$ and have a multivariable quadratic equation equal to zero. How do I conclude that the function is one to one?
real-analysis rational-functions
edited Sep 3 at 7:56
Asaf Karagila♦
294k31410737
asked Sep 2 at 17:47
Bret Hisey
1006
How do I prove that the function
$$f(x) = frac-2x+1(2x-1)^2-1$$
is one-to-one on the interval $(0,1)$? I have simplified the function $f(x)=f(y)$ and have a multivariable quadratic equation equal to zero. How do I conclude that the function is one to one?
real-analysis rational-functions
edited Sep 3 at 7:56
Asaf Karagila♦
294k31410737
asked Sep 2 at 17:47
Bret Hisey
1006
edited Sep 3 at 7:56
Asaf Karagila♦
294k31410737
edited Sep 3 at 7:56
Asaf Karagila♦
294k31410737
edited Sep 3 at 7:56
Asaf Karagila♦
294k31410737
294k31410737
asked Sep 2 at 17:47
Bret Hisey
1006
asked Sep 2 at 17:47
Bret Hisey
1006
asked Sep 2 at 17:47
Bret Hisey
1006
1006
I apologize for leaving out a piece of information, I only need to know if the function is one to one from the interval (a,b).
– Bret Hisey
Sep 2 at 18:47
add a comment |Â
I apologize for leaving out a piece of information, I only need to know if the function is one to one from the interval (a,b).
– Bret Hisey
Sep 2 at 18:47
I apologize for leaving out a piece of information, I only need to know if the function is one to one from the interval (a,b).
– Bret Hisey
Sep 2 at 18:47
I apologize for leaving out a piece of information, I only need to know if the function is one to one from the interval (a,b).
– Bret Hisey
Sep 2 at 18:47
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
10
down vote
You can't prove that, because it is not true. For instance,$$fleft(frac1-sqrt54right)=fleft(frac1+sqrt54right)=1.$$
answered Sep 2 at 17:57
José Carlos Santos
120k16101182
The question was changed/clarified, so now it is true.
– Don Hatch
Sep 3 at 6:28
add a comment |Â
up vote
6
down vote
Alt. hint: $,h(x)=1-2x,$ is a bijection, and $,g(x) = dfracxx^2-1,$ is not injective on $,Bbb R,$ (though it is injective on $,(-1,1),$ and $Bbb R setminus [-1,1]$ respectively). It follows that $,f = g circ h,$ is not injective on $,Bbb R,$.
[ EDIT ] Â Addressing OP's clarification that the question refers to
is one-to-one from the interval (0,1), note that $,g(x),$ is indeed injective on $,(-1,1),$ since $,a,b in (-1,1) implies ab+1 ne 0,$ and:$$
0 = g(a) - g(b) = fracaa^2-1 - fracbb^2-1= fracab^2-a-a^2b+b(a^2-1)(b^2-1) \[10px] = frac(ab+1)(b-a)(a^2-1)(b^2-1) quadimpliesquad a = b
$$
Given that $,hbig((0,1)big)=(-1,1),$ it follows that $,f = g circ h,$ is injective on $,(0,1),$.
answered Sep 2 at 18:28
dxiv
55.8k64798
add a comment |Â
up vote
4
down vote
Note that $f(x)$ has $2$ vertical asymptotes at
$$(2x-1)^2=1 implies 2x-1=pm 1 implies x=0,1$$
and it easy to check that
- $lim_xto -infty f(x)=0$
- $lim_xto 0^- f(x)=infty$
- $lim_xto 0^+ f(x)=-infty$
- $lim_xto 1^- f(x)=infty$
- $lim_xto 1^+ f(x)=+infty$
- $lim_xto infty f(x)=0$
and therefore, since the function is continuous in its domain, by intermediate value theorem it can't be one to one.
edited Sep 2 at 20:23
Yashas
172115
answered Sep 2 at 18:44
gimusi
71k73786
add a comment |Â
up vote
3
down vote
$f(x) = frac-2x+1(2x-1)^2-1$
$f'(x) = frac (-2x+1)'[(2x-1)^2 -1]- (-2x+1)[(2x-1)^2 - 1]'[(2x-1)^2 - 1]^2$
$=frac -2[(2x-1)^2 - 1]-(-2x + 1)2(2x-1)2[(2x-1)^2 - 1]^2=frac 2[1-(2x-1)^2] +4(2x -1)^2[(2x-1)^2 - 1]^2$
Now if $0 < x < 1$ then $-1 < 2x -1 < 1$ and $0le (2x -1)^2 < 1$ and $0< 1-(2x-1)^2 le 1$, while $[(2x-1)^2 - 1]^2 >0$ and $(2x -1)^2 > 0$.
So for $0 < x < 1$, $f'(x) > 0$ and $f$ is therefore strictly increasing and $1-1$.
answered Sep 2 at 22:43
fleablood
61.1k22676
add a comment |Â
up vote
0
down vote
One other pair of $x, y$ that give the same image are of the form: $$2xy-x-y+1$$ , with $(2x-1)^2-1 neq0$ and $(2y-1)^2-1 neq0$
answered Sep 2 at 18:21
dmtri
810317
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
You can't prove that, because it is not true. For instance,$$fleft(frac1-sqrt54right)=fleft(frac1+sqrt54right)=1.$$
answered Sep 2 at 17:57
José Carlos Santos
120k16101182
The question was changed/clarified, so now it is true.
– Don Hatch
Sep 3 at 6:28
add a comment |Â
up vote
10
down vote
You can't prove that, because it is not true. For instance,$$fleft(frac1-sqrt54right)=fleft(frac1+sqrt54right)=1.$$
answered Sep 2 at 17:57
José Carlos Santos
120k16101182
The question was changed/clarified, so now it is true.
– Don Hatch
Sep 3 at 6:28
add a comment |Â
up vote
10
down vote
up vote
10
down vote
You can't prove that, because it is not true. For instance,$$fleft(frac1-sqrt54right)=fleft(frac1+sqrt54right)=1.$$
answered Sep 2 at 17:57
José Carlos Santos
120k16101182
You can't prove that, because it is not true. For instance,$$fleft(frac1-sqrt54right)=fleft(frac1+sqrt54right)=1.$$
answered Sep 2 at 17:57
José Carlos Santos
120k16101182
answered Sep 2 at 17:57
José Carlos Santos
120k16101182
answered Sep 2 at 17:57
José Carlos Santos
120k16101182
answered Sep 2 at 17:57
José Carlos Santos
120k16101182
120k16101182
The question was changed/clarified, so now it is true.
– Don Hatch
Sep 3 at 6:28
add a comment |Â
The question was changed/clarified, so now it is true.
– Don Hatch
Sep 3 at 6:28
The question was changed/clarified, so now it is true.
– Don Hatch
Sep 3 at 6:28
The question was changed/clarified, so now it is true.
– Don Hatch
Sep 3 at 6:28
add a comment |Â
up vote
6
down vote
Alt. hint: $,h(x)=1-2x,$ is a bijection, and $,g(x) = dfracxx^2-1,$ is not injective on $,Bbb R,$ (though it is injective on $,(-1,1),$ and $Bbb R setminus [-1,1]$ respectively). It follows that $,f = g circ h,$ is not injective on $,Bbb R,$.
[ EDIT ] Â Addressing OP's clarification that the question refers to
is one-to-one from the interval (0,1), note that $,g(x),$ is indeed injective on $,(-1,1),$ since $,a,b in (-1,1) implies ab+1 ne 0,$ and:$$
0 = g(a) - g(b) = fracaa^2-1 - fracbb^2-1= fracab^2-a-a^2b+b(a^2-1)(b^2-1) \[10px] = frac(ab+1)(b-a)(a^2-1)(b^2-1) quadimpliesquad a = b
$$
Given that $,hbig((0,1)big)=(-1,1),$ it follows that $,f = g circ h,$ is injective on $,(0,1),$.
answered Sep 2 at 18:28
dxiv
55.8k64798
add a comment |Â
up vote
6
down vote
Alt. hint: $,h(x)=1-2x,$ is a bijection, and $,g(x) = dfracxx^2-1,$ is not injective on $,Bbb R,$ (though it is injective on $,(-1,1),$ and $Bbb R setminus [-1,1]$ respectively). It follows that $,f = g circ h,$ is not injective on $,Bbb R,$.
[ EDIT ] Â Addressing OP's clarification that the question refers to
is one-to-one from the interval (0,1), note that $,g(x),$ is indeed injective on $,(-1,1),$ since $,a,b in (-1,1) implies ab+1 ne 0,$ and:$$
0 = g(a) - g(b) = fracaa^2-1 - fracbb^2-1= fracab^2-a-a^2b+b(a^2-1)(b^2-1) \[10px] = frac(ab+1)(b-a)(a^2-1)(b^2-1) quadimpliesquad a = b
$$
Given that $,hbig((0,1)big)=(-1,1),$ it follows that $,f = g circ h,$ is injective on $,(0,1),$.
answered Sep 2 at 18:28
dxiv
55.8k64798
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Alt. hint: $,h(x)=1-2x,$ is a bijection, and $,g(x) = dfracxx^2-1,$ is not injective on $,Bbb R,$ (though it is injective on $,(-1,1),$ and $Bbb R setminus [-1,1]$ respectively). It follows that $,f = g circ h,$ is not injective on $,Bbb R,$.
[ EDIT ] Â Addressing OP's clarification that the question refers to
is one-to-one from the interval (0,1), note that $,g(x),$ is indeed injective on $,(-1,1),$ since $,a,b in (-1,1) implies ab+1 ne 0,$ and:$$
0 = g(a) - g(b) = fracaa^2-1 - fracbb^2-1= fracab^2-a-a^2b+b(a^2-1)(b^2-1) \[10px] = frac(ab+1)(b-a)(a^2-1)(b^2-1) quadimpliesquad a = b
$$
Given that $,hbig((0,1)big)=(-1,1),$ it follows that $,f = g circ h,$ is injective on $,(0,1),$.
answered Sep 2 at 18:28
dxiv
55.8k64798
Alt. hint: $,h(x)=1-2x,$ is a bijection, and $,g(x) = dfracxx^2-1,$ is not injective on $,Bbb R,$ (though it is injective on $,(-1,1),$ and $Bbb R setminus [-1,1]$ respectively). It follows that $,f = g circ h,$ is not injective on $,Bbb R,$.
[ EDIT ] Â Addressing OP's clarification that the question refers to
is one-to-one from the interval (0,1), note that $,g(x),$ is indeed injective on $,(-1,1),$ since $,a,b in (-1,1) implies ab+1 ne 0,$ and:$$
0 = g(a) - g(b) = fracaa^2-1 - fracbb^2-1= fracab^2-a-a^2b+b(a^2-1)(b^2-1) \[10px] = frac(ab+1)(b-a)(a^2-1)(b^2-1) quadimpliesquad a = b
$$
Given that $,hbig((0,1)big)=(-1,1),$ it follows that $,f = g circ h,$ is injective on $,(0,1),$.
answered Sep 2 at 18:28
dxiv
55.8k64798
edited Sep 2 at 19:08
answered Sep 2 at 18:28
dxiv
55.8k64798
answered Sep 2 at 18:28
dxiv
55.8k64798
answered Sep 2 at 18:28
dxiv
55.8k64798
55.8k64798
add a comment |Â
add a comment |Â
up vote
4
down vote
Note that $f(x)$ has $2$ vertical asymptotes at
$$(2x-1)^2=1 implies 2x-1=pm 1 implies x=0,1$$
and it easy to check that
- $lim_xto -infty f(x)=0$
- $lim_xto 0^- f(x)=infty$
- $lim_xto 0^+ f(x)=-infty$
- $lim_xto 1^- f(x)=infty$
- $lim_xto 1^+ f(x)=+infty$
- $lim_xto infty f(x)=0$
and therefore, since the function is continuous in its domain, by intermediate value theorem it can't be one to one.
edited Sep 2 at 20:23
Yashas
172115
answered Sep 2 at 18:44
gimusi
71k73786
add a comment |Â
up vote
4
down vote
Note that $f(x)$ has $2$ vertical asymptotes at
$$(2x-1)^2=1 implies 2x-1=pm 1 implies x=0,1$$
and it easy to check that
- $lim_xto -infty f(x)=0$
- $lim_xto 0^- f(x)=infty$
- $lim_xto 0^+ f(x)=-infty$
- $lim_xto 1^- f(x)=infty$
- $lim_xto 1^+ f(x)=+infty$
- $lim_xto infty f(x)=0$
and therefore, since the function is continuous in its domain, by intermediate value theorem it can't be one to one.
edited Sep 2 at 20:23
Yashas
172115
answered Sep 2 at 18:44
gimusi
71k73786
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Note that $f(x)$ has $2$ vertical asymptotes at
$$(2x-1)^2=1 implies 2x-1=pm 1 implies x=0,1$$
and it easy to check that
- $lim_xto -infty f(x)=0$
- $lim_xto 0^- f(x)=infty$
- $lim_xto 0^+ f(x)=-infty$
- $lim_xto 1^- f(x)=infty$
- $lim_xto 1^+ f(x)=+infty$
- $lim_xto infty f(x)=0$
and therefore, since the function is continuous in its domain, by intermediate value theorem it can't be one to one.
edited Sep 2 at 20:23
Yashas
172115
answered Sep 2 at 18:44
gimusi
71k73786
Note that $f(x)$ has $2$ vertical asymptotes at
$$(2x-1)^2=1 implies 2x-1=pm 1 implies x=0,1$$
and it easy to check that
- $lim_xto -infty f(x)=0$
- $lim_xto 0^- f(x)=infty$
- $lim_xto 0^+ f(x)=-infty$
- $lim_xto 1^- f(x)=infty$
- $lim_xto 1^+ f(x)=+infty$
- $lim_xto infty f(x)=0$
and therefore, since the function is continuous in its domain, by intermediate value theorem it can't be one to one.
edited Sep 2 at 20:23
Yashas
172115
answered Sep 2 at 18:44
gimusi
71k73786
edited Sep 2 at 20:23
Yashas
172115
edited Sep 2 at 20:23
Yashas
172115
edited Sep 2 at 20:23
Yashas
172115
172115
answered Sep 2 at 18:44
gimusi
71k73786
answered Sep 2 at 18:44
gimusi
71k73786
answered Sep 2 at 18:44
gimusi
71k73786
71k73786
add a comment |Â
add a comment |Â
up vote
3
down vote
$f(x) = frac-2x+1(2x-1)^2-1$
$f'(x) = frac (-2x+1)'[(2x-1)^2 -1]- (-2x+1)[(2x-1)^2 - 1]'[(2x-1)^2 - 1]^2$
$=frac -2[(2x-1)^2 - 1]-(-2x + 1)2(2x-1)2[(2x-1)^2 - 1]^2=frac 2[1-(2x-1)^2] +4(2x -1)^2[(2x-1)^2 - 1]^2$
Now if $0 < x < 1$ then $-1 < 2x -1 < 1$ and $0le (2x -1)^2 < 1$ and $0< 1-(2x-1)^2 le 1$, while $[(2x-1)^2 - 1]^2 >0$ and $(2x -1)^2 > 0$.
So for $0 < x < 1$, $f'(x) > 0$ and $f$ is therefore strictly increasing and $1-1$.
answered Sep 2 at 22:43
fleablood
61.1k22676
add a comment |Â
up vote
3
down vote
$f(x) = frac-2x+1(2x-1)^2-1$
$f'(x) = frac (-2x+1)'[(2x-1)^2 -1]- (-2x+1)[(2x-1)^2 - 1]'[(2x-1)^2 - 1]^2$
$=frac -2[(2x-1)^2 - 1]-(-2x + 1)2(2x-1)2[(2x-1)^2 - 1]^2=frac 2[1-(2x-1)^2] +4(2x -1)^2[(2x-1)^2 - 1]^2$
Now if $0 < x < 1$ then $-1 < 2x -1 < 1$ and $0le (2x -1)^2 < 1$ and $0< 1-(2x-1)^2 le 1$, while $[(2x-1)^2 - 1]^2 >0$ and $(2x -1)^2 > 0$.
So for $0 < x < 1$, $f'(x) > 0$ and $f$ is therefore strictly increasing and $1-1$.
answered Sep 2 at 22:43
fleablood
61.1k22676
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$f(x) = frac-2x+1(2x-1)^2-1$
$f'(x) = frac (-2x+1)'[(2x-1)^2 -1]- (-2x+1)[(2x-1)^2 - 1]'[(2x-1)^2 - 1]^2$
$=frac -2[(2x-1)^2 - 1]-(-2x + 1)2(2x-1)2[(2x-1)^2 - 1]^2=frac 2[1-(2x-1)^2] +4(2x -1)^2[(2x-1)^2 - 1]^2$
Now if $0 < x < 1$ then $-1 < 2x -1 < 1$ and $0le (2x -1)^2 < 1$ and $0< 1-(2x-1)^2 le 1$, while $[(2x-1)^2 - 1]^2 >0$ and $(2x -1)^2 > 0$.
So for $0 < x < 1$, $f'(x) > 0$ and $f$ is therefore strictly increasing and $1-1$.
answered Sep 2 at 22:43
fleablood
61.1k22676
$f(x) = frac-2x+1(2x-1)^2-1$
$f'(x) = frac (-2x+1)'[(2x-1)^2 -1]- (-2x+1)[(2x-1)^2 - 1]'[(2x-1)^2 - 1]^2$
$=frac -2[(2x-1)^2 - 1]-(-2x + 1)2(2x-1)2[(2x-1)^2 - 1]^2=frac 2[1-(2x-1)^2] +4(2x -1)^2[(2x-1)^2 - 1]^2$
Now if $0 < x < 1$ then $-1 < 2x -1 < 1$ and $0le (2x -1)^2 < 1$ and $0< 1-(2x-1)^2 le 1$, while $[(2x-1)^2 - 1]^2 >0$ and $(2x -1)^2 > 0$.
So for $0 < x < 1$, $f'(x) > 0$ and $f$ is therefore strictly increasing and $1-1$.
answered Sep 2 at 22:43
fleablood
61.1k22676
answered Sep 2 at 22:43
fleablood
61.1k22676
answered Sep 2 at 22:43
fleablood
61.1k22676
answered Sep 2 at 22:43
fleablood
61.1k22676
61.1k22676
add a comment |Â
add a comment |Â
up vote
0
down vote
One other pair of $x, y$ that give the same image are of the form: $$2xy-x-y+1$$ , with $(2x-1)^2-1 neq0$ and $(2y-1)^2-1 neq0$
answered Sep 2 at 18:21
dmtri
810317
add a comment |Â
up vote
0
down vote
One other pair of $x, y$ that give the same image are of the form: $$2xy-x-y+1$$ , with $(2x-1)^2-1 neq0$ and $(2y-1)^2-1 neq0$
answered Sep 2 at 18:21
dmtri
810317
add a comment |Â
up vote
0
down vote
up vote
0
down vote
One other pair of $x, y$ that give the same image are of the form: $$2xy-x-y+1$$ , with $(2x-1)^2-1 neq0$ and $(2y-1)^2-1 neq0$
answered Sep 2 at 18:21
dmtri
810317
One other pair of $x, y$ that give the same image are of the form: $$2xy-x-y+1$$ , with $(2x-1)^2-1 neq0$ and $(2y-1)^2-1 neq0$
answered Sep 2 at 18:21
dmtri
810317
answered Sep 2 at 18:21
dmtri
810317
answered Sep 2 at 18:21
dmtri
810317
answered Sep 2 at 18:21
dmtri
810317
810317
add a comment |Â
add a comment |Â
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I apologize for leaving out a piece of information, I only need to know if the function is one to one from the interval (a,b).
– Bret Hisey
Sep 2 at 18:47