$xsqrt{x-1}$ bijectivity
So I know to prove that in order to prove the function is a bi-jectivity I have to solve the equation $xsqrt{x-1}= y$ and find one and only one solution, and I am having trouble doing it. The function is from $[1 ; +infty[ to [0 ; +infty[$
functions
asked Nov 22 '18 at 2:26
Yassine Sama
162
add a comment |
So I know to prove that in order to prove the function is a bi-jectivity I have to solve the equation $xsqrt{x-1}= y$ and find one and only one solution, and I am having trouble doing it. The function is from $[1 ; +infty[ to [0 ; +infty[$
functions
asked Nov 22 '18 at 2:26
Yassine Sama
162
2
$y$ cannot be negative. So if you are considering this as a function from ${0} cup [1,infty) longrightarrow Bbb{R}$, then it cannot be surjective. You need to specify the co-domain before we discuss surjectivity.
– Anurag A
Nov 22 '18 at 2:46
1
Oh i forgot I'm going to edit it.
– Yassine Sama
Nov 22 '18 at 2:56
1
NB: It's "bijectivity," not "bi-jectivity".
– Shaun
Nov 22 '18 at 3:44
add a comment |
So I know to prove that in order to prove the function is a bi-jectivity I have to solve the equation $xsqrt{x-1}= y$ and find one and only one solution, and I am having trouble doing it. The function is from $[1 ; +infty[ to [0 ; +infty[$
functions
asked Nov 22 '18 at 2:26
Yassine Sama
162
So I know to prove that in order to prove the function is a bi-jectivity I have to solve the equation $xsqrt{x-1}= y$ and find one and only one solution, and I am having trouble doing it. The function is from $[1 ; +infty[ to [0 ; +infty[$
functions
functions
asked Nov 22 '18 at 2:26
Yassine Sama
162
asked Nov 22 '18 at 2:26
Yassine Sama
162
edited Nov 22 '18 at 3:47
asked Nov 22 '18 at 2:26
Yassine Sama
162
asked Nov 22 '18 at 2:26
Yassine Sama
162
asked Nov 22 '18 at 2:26
Yassine Sama
162
162
2
$y$ cannot be negative. So if you are considering this as a function from ${0} cup [1,infty) longrightarrow Bbb{R}$, then it cannot be surjective. You need to specify the co-domain before we discuss surjectivity.
– Anurag A
Nov 22 '18 at 2:46
1
Oh i forgot I'm going to edit it.
– Yassine Sama
Nov 22 '18 at 2:56
1
NB: It's "bijectivity," not "bi-jectivity".
– Shaun
Nov 22 '18 at 3:44
add a comment |
2
$y$ cannot be negative. So if you are considering this as a function from ${0} cup [1,infty) longrightarrow Bbb{R}$, then it cannot be surjective. You need to specify the co-domain before we discuss surjectivity.
– Anurag A
Nov 22 '18 at 2:46
1
Oh i forgot I'm going to edit it.
– Yassine Sama
Nov 22 '18 at 2:56
1
NB: It's "bijectivity," not "bi-jectivity".
– Shaun
Nov 22 '18 at 3:44
2
2
$y$ cannot be negative. So if you are considering this as a function from ${0} cup [1,infty) longrightarrow Bbb{R}$, then it cannot be surjective. You need to specify the co-domain before we discuss surjectivity.
– Anurag A
Nov 22 '18 at 2:46
$y$ cannot be negative. So if you are considering this as a function from ${0} cup [1,infty) longrightarrow Bbb{R}$, then it cannot be surjective. You need to specify the co-domain before we discuss surjectivity.
– Anurag A
Nov 22 '18 at 2:46
1
1
Oh i forgot I'm going to edit it.
– Yassine Sama
Nov 22 '18 at 2:56
Oh i forgot I'm going to edit it.
– Yassine Sama
Nov 22 '18 at 2:56
1
1
NB: It's "bijectivity," not "bi-jectivity".
– Shaun
Nov 22 '18 at 3:44
NB: It's "bijectivity," not "bi-jectivity".
– Shaun
Nov 22 '18 at 3:44
add a comment |
2 Answers
2
active
oldest
votes
Finding an explicit inverse will be difficult since you ultimately have to solve the cubic equation $x^2(x-1)=y^2$ (after you have justified why you can square without losing any information in this case). Though you can make a computer do this, for instance you can check out this solution. But then you have to compose them both ways to show that this works, which is equally a nightmare. Instead, I would suggest the following:
Let $f(x)= x sqrt{x-1}$.
Surjectivity: First, what is $f(1)$? What is $lim_{x to infty} f(x)$? Now you can use a nifty three word theorem you learned in Calculus to justify that $f(x)$ 'hits' every real number from $0$ to $infty$. [For this you will need continuity, but by the work you will do below, your function is differentiable, hence continuous.]
Injectivity: Find $f'(x)$? Given your domain is $[1,infty)$, show or explain why $f'(x)>0$ on this interval. But then $f(x)$ is strictly increasing. Then you are done by a little lemma you need to prove:
Lemma: Suppose $f(x)$ is differentiable with $f'(x)>0$. Then if $f(a)=f(b)$, then it must be that $a=b$ (so that $f$ is injective). [Hint: Assume $f(a)=f(b)$. Then $f(x)$ is differentiable with two values equal, what theorem can you use to justify that there must be a point where $f$ 'turns around' so that $f'(x)=0$? Think about this for a bit and it must just rolle through your mind.]
Once all this is done, you have shown $f(x)$ is injective and surjective! But then $f(x)$ is a bijection from $[1,infty)$ to $[0,infty)$.
answered Nov 22 '18 at 4:03
mathematics2x2life
8,05121738
Thank you so much for your answer but the problem is I can't use these theoreme since I haven't studied them yet
– Yassine Sama
Nov 22 '18 at 6:37
add a comment |
1) Injective.
$f(x)=x(x-1)^{1/2}$ is strictly monotonically increasing.
Let $1 le x_1 < x_2.$
$g(x):=√x$ is stricly increasing.
For $a,b ge 0:$
$a-b=(√a-√b)(√a+√b).$
$a-b >0$ implies $√a-√b >0$ (why?).
Hence
$(x_1-1)^{1/2} lt (x_2-1)^{1/2}$, and
$x_1(x_1)^{1/2} lt x_2(x_2-1)^{1/2}$(why?), i.e
strictly increasing, injective.
2) Bijective
Let $y > 0 =f(0). $
Since $lim_{x rightarrow infty}f(x)= infty$ , $f$ is not bounded above . There is $b>0$ s.t.
$y < f(b)$.
Consider the continuos function $f$ on
$[0,b]$, with $f(0)lt y lt f(b)$.
Intermediate Value Theorem(Corollary):
There is a $p in [0,b]$ with $f(p)=y$,
hence surjective.
answered Nov 22 '18 at 10:41
Peter Szilas
10.7k2720
add a comment |
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2 Answers
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Finding an explicit inverse will be difficult since you ultimately have to solve the cubic equation $x^2(x-1)=y^2$ (after you have justified why you can square without losing any information in this case). Though you can make a computer do this, for instance you can check out this solution. But then you have to compose them both ways to show that this works, which is equally a nightmare. Instead, I would suggest the following:
Let $f(x)= x sqrt{x-1}$.
Surjectivity: First, what is $f(1)$? What is $lim_{x to infty} f(x)$? Now you can use a nifty three word theorem you learned in Calculus to justify that $f(x)$ 'hits' every real number from $0$ to $infty$. [For this you will need continuity, but by the work you will do below, your function is differentiable, hence continuous.]
Injectivity: Find $f'(x)$? Given your domain is $[1,infty)$, show or explain why $f'(x)>0$ on this interval. But then $f(x)$ is strictly increasing. Then you are done by a little lemma you need to prove:
Lemma: Suppose $f(x)$ is differentiable with $f'(x)>0$. Then if $f(a)=f(b)$, then it must be that $a=b$ (so that $f$ is injective). [Hint: Assume $f(a)=f(b)$. Then $f(x)$ is differentiable with two values equal, what theorem can you use to justify that there must be a point where $f$ 'turns around' so that $f'(x)=0$? Think about this for a bit and it must just rolle through your mind.]
Once all this is done, you have shown $f(x)$ is injective and surjective! But then $f(x)$ is a bijection from $[1,infty)$ to $[0,infty)$.
answered Nov 22 '18 at 4:03
mathematics2x2life
8,05121738
Thank you so much for your answer but the problem is I can't use these theoreme since I haven't studied them yet
– Yassine Sama
Nov 22 '18 at 6:37
add a comment |
Finding an explicit inverse will be difficult since you ultimately have to solve the cubic equation $x^2(x-1)=y^2$ (after you have justified why you can square without losing any information in this case). Though you can make a computer do this, for instance you can check out this solution. But then you have to compose them both ways to show that this works, which is equally a nightmare. Instead, I would suggest the following:
Let $f(x)= x sqrt{x-1}$.
Surjectivity: First, what is $f(1)$? What is $lim_{x to infty} f(x)$? Now you can use a nifty three word theorem you learned in Calculus to justify that $f(x)$ 'hits' every real number from $0$ to $infty$. [For this you will need continuity, but by the work you will do below, your function is differentiable, hence continuous.]
Injectivity: Find $f'(x)$? Given your domain is $[1,infty)$, show or explain why $f'(x)>0$ on this interval. But then $f(x)$ is strictly increasing. Then you are done by a little lemma you need to prove:
Lemma: Suppose $f(x)$ is differentiable with $f'(x)>0$. Then if $f(a)=f(b)$, then it must be that $a=b$ (so that $f$ is injective). [Hint: Assume $f(a)=f(b)$. Then $f(x)$ is differentiable with two values equal, what theorem can you use to justify that there must be a point where $f$ 'turns around' so that $f'(x)=0$? Think about this for a bit and it must just rolle through your mind.]
Once all this is done, you have shown $f(x)$ is injective and surjective! But then $f(x)$ is a bijection from $[1,infty)$ to $[0,infty)$.
answered Nov 22 '18 at 4:03
mathematics2x2life
8,05121738
Thank you so much for your answer but the problem is I can't use these theoreme since I haven't studied them yet
– Yassine Sama
Nov 22 '18 at 6:37
add a comment |
Finding an explicit inverse will be difficult since you ultimately have to solve the cubic equation $x^2(x-1)=y^2$ (after you have justified why you can square without losing any information in this case). Though you can make a computer do this, for instance you can check out this solution. But then you have to compose them both ways to show that this works, which is equally a nightmare. Instead, I would suggest the following:
Let $f(x)= x sqrt{x-1}$.
Surjectivity: First, what is $f(1)$? What is $lim_{x to infty} f(x)$? Now you can use a nifty three word theorem you learned in Calculus to justify that $f(x)$ 'hits' every real number from $0$ to $infty$. [For this you will need continuity, but by the work you will do below, your function is differentiable, hence continuous.]
Injectivity: Find $f'(x)$? Given your domain is $[1,infty)$, show or explain why $f'(x)>0$ on this interval. But then $f(x)$ is strictly increasing. Then you are done by a little lemma you need to prove:
Lemma: Suppose $f(x)$ is differentiable with $f'(x)>0$. Then if $f(a)=f(b)$, then it must be that $a=b$ (so that $f$ is injective). [Hint: Assume $f(a)=f(b)$. Then $f(x)$ is differentiable with two values equal, what theorem can you use to justify that there must be a point where $f$ 'turns around' so that $f'(x)=0$? Think about this for a bit and it must just rolle through your mind.]
Once all this is done, you have shown $f(x)$ is injective and surjective! But then $f(x)$ is a bijection from $[1,infty)$ to $[0,infty)$.
answered Nov 22 '18 at 4:03
mathematics2x2life
8,05121738
Finding an explicit inverse will be difficult since you ultimately have to solve the cubic equation $x^2(x-1)=y^2$ (after you have justified why you can square without losing any information in this case). Though you can make a computer do this, for instance you can check out this solution. But then you have to compose them both ways to show that this works, which is equally a nightmare. Instead, I would suggest the following:
Let $f(x)= x sqrt{x-1}$.
Surjectivity: First, what is $f(1)$? What is $lim_{x to infty} f(x)$? Now you can use a nifty three word theorem you learned in Calculus to justify that $f(x)$ 'hits' every real number from $0$ to $infty$. [For this you will need continuity, but by the work you will do below, your function is differentiable, hence continuous.]
Injectivity: Find $f'(x)$? Given your domain is $[1,infty)$, show or explain why $f'(x)>0$ on this interval. But then $f(x)$ is strictly increasing. Then you are done by a little lemma you need to prove:
Lemma: Suppose $f(x)$ is differentiable with $f'(x)>0$. Then if $f(a)=f(b)$, then it must be that $a=b$ (so that $f$ is injective). [Hint: Assume $f(a)=f(b)$. Then $f(x)$ is differentiable with two values equal, what theorem can you use to justify that there must be a point where $f$ 'turns around' so that $f'(x)=0$? Think about this for a bit and it must just rolle through your mind.]
Once all this is done, you have shown $f(x)$ is injective and surjective! But then $f(x)$ is a bijection from $[1,infty)$ to $[0,infty)$.
answered Nov 22 '18 at 4:03
mathematics2x2life
8,05121738
answered Nov 22 '18 at 4:03
mathematics2x2life
8,05121738
answered Nov 22 '18 at 4:03
mathematics2x2life
8,05121738
answered Nov 22 '18 at 4:03
mathematics2x2life
8,05121738
8,05121738
Thank you so much for your answer but the problem is I can't use these theoreme since I haven't studied them yet
– Yassine Sama
Nov 22 '18 at 6:37
add a comment |
Thank you so much for your answer but the problem is I can't use these theoreme since I haven't studied them yet
– Yassine Sama
Nov 22 '18 at 6:37
Thank you so much for your answer but the problem is I can't use these theoreme since I haven't studied them yet
– Yassine Sama
Nov 22 '18 at 6:37
Thank you so much for your answer but the problem is I can't use these theoreme since I haven't studied them yet
– Yassine Sama
Nov 22 '18 at 6:37
add a comment |
1) Injective.
$f(x)=x(x-1)^{1/2}$ is strictly monotonically increasing.
Let $1 le x_1 < x_2.$
$g(x):=√x$ is stricly increasing.
For $a,b ge 0:$
$a-b=(√a-√b)(√a+√b).$
$a-b >0$ implies $√a-√b >0$ (why?).
Hence
$(x_1-1)^{1/2} lt (x_2-1)^{1/2}$, and
$x_1(x_1)^{1/2} lt x_2(x_2-1)^{1/2}$(why?), i.e
strictly increasing, injective.
2) Bijective
Let $y > 0 =f(0). $
Since $lim_{x rightarrow infty}f(x)= infty$ , $f$ is not bounded above . There is $b>0$ s.t.
$y < f(b)$.
Consider the continuos function $f$ on
$[0,b]$, with $f(0)lt y lt f(b)$.
Intermediate Value Theorem(Corollary):
There is a $p in [0,b]$ with $f(p)=y$,
hence surjective.
answered Nov 22 '18 at 10:41
Peter Szilas
10.7k2720
add a comment |
1) Injective.
$f(x)=x(x-1)^{1/2}$ is strictly monotonically increasing.
Let $1 le x_1 < x_2.$
$g(x):=√x$ is stricly increasing.
For $a,b ge 0:$
$a-b=(√a-√b)(√a+√b).$
$a-b >0$ implies $√a-√b >0$ (why?).
Hence
$(x_1-1)^{1/2} lt (x_2-1)^{1/2}$, and
$x_1(x_1)^{1/2} lt x_2(x_2-1)^{1/2}$(why?), i.e
strictly increasing, injective.
2) Bijective
Let $y > 0 =f(0). $
Since $lim_{x rightarrow infty}f(x)= infty$ , $f$ is not bounded above . There is $b>0$ s.t.
$y < f(b)$.
Consider the continuos function $f$ on
$[0,b]$, with $f(0)lt y lt f(b)$.
Intermediate Value Theorem(Corollary):
There is a $p in [0,b]$ with $f(p)=y$,
hence surjective.
answered Nov 22 '18 at 10:41
Peter Szilas
10.7k2720
add a comment |
1) Injective.
$f(x)=x(x-1)^{1/2}$ is strictly monotonically increasing.
Let $1 le x_1 < x_2.$
$g(x):=√x$ is stricly increasing.
For $a,b ge 0:$
$a-b=(√a-√b)(√a+√b).$
$a-b >0$ implies $√a-√b >0$ (why?).
Hence
$(x_1-1)^{1/2} lt (x_2-1)^{1/2}$, and
$x_1(x_1)^{1/2} lt x_2(x_2-1)^{1/2}$(why?), i.e
strictly increasing, injective.
2) Bijective
Let $y > 0 =f(0). $
Since $lim_{x rightarrow infty}f(x)= infty$ , $f$ is not bounded above . There is $b>0$ s.t.
$y < f(b)$.
Consider the continuos function $f$ on
$[0,b]$, with $f(0)lt y lt f(b)$.
Intermediate Value Theorem(Corollary):
There is a $p in [0,b]$ with $f(p)=y$,
hence surjective.
answered Nov 22 '18 at 10:41
Peter Szilas
10.7k2720
1) Injective.
$f(x)=x(x-1)^{1/2}$ is strictly monotonically increasing.
Let $1 le x_1 < x_2.$
$g(x):=√x$ is stricly increasing.
For $a,b ge 0:$
$a-b=(√a-√b)(√a+√b).$
$a-b >0$ implies $√a-√b >0$ (why?).
Hence
$(x_1-1)^{1/2} lt (x_2-1)^{1/2}$, and
$x_1(x_1)^{1/2} lt x_2(x_2-1)^{1/2}$(why?), i.e
strictly increasing, injective.
2) Bijective
Let $y > 0 =f(0). $
Since $lim_{x rightarrow infty}f(x)= infty$ , $f$ is not bounded above . There is $b>0$ s.t.
$y < f(b)$.
Consider the continuos function $f$ on
$[0,b]$, with $f(0)lt y lt f(b)$.
Intermediate Value Theorem(Corollary):
There is a $p in [0,b]$ with $f(p)=y$,
hence surjective.
answered Nov 22 '18 at 10:41
Peter Szilas
10.7k2720
edited Nov 22 '18 at 11:12
answered Nov 22 '18 at 10:41
Peter Szilas
10.7k2720
answered Nov 22 '18 at 10:41
Peter Szilas
10.7k2720
answered Nov 22 '18 at 10:41
Peter Szilas
10.7k2720
10.7k2720
add a comment |
add a comment |
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2
$y$ cannot be negative. So if you are considering this as a function from ${0} cup [1,infty) longrightarrow Bbb{R}$, then it cannot be surjective. You need to specify the co-domain before we discuss surjectivity.
– Anurag A
Nov 22 '18 at 2:46
1
Oh i forgot I'm going to edit it.
– Yassine Sama
Nov 22 '18 at 2:56
1
NB: It's "bijectivity," not "bi-jectivity".
– Shaun
Nov 22 '18 at 3:44