Prove that the function is bijective












3














A function $f:mathbb{R}tomathbb{R}$ is defined by, $$f(x)={x+arctan(x)}.$$
Prove that this function is bijective.



Here's my attempt:-
Since $x$ and $arctan(x)$ are continuous, $f(x)={x+arctan(x)}$ is also continuous. And further, $$ f'(x) = 1+ frac1{1+x^2}gt 0 quadforall x inmathbb{R}$$



Therefore the function is strictly increasing and hence, is strictly monotonic. Therefore this function is bijective. Would this be correct? Can you show me a better way (that doesn't involve derivative) of proving this? Thank you!






calculus functions







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  • 6




    Well...$e^x$ is strictly increasing but it is not a bijection. Your argument is good, but incomplete,
    – lulu
    Nov 28 '18 at 11:50






  • 1




    You are implicitly using the Intermediate Value Theorem in your argument. If you want to show the function reaches every value in the interval $[-M, M]$, you first need to show that it reaches $M$ and $-M$.
    – N. F. Taussig
    Nov 28 '18 at 11:53






  • 1




    Your argument applies without change to the function $arctan(x)$. But $arctan(x)$ isn't bijective! So you must have gone wrong somewhere.
    – TonyK
    Nov 28 '18 at 13:59


















3














A function $f:mathbb{R}tomathbb{R}$ is defined by, $$f(x)={x+arctan(x)}.$$
Prove that this function is bijective.



Here's my attempt:-
Since $x$ and $arctan(x)$ are continuous, $f(x)={x+arctan(x)}$ is also continuous. And further, $$ f'(x) = 1+ frac1{1+x^2}gt 0 quadforall x inmathbb{R}$$



Therefore the function is strictly increasing and hence, is strictly monotonic. Therefore this function is bijective. Would this be correct? Can you show me a better way (that doesn't involve derivative) of proving this? Thank you!






calculus functions







share|cite|improve this question




















  • 6




    Well...$e^x$ is strictly increasing but it is not a bijection. Your argument is good, but incomplete,
    – lulu
    Nov 28 '18 at 11:50






  • 1




    You are implicitly using the Intermediate Value Theorem in your argument. If you want to show the function reaches every value in the interval $[-M, M]$, you first need to show that it reaches $M$ and $-M$.
    – N. F. Taussig
    Nov 28 '18 at 11:53






  • 1




    Your argument applies without change to the function $arctan(x)$. But $arctan(x)$ isn't bijective! So you must have gone wrong somewhere.
    – TonyK
    Nov 28 '18 at 13:59
















3












3








3







A function $f:mathbb{R}tomathbb{R}$ is defined by, $$f(x)={x+arctan(x)}.$$
Prove that this function is bijective.



Here's my attempt:-
Since $x$ and $arctan(x)$ are continuous, $f(x)={x+arctan(x)}$ is also continuous. And further, $$ f'(x) = 1+ frac1{1+x^2}gt 0 quadforall x inmathbb{R}$$



Therefore the function is strictly increasing and hence, is strictly monotonic. Therefore this function is bijective. Would this be correct? Can you show me a better way (that doesn't involve derivative) of proving this? Thank you!






calculus functions







share|cite|improve this question















A function $f:mathbb{R}tomathbb{R}$ is defined by, $$f(x)={x+arctan(x)}.$$
Prove that this function is bijective.



Here's my attempt:-
Since $x$ and $arctan(x)$ are continuous, $f(x)={x+arctan(x)}$ is also continuous. And further, $$ f'(x) = 1+ frac1{1+x^2}gt 0 quadforall x inmathbb{R}$$



Therefore the function is strictly increasing and hence, is strictly monotonic. Therefore this function is bijective. Would this be correct? Can you show me a better way (that doesn't involve derivative) of proving this? Thank you!






calculus functions




calculus functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 11:58









Lorenzo B.

1,8322520




1,8322520










asked Nov 28 '18 at 11:47







user612946















  • 6




    Well...$e^x$ is strictly increasing but it is not a bijection. Your argument is good, but incomplete,
    – lulu
    Nov 28 '18 at 11:50






  • 1




    You are implicitly using the Intermediate Value Theorem in your argument. If you want to show the function reaches every value in the interval $[-M, M]$, you first need to show that it reaches $M$ and $-M$.
    – N. F. Taussig
    Nov 28 '18 at 11:53






  • 1




    Your argument applies without change to the function $arctan(x)$. But $arctan(x)$ isn't bijective! So you must have gone wrong somewhere.
    – TonyK
    Nov 28 '18 at 13:59
















  • 6




    Well...$e^x$ is strictly increasing but it is not a bijection. Your argument is good, but incomplete,
    – lulu
    Nov 28 '18 at 11:50






  • 1




    You are implicitly using the Intermediate Value Theorem in your argument. If you want to show the function reaches every value in the interval $[-M, M]$, you first need to show that it reaches $M$ and $-M$.
    – N. F. Taussig
    Nov 28 '18 at 11:53






  • 1




    Your argument applies without change to the function $arctan(x)$. But $arctan(x)$ isn't bijective! So you must have gone wrong somewhere.
    – TonyK
    Nov 28 '18 at 13:59










6




6




Well...$e^x$ is strictly increasing but it is not a bijection. Your argument is good, but incomplete,
– lulu
Nov 28 '18 at 11:50




Well...$e^x$ is strictly increasing but it is not a bijection. Your argument is good, but incomplete,
– lulu
Nov 28 '18 at 11:50




1




1




You are implicitly using the Intermediate Value Theorem in your argument. If you want to show the function reaches every value in the interval $[-M, M]$, you first need to show that it reaches $M$ and $-M$.
– N. F. Taussig
Nov 28 '18 at 11:53




You are implicitly using the Intermediate Value Theorem in your argument. If you want to show the function reaches every value in the interval $[-M, M]$, you first need to show that it reaches $M$ and $-M$.
– N. F. Taussig
Nov 28 '18 at 11:53




1




1




Your argument applies without change to the function $arctan(x)$. But $arctan(x)$ isn't bijective! So you must have gone wrong somewhere.
– TonyK
Nov 28 '18 at 13:59






Your argument applies without change to the function $arctan(x)$. But $arctan(x)$ isn't bijective! So you must have gone wrong somewhere.
– TonyK
Nov 28 '18 at 13:59












2 Answers
2






active

oldest

votes


















6














You only have shown that $f$ is injective.



It remains to show that $f$ is surjective: to this end let $y_0 in mathbb R$.



Since $f(x) to infty$ as $ xto infty$ and $f(x) to -infty$ as $ xto -infty$, there are $a,b in mathbb R$ such that $a<b, f(b)> y_0$ and $f(a) <y_0$.



The intermediate value theorem shows now that $f(x_0)=y_0$ for some $x_0 in [a,b].$






share|cite|improve this answer





















  • Is $x sinx$ bijective over $mathbb{R}$ ?
    – user612946
    Nov 28 '18 at 13:10






  • 3




    No, $xsin x$ has a lot of zeros.
    – Fred
    Nov 28 '18 at 14:01





















3














A strictly monotonic function need not be surjective. In this case $-pi /2 <arctan x< pi /2$ so $f(x) to infty$ as $x to infty$ and $f(x) to -infty$ as $x to -infty$. From this it follows that the range (which is necessarily an interval by IVP) is the whole real line.






share|cite|improve this answer





















  • Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
    – user612946
    Nov 28 '18 at 12:03






  • 1




    Over which domain is $xsinx$ bijective? I don't think it is bijective over $mathbb{R}$ ,as it has zeroes in $mathbb{R}$ for multiple $x$ in $mathbb{R}$
    – user612946
    Nov 28 '18 at 12:32








  • 1




    @Samurai You're right, and Kavi is wrong. Bijective, continuous functions must be monotonic as bijective must be one-to-one, so the function cannot attain any particular value more than once.
    – Shufflepants
    Nov 28 '18 at 16:34












  • Thank you @Shufflepants :-))
    – user612946
    Nov 28 '18 at 16:35






  • 1




    But of course, it is possible to have a discontinuous function that is not monotonic but bijective. Though, it must be monotonic on every interval on which it is continuous.
    – Shufflepants
    Nov 28 '18 at 16:39











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6














You only have shown that $f$ is injective.



It remains to show that $f$ is surjective: to this end let $y_0 in mathbb R$.



Since $f(x) to infty$ as $ xto infty$ and $f(x) to -infty$ as $ xto -infty$, there are $a,b in mathbb R$ such that $a<b, f(b)> y_0$ and $f(a) <y_0$.



The intermediate value theorem shows now that $f(x_0)=y_0$ for some $x_0 in [a,b].$






share|cite|improve this answer





















  • Is $x sinx$ bijective over $mathbb{R}$ ?
    – user612946
    Nov 28 '18 at 13:10






  • 3




    No, $xsin x$ has a lot of zeros.
    – Fred
    Nov 28 '18 at 14:01


















6














You only have shown that $f$ is injective.



It remains to show that $f$ is surjective: to this end let $y_0 in mathbb R$.



Since $f(x) to infty$ as $ xto infty$ and $f(x) to -infty$ as $ xto -infty$, there are $a,b in mathbb R$ such that $a<b, f(b)> y_0$ and $f(a) <y_0$.



The intermediate value theorem shows now that $f(x_0)=y_0$ for some $x_0 in [a,b].$






share|cite|improve this answer





















  • Is $x sinx$ bijective over $mathbb{R}$ ?
    – user612946
    Nov 28 '18 at 13:10






  • 3




    No, $xsin x$ has a lot of zeros.
    – Fred
    Nov 28 '18 at 14:01
















6












6








6






You only have shown that $f$ is injective.



It remains to show that $f$ is surjective: to this end let $y_0 in mathbb R$.



Since $f(x) to infty$ as $ xto infty$ and $f(x) to -infty$ as $ xto -infty$, there are $a,b in mathbb R$ such that $a<b, f(b)> y_0$ and $f(a) <y_0$.



The intermediate value theorem shows now that $f(x_0)=y_0$ for some $x_0 in [a,b].$






share|cite|improve this answer












You only have shown that $f$ is injective.



It remains to show that $f$ is surjective: to this end let $y_0 in mathbb R$.



Since $f(x) to infty$ as $ xto infty$ and $f(x) to -infty$ as $ xto -infty$, there are $a,b in mathbb R$ such that $a<b, f(b)> y_0$ and $f(a) <y_0$.



The intermediate value theorem shows now that $f(x_0)=y_0$ for some $x_0 in [a,b].$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 28 '18 at 11:53









Fred

44.2k1845




44.2k1845












  • Is $x sinx$ bijective over $mathbb{R}$ ?
    – user612946
    Nov 28 '18 at 13:10






  • 3




    No, $xsin x$ has a lot of zeros.
    – Fred
    Nov 28 '18 at 14:01




















  • Is $x sinx$ bijective over $mathbb{R}$ ?
    – user612946
    Nov 28 '18 at 13:10






  • 3




    No, $xsin x$ has a lot of zeros.
    – Fred
    Nov 28 '18 at 14:01


















Is $x sinx$ bijective over $mathbb{R}$ ?
– user612946
Nov 28 '18 at 13:10




Is $x sinx$ bijective over $mathbb{R}$ ?
– user612946
Nov 28 '18 at 13:10




3




3




No, $xsin x$ has a lot of zeros.
– Fred
Nov 28 '18 at 14:01






No, $xsin x$ has a lot of zeros.
– Fred
Nov 28 '18 at 14:01













3














A strictly monotonic function need not be surjective. In this case $-pi /2 <arctan x< pi /2$ so $f(x) to infty$ as $x to infty$ and $f(x) to -infty$ as $x to -infty$. From this it follows that the range (which is necessarily an interval by IVP) is the whole real line.






share|cite|improve this answer





















  • Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
    – user612946
    Nov 28 '18 at 12:03






  • 1




    Over which domain is $xsinx$ bijective? I don't think it is bijective over $mathbb{R}$ ,as it has zeroes in $mathbb{R}$ for multiple $x$ in $mathbb{R}$
    – user612946
    Nov 28 '18 at 12:32








  • 1




    @Samurai You're right, and Kavi is wrong. Bijective, continuous functions must be monotonic as bijective must be one-to-one, so the function cannot attain any particular value more than once.
    – Shufflepants
    Nov 28 '18 at 16:34












  • Thank you @Shufflepants :-))
    – user612946
    Nov 28 '18 at 16:35






  • 1




    But of course, it is possible to have a discontinuous function that is not monotonic but bijective. Though, it must be monotonic on every interval on which it is continuous.
    – Shufflepants
    Nov 28 '18 at 16:39
















3














A strictly monotonic function need not be surjective. In this case $-pi /2 <arctan x< pi /2$ so $f(x) to infty$ as $x to infty$ and $f(x) to -infty$ as $x to -infty$. From this it follows that the range (which is necessarily an interval by IVP) is the whole real line.






share|cite|improve this answer





















  • Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
    – user612946
    Nov 28 '18 at 12:03






  • 1




    Over which domain is $xsinx$ bijective? I don't think it is bijective over $mathbb{R}$ ,as it has zeroes in $mathbb{R}$ for multiple $x$ in $mathbb{R}$
    – user612946
    Nov 28 '18 at 12:32








  • 1




    @Samurai You're right, and Kavi is wrong. Bijective, continuous functions must be monotonic as bijective must be one-to-one, so the function cannot attain any particular value more than once.
    – Shufflepants
    Nov 28 '18 at 16:34












  • Thank you @Shufflepants :-))
    – user612946
    Nov 28 '18 at 16:35






  • 1




    But of course, it is possible to have a discontinuous function that is not monotonic but bijective. Though, it must be monotonic on every interval on which it is continuous.
    – Shufflepants
    Nov 28 '18 at 16:39














3












3








3






A strictly monotonic function need not be surjective. In this case $-pi /2 <arctan x< pi /2$ so $f(x) to infty$ as $x to infty$ and $f(x) to -infty$ as $x to -infty$. From this it follows that the range (which is necessarily an interval by IVP) is the whole real line.






share|cite|improve this answer












A strictly monotonic function need not be surjective. In this case $-pi /2 <arctan x< pi /2$ so $f(x) to infty$ as $x to infty$ and $f(x) to -infty$ as $x to -infty$. From this it follows that the range (which is necessarily an interval by IVP) is the whole real line.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 28 '18 at 11:53









Kavi Rama Murthy

50.5k31854




50.5k31854












  • Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
    – user612946
    Nov 28 '18 at 12:03






  • 1




    Over which domain is $xsinx$ bijective? I don't think it is bijective over $mathbb{R}$ ,as it has zeroes in $mathbb{R}$ for multiple $x$ in $mathbb{R}$
    – user612946
    Nov 28 '18 at 12:32








  • 1




    @Samurai You're right, and Kavi is wrong. Bijective, continuous functions must be monotonic as bijective must be one-to-one, so the function cannot attain any particular value more than once.
    – Shufflepants
    Nov 28 '18 at 16:34












  • Thank you @Shufflepants :-))
    – user612946
    Nov 28 '18 at 16:35






  • 1




    But of course, it is possible to have a discontinuous function that is not monotonic but bijective. Though, it must be monotonic on every interval on which it is continuous.
    – Shufflepants
    Nov 28 '18 at 16:39


















  • Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
    – user612946
    Nov 28 '18 at 12:03






  • 1




    Over which domain is $xsinx$ bijective? I don't think it is bijective over $mathbb{R}$ ,as it has zeroes in $mathbb{R}$ for multiple $x$ in $mathbb{R}$
    – user612946
    Nov 28 '18 at 12:32








  • 1




    @Samurai You're right, and Kavi is wrong. Bijective, continuous functions must be monotonic as bijective must be one-to-one, so the function cannot attain any particular value more than once.
    – Shufflepants
    Nov 28 '18 at 16:34












  • Thank you @Shufflepants :-))
    – user612946
    Nov 28 '18 at 16:35






  • 1




    But of course, it is possible to have a discontinuous function that is not monotonic but bijective. Though, it must be monotonic on every interval on which it is continuous.
    – Shufflepants
    Nov 28 '18 at 16:39
















Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
– user612946
Nov 28 '18 at 12:03




Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
– user612946
Nov 28 '18 at 12:03




1




1




Over which domain is $xsinx$ bijective? I don't think it is bijective over $mathbb{R}$ ,as it has zeroes in $mathbb{R}$ for multiple $x$ in $mathbb{R}$
– user612946
Nov 28 '18 at 12:32






Over which domain is $xsinx$ bijective? I don't think it is bijective over $mathbb{R}$ ,as it has zeroes in $mathbb{R}$ for multiple $x$ in $mathbb{R}$
– user612946
Nov 28 '18 at 12:32






1




1




@Samurai You're right, and Kavi is wrong. Bijective, continuous functions must be monotonic as bijective must be one-to-one, so the function cannot attain any particular value more than once.
– Shufflepants
Nov 28 '18 at 16:34






@Samurai You're right, and Kavi is wrong. Bijective, continuous functions must be monotonic as bijective must be one-to-one, so the function cannot attain any particular value more than once.
– Shufflepants
Nov 28 '18 at 16:34














Thank you @Shufflepants :-))
– user612946
Nov 28 '18 at 16:35




Thank you @Shufflepants :-))
– user612946
Nov 28 '18 at 16:35




1




1




But of course, it is possible to have a discontinuous function that is not monotonic but bijective. Though, it must be monotonic on every interval on which it is continuous.
– Shufflepants
Nov 28 '18 at 16:39




But of course, it is possible to have a discontinuous function that is not monotonic but bijective. Though, it must be monotonic on every interval on which it is continuous.
– Shufflepants
Nov 28 '18 at 16:39


















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