Prove that the function is bijective
up vote
3
down vote
favorite
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
edited Nov 28 at 11:58
Lorenzo B.
1,7472519
add a comment |
up vote
3
down vote
favorite
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
edited Nov 28 at 11:58
Lorenzo B.
1,7472519
6
Well...$e^x$ is strictly increasing but it is not a bijection. Your argument is good, but incomplete,
– lulu
Nov 28 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 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 at 13:59
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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
edited Nov 28 at 11:58
Lorenzo B.
1,7472519
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
edited Nov 28 at 11:58
Lorenzo B.
1,7472519
edited Nov 28 at 11:58
Lorenzo B.
1,7472519
edited Nov 28 at 11:58
Lorenzo B.
1,7472519
edited Nov 28 at 11:58
Lorenzo B.
1,7472519
edited Nov 28 at 11:58
Lorenzo B.
1,7472519
1,7472519
asked Nov 28 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 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 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 at 13:59
add a comment |
6
Well...$e^x$ is strictly increasing but it is not a bijection. Your argument is good, but incomplete,
– lulu
Nov 28 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 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 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 at 11:50
Well...$e^x$ is strictly increasing but it is not a bijection. Your argument is good, but incomplete,
– lulu
Nov 28 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 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 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 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 at 13:59
add a comment |
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
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 $x_0 in [a,b].$
answered Nov 28 at 11:53
Fred
43.4k1644
Is $x sinx$ bijective over $mathbb{R}$ ?
– user612946
Nov 28 at 13:10
3
No, $xsin x$ has a lot of zeros.
– Fred
Nov 28 at 14:01
add a comment |
up vote
3
down vote
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.
answered Nov 28 at 11:53
Kavi Rama Murthy
45.8k31853
Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
– user612946
Nov 28 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 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 at 16:34
Thank you @Shufflepants :-))
– user612946
Nov 28 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 at 16:39
|
show 4 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
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 $x_0 in [a,b].$
answered Nov 28 at 11:53
Fred
43.4k1644
Is $x sinx$ bijective over $mathbb{R}$ ?
– user612946
Nov 28 at 13:10
3
No, $xsin x$ has a lot of zeros.
– Fred
Nov 28 at 14:01
add a comment |
up vote
6
down vote
accepted
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 $x_0 in [a,b].$
answered Nov 28 at 11:53
Fred
43.4k1644
Is $x sinx$ bijective over $mathbb{R}$ ?
– user612946
Nov 28 at 13:10
3
No, $xsin x$ has a lot of zeros.
– Fred
Nov 28 at 14:01
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
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 $x_0 in [a,b].$
answered Nov 28 at 11:53
Fred
43.4k1644
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 $x_0 in [a,b].$
answered Nov 28 at 11:53
Fred
43.4k1644
answered Nov 28 at 11:53
Fred
43.4k1644
answered Nov 28 at 11:53
Fred
43.4k1644
answered Nov 28 at 11:53
Fred
43.4k1644
43.4k1644
Is $x sinx$ bijective over $mathbb{R}$ ?
– user612946
Nov 28 at 13:10
3
No, $xsin x$ has a lot of zeros.
– Fred
Nov 28 at 14:01
add a comment |
Is $x sinx$ bijective over $mathbb{R}$ ?
– user612946
Nov 28 at 13:10
3
No, $xsin x$ has a lot of zeros.
– Fred
Nov 28 at 14:01
Is $x sinx$ bijective over $mathbb{R}$ ?
– user612946
Nov 28 at 13:10
Is $x sinx$ bijective over $mathbb{R}$ ?
– user612946
Nov 28 at 13:10
3
3
No, $xsin x$ has a lot of zeros.
– Fred
Nov 28 at 14:01
No, $xsin x$ has a lot of zeros.
– Fred
Nov 28 at 14:01
add a comment |
up vote
3
down vote
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.
answered Nov 28 at 11:53
Kavi Rama Murthy
45.8k31853
Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
– user612946
Nov 28 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 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 at 16:34
Thank you @Shufflepants :-))
– user612946
Nov 28 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 at 16:39
|
show 4 more comments
up vote
3
down vote
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.
answered Nov 28 at 11:53
Kavi Rama Murthy
45.8k31853
Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
– user612946
Nov 28 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 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 at 16:34
Thank you @Shufflepants :-))
– user612946
Nov 28 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 at 16:39
|
show 4 more comments
up vote
3
down vote
up vote
3
down vote
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.
answered Nov 28 at 11:53
Kavi Rama Murthy
45.8k31853
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.
answered Nov 28 at 11:53
Kavi Rama Murthy
45.8k31853
answered Nov 28 at 11:53
Kavi Rama Murthy
45.8k31853
answered Nov 28 at 11:53
Kavi Rama Murthy
45.8k31853
answered Nov 28 at 11:53
Kavi Rama Murthy
45.8k31853
45.8k31853
Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
– user612946
Nov 28 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 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 at 16:34
Thank you @Shufflepants :-))
– user612946
Nov 28 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 at 16:39
|
show 4 more comments
Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
– user612946
Nov 28 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 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 at 16:34
Thank you @Shufflepants :-))
– user612946
Nov 28 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 at 16:39
Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
– user612946
Nov 28 at 12:03
Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
– user612946
Nov 28 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 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 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 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 at 16:34
Thank you @Shufflepants :-))
– user612946
Nov 28 at 16:35
Thank you @Shufflepants :-))
– user612946
Nov 28 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 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 at 16:39
|
show 4 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017051%2fprove-that-the-function-is-bijective%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
6
Well...$e^x$ is strictly increasing but it is not a bijection. Your argument is good, but incomplete,
– lulu
Nov 28 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 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 at 13:59