Prove that the function is bijective
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 '18 at 11:58
Lorenzo B.
1,8322520
add a comment |
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 '18 at 11:58
Lorenzo B.
1,8322520
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
add a comment |
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 '18 at 11:58
Lorenzo B.
1,8322520
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 '18 at 11:58
Lorenzo B.
1,8322520
edited Nov 28 '18 at 11:58
Lorenzo B.
1,8322520
edited Nov 28 '18 at 11:58
Lorenzo B.
1,8322520
edited Nov 28 '18 at 11:58
Lorenzo B.
1,8322520
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
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 '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
add a comment |
2 Answers
2
active
oldest
votes
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 '18 at 11:53
Fred
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
add a comment |
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 '18 at 11:53
Kavi Rama Murthy
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
|
show 4 more comments
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 '18 at 11:53
Fred
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
add a comment |
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 '18 at 11:53
Fred
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
add a comment |
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 '18 at 11:53
Fred
44.2k1845
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 '18 at 11:53
Fred
44.2k1845
answered Nov 28 '18 at 11:53
Fred
44.2k1845
answered Nov 28 '18 at 11:53
Fred
44.2k1845
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
add a comment |
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
add a comment |
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 '18 at 11:53
Kavi Rama Murthy
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
|
show 4 more comments
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 '18 at 11:53
Kavi Rama Murthy
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
|
show 4 more comments
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 '18 at 11:53
Kavi Rama Murthy
50.5k31854
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 '18 at 11:53
Kavi Rama Murthy
50.5k31854
answered Nov 28 '18 at 11:53
Kavi Rama Murthy
50.5k31854
answered Nov 28 '18 at 11:53
Kavi Rama Murthy
50.5k31854
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
|
show 4 more comments
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
|
show 4 more comments
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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