Find the values of $p$ where $cos x - 2px$ is invertible

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I'm sorry that I posted a homework question.I just couldn't find answers properly.

I attempted in the following way:

• To prove that it is invertible it's enough to prove that it is bijective


• To prove that it is bijective we have to prove that it is both injective and surjective


• I can prove injectivity by proving that it is a strictly monotonous function( Not sure about this concept also) and then get values of $ p $


• I don't know if showing that it is monotonous for certain values of $p$ is good enough to prove it's surjectivity

Could you help me in proving it's surjectivity and also help me understand if proving strictly monotonocity is enough for injectivity or not

Any little help is appreciated!

calculus functions

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edited Jul 19 at 3:05

Key Flex

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asked Jul 18 at 10:45

Banchin

162

add a comment | 

up vote
0
down vote

favorite

I'm sorry that I posted a homework question.I just couldn't find answers properly.

I attempted in the following way:

• To prove that it is invertible it's enough to prove that it is bijective


• To prove that it is bijective we have to prove that it is both injective and surjective


• I can prove injectivity by proving that it is a strictly monotonous function( Not sure about this concept also) and then get values of $ p $


• I don't know if showing that it is monotonous for certain values of $p$ is good enough to prove it's surjectivity

Could you help me in proving it's surjectivity and also help me understand if proving strictly monotonocity is enough for injectivity or not

Any little help is appreciated!

calculus functions

share|cite|improve this question

edited Jul 19 at 3:05

Key Flex

4,346425

asked Jul 18 at 10:45

Banchin

162

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I'm sorry that I posted a homework question.I just couldn't find answers properly.

I attempted in the following way:

• To prove that it is invertible it's enough to prove that it is bijective


• To prove that it is bijective we have to prove that it is both injective and surjective


• I can prove injectivity by proving that it is a strictly monotonous function( Not sure about this concept also) and then get values of $ p $


• I don't know if showing that it is monotonous for certain values of $p$ is good enough to prove it's surjectivity

Could you help me in proving it's surjectivity and also help me understand if proving strictly monotonocity is enough for injectivity or not

Any little help is appreciated!

calculus functions

share|cite|improve this question

edited Jul 19 at 3:05

Key Flex

4,346425

asked Jul 18 at 10:45

Banchin

162

I'm sorry that I posted a homework question.I just couldn't find answers properly.

I attempted in the following way:

• To prove that it is invertible it's enough to prove that it is bijective


• To prove that it is bijective we have to prove that it is both injective and surjective


• I can prove injectivity by proving that it is a strictly monotonous function( Not sure about this concept also) and then get values of $ p $


• I don't know if showing that it is monotonous for certain values of $p$ is good enough to prove it's surjectivity

Could you help me in proving it's surjectivity and also help me understand if proving strictly monotonocity is enough for injectivity or not

Any little help is appreciated!

calculus functions

share|cite|improve this question

edited Jul 19 at 3:05

Key Flex

4,346425

asked Jul 18 at 10:45

Banchin

162

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 3:05

Key Flex

4,346425

edited Jul 19 at 3:05

Key Flex

4,346425

edited Jul 19 at 3:05

Key Flex

4,346425

4,346425

asked Jul 18 at 10:45

Banchin

162

asked Jul 18 at 10:45

Banchin

162

asked Jul 18 at 10:45

Banchin

162

162

add a comment | 

add a comment | 

1 Answer
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If $f$ is strictly monotonous then for all $x,yinBbbR$ you have
$$x>y Rightarrow f(x)>f(y),$$
from which it easily follows that $f(x)=f(y) Rightarrow x=y$, so indeed $f$ is injective.

So $f$ is invertible whenever $p$ is such that $f'(x)>0$ for all $x$, or such that $f'(x)<0$ for all $x$. This still leaves some values of $p$ unresolved though. As this is homework, I'll leave it to you (for now) to figure out what happens then.

EDIT: Because $f'(x)=sin(x)-2p$, if $p>frac12$ then $f'(x)<0$, and if $p<frac12$ then $f'(x)>0$. Hence $f(x)$ is invertible for these values of $p$.

If $-frac12<p<frac12$ then $f'(0)>0$ and $f'(pi)<0$. This means $f$ is not invertible (why?).

Then the only remaining cases are $p=frac12$ and $p=-frac12$.

share|cite|improve this answer

edited Jul 24 at 17:40

answered Jul 18 at 10:49

Servaes

20.2k33484

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm sorry, I still couldn't find a way for checking it's surjectivity...could you help me with that?
  – Banchin
  Jul 21 at 1:16
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

If $f$ is strictly monotonous then for all $x,yinBbbR$ you have
$$x>y Rightarrow f(x)>f(y),$$
from which it easily follows that $f(x)=f(y) Rightarrow x=y$, so indeed $f$ is injective.

So $f$ is invertible whenever $p$ is such that $f'(x)>0$ for all $x$, or such that $f'(x)<0$ for all $x$. This still leaves some values of $p$ unresolved though. As this is homework, I'll leave it to you (for now) to figure out what happens then.

EDIT: Because $f'(x)=sin(x)-2p$, if $p>frac12$ then $f'(x)<0$, and if $p<frac12$ then $f'(x)>0$. Hence $f(x)$ is invertible for these values of $p$.

If $-frac12<p<frac12$ then $f'(0)>0$ and $f'(pi)<0$. This means $f$ is not invertible (why?).

Then the only remaining cases are $p=frac12$ and $p=-frac12$.

share|cite|improve this answer

edited Jul 24 at 17:40

answered Jul 18 at 10:49

Servaes

20.2k33484

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm sorry, I still couldn't find a way for checking it's surjectivity...could you help me with that?
  – Banchin
  Jul 21 at 1:16
  

  
  
  
  

add a comment | 

up vote
0
down vote

If $f$ is strictly monotonous then for all $x,yinBbbR$ you have
$$x>y Rightarrow f(x)>f(y),$$
from which it easily follows that $f(x)=f(y) Rightarrow x=y$, so indeed $f$ is injective.

So $f$ is invertible whenever $p$ is such that $f'(x)>0$ for all $x$, or such that $f'(x)<0$ for all $x$. This still leaves some values of $p$ unresolved though. As this is homework, I'll leave it to you (for now) to figure out what happens then.

EDIT: Because $f'(x)=sin(x)-2p$, if $p>frac12$ then $f'(x)<0$, and if $p<frac12$ then $f'(x)>0$. Hence $f(x)$ is invertible for these values of $p$.

If $-frac12<p<frac12$ then $f'(0)>0$ and $f'(pi)<0$. This means $f$ is not invertible (why?).

Then the only remaining cases are $p=frac12$ and $p=-frac12$.

share|cite|improve this answer

edited Jul 24 at 17:40

answered Jul 18 at 10:49

Servaes

20.2k33484

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm sorry, I still couldn't find a way for checking it's surjectivity...could you help me with that?
  – Banchin
  Jul 21 at 1:16
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

If $f$ is strictly monotonous then for all $x,yinBbbR$ you have
$$x>y Rightarrow f(x)>f(y),$$
from which it easily follows that $f(x)=f(y) Rightarrow x=y$, so indeed $f$ is injective.

So $f$ is invertible whenever $p$ is such that $f'(x)>0$ for all $x$, or such that $f'(x)<0$ for all $x$. This still leaves some values of $p$ unresolved though. As this is homework, I'll leave it to you (for now) to figure out what happens then.

EDIT: Because $f'(x)=sin(x)-2p$, if $p>frac12$ then $f'(x)<0$, and if $p<frac12$ then $f'(x)>0$. Hence $f(x)$ is invertible for these values of $p$.

If $-frac12<p<frac12$ then $f'(0)>0$ and $f'(pi)<0$. This means $f$ is not invertible (why?).

Then the only remaining cases are $p=frac12$ and $p=-frac12$.

share|cite|improve this answer

edited Jul 24 at 17:40

answered Jul 18 at 10:49

Servaes

20.2k33484

If $f$ is strictly monotonous then for all $x,yinBbbR$ you have
$$x>y Rightarrow f(x)>f(y),$$
from which it easily follows that $f(x)=f(y) Rightarrow x=y$, so indeed $f$ is injective.

So $f$ is invertible whenever $p$ is such that $f'(x)>0$ for all $x$, or such that $f'(x)<0$ for all $x$. This still leaves some values of $p$ unresolved though. As this is homework, I'll leave it to you (for now) to figure out what happens then.

EDIT: Because $f'(x)=sin(x)-2p$, if $p>frac12$ then $f'(x)<0$, and if $p<frac12$ then $f'(x)>0$. Hence $f(x)$ is invertible for these values of $p$.

If $-frac12<p<frac12$ then $f'(0)>0$ and $f'(pi)<0$. This means $f$ is not invertible (why?).

Then the only remaining cases are $p=frac12$ and $p=-frac12$.

share|cite|improve this answer

edited Jul 24 at 17:40

answered Jul 18 at 10:49

Servaes

20.2k33484

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 17:40

edited Jul 24 at 17:40

edited Jul 24 at 17:40

answered Jul 18 at 10:49

Servaes

20.2k33484

answered Jul 18 at 10:49

Servaes

20.2k33484

answered Jul 18 at 10:49

Servaes

20.2k33484

20.2k33484

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm sorry, I still couldn't find a way for checking it's surjectivity...could you help me with that?
  – Banchin
  Jul 21 at 1:16
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm sorry, I still couldn't find a way for checking it's surjectivity...could you help me with that?
  – Banchin
  Jul 21 at 1:16
  

  
  
  
  

I'm sorry, I still couldn't find a way for checking it's surjectivity...could you help me with that?
– Banchin
Jul 21 at 1:16

I'm sorry, I still couldn't find a way for checking it's surjectivity...could you help me with that?
– Banchin
Jul 21 at 1:16

add a comment | 

 

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