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0

$begingroup$

If $fleft(dfrac{x-1}{x+1}right)=x^2+x-1$. What is $x=?$

A) $0quad$ B) $1quad$ C)$-1quad$ D)$2quad$ E)$-2$

I believe, the problem is missing something, because for every $x$ I plug in, there exists a value of the function.

functions

share|cite|improve this question

edited 14 hours ago

Eldar Rahimli

asked 16 hours ago

Eldar RahimliEldar Rahimli

21510

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This depends obviously of which function $f$ you are considering.
  $endgroup$
  – TheSilverDoe
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is the only information I've been given in the exam. Can you please explain what do you mean by which function $f$?
  $endgroup$
  – Eldar Rahimli
  16 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $x$ can be $0$ is $f$ is constant equal to $-1$ for example. It can be $x=1$ if $f$ is constant equal to $1$. It can be $x=2$ if $f$ is constant equal to $5$. It can be $x=-2$ if $f$ is constant equal to $1$. The only answer $x$ cannot be is $-1$, because in that case, the quotient is not defined.
  $endgroup$
  – TheSilverDoe
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is dubious that the question was reproduced verbatim/in extenso.
  $endgroup$
  – Yves Daoust
  16 hours ago
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

If $fleft(dfrac{x-1}{x+1}right)=x^2+x-1$. What is $x=?$

A) $0quad$ B) $1quad$ C)$-1quad$ D)$2quad$ E)$-2$

I believe, the problem is missing something, because for every $x$ I plug in, there exists a value of the function.

functions

share|cite|improve this question

edited 14 hours ago

Eldar Rahimli

asked 16 hours ago

Eldar RahimliEldar Rahimli

21510

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This depends obviously of which function $f$ you are considering.
  $endgroup$
  – TheSilverDoe
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is the only information I've been given in the exam. Can you please explain what do you mean by which function $f$?
  $endgroup$
  – Eldar Rahimli
  16 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $x$ can be $0$ is $f$ is constant equal to $-1$ for example. It can be $x=1$ if $f$ is constant equal to $1$. It can be $x=2$ if $f$ is constant equal to $5$. It can be $x=-2$ if $f$ is constant equal to $1$. The only answer $x$ cannot be is $-1$, because in that case, the quotient is not defined.
  $endgroup$
  – TheSilverDoe
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is dubious that the question was reproduced verbatim/in extenso.
  $endgroup$
  – Yves Daoust
  16 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

If $fleft(dfrac{x-1}{x+1}right)=x^2+x-1$. What is $x=?$

A) $0quad$ B) $1quad$ C)$-1quad$ D)$2quad$ E)$-2$

I believe, the problem is missing something, because for every $x$ I plug in, there exists a value of the function.

functions

share|cite|improve this question

edited 14 hours ago

Eldar Rahimli

asked 16 hours ago

Eldar RahimliEldar Rahimli

21510

$endgroup$

share|cite|improve this question

edited 14 hours ago

Eldar Rahimli

asked 16 hours ago

Eldar RahimliEldar Rahimli

21510

share|cite|improve this question

edited 14 hours ago

Eldar Rahimli

asked 16 hours ago

Eldar RahimliEldar Rahimli

21510

asked 16 hours ago

Eldar RahimliEldar Rahimli

21510

asked 16 hours ago

Eldar RahimliEldar Rahimli

21510