Question

Consider the function $f\left(x\right)=\frac{x-1}{x+1}$. What is $\frac{f\left(x\right)+1}{f\left(x\right)-1}+x$ equal to?

• A. $0$
• B. $1$
• C. $2x$
• D. $4x$

Answer 1

The correct option is A $0$

$f\left(x\right)=\frac{x-1}{x+1}$

$\Rightarrow \frac{\frac{x-1}{x+1}+1}{\frac{x-1}{x+1}-1}+x$

$\Rightarrow \frac{2x}{-2}+x$

hence, $f\left(x\right)=\frac{x-1}{x+1}=0$