Question

Let $f:R-\left\{1\right\}\to R$ be defined $f\left(x\right)=\frac{x+1}{x-1}$, show that $f\left(x\right)+f\left(\frac{1}{x}\right)=0$ for $(x\ne 0)$.

Answer 1

Given: $f\left(x\right)=\frac{x+1}{x-1}$ where, $f:R-\left\{1\right\}\to R$

To Show: $f\left(x\right)+f\left(\frac{1}{x}\right)=0$ for$(x\ne 0)$

Solution:

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

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

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