Question

Let $f:R-\left\{1\right\}\to R$ is defined by $f\left(x\right)=x+1$ and $g\left(x\right):R-\left\{1\right\}\to$ be defined by $g\left(x\right)=\frac{x^2-1}{x-1}$. show that $f=g$.

Answer 1

$g\left(x\right)=\frac{x^2-1}{x-1}$

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

$\sin x\ne 1$

$g\left(x\right)=x+1=f\left(x\right)$

$f=g$