Question

Let$f:R\to R$ be defined as $f\left(x\right)=x-1$ and $g:R-\{1,-1\}\to R$ be defined as $g\left(x\right)=\frac{x^2}{x^2-1}$Then the function $fog$ is:

Answer 1

$f:R\to R$ defined as
$f\left(x\right)=x-1\text{and}g:R\to \{1,-1\}\to R,g\left(x\right)=\frac{x^2}{x^2-1}$
Now $f\circ g\left(x\right)=\frac{x^2}{x^2-1}-1=\frac{1}{x^2-1}$
$\therefore$ Domain of $f\circ g\left(x\right)=R-\{-1,1\}$
And range of $f\circ g\left(x\right)=(-\infty ,-1]\cup (0,\infty )$
Now,$\frac{d}{dx}(f\circ g(x\left)\right)=\frac{-1}{{(x^2-1)}^2}\cdot 2x=\frac{-2x}{{(1-x^2)}^2}$
$\therefore \frac{d}{dx}(f\circ g(x\left)\right)>0forx\in (-\infty ,0)$
and $\frac{d}{dx}\left(\mathrm{fog}\right(x\left)\right)<0)forx\in (0,\infty )$
$\therefore f\circ g\left(x\right)$ is neither one-one nor onto.