Let f : R $\to$ R be defined as f (x) = x $-$ 1 and g : R $-$ {1, $-$1} $\to$ R be defined as $g(x) = {{{x^2}} \over {{x^2} - 1}}$.

Then the function fog is :

<p>$f:R \to R$ defined as</p> <p>$f(x) = x - 1$ and $g:R \to \{ 1, - 1\} \to R,\,g(x) = {{{x^2}} \over {{x^2} - 1}}$</p> <p>Now $fog(x) = {{{x^2}} \over {{x^2} - 1}} - 1 = {1 \over {{x^2} - 1}}$</p> <p>$\therefore$ Domain of $fog(x) = R - \{ - 1,1\}$</p> <p>And range of $fog(x) = ( - \infty , - 1] \cup (0,\infty )$</p> <p>Now, $${d \over {dx}}(fog(x)) = {{ - 1} \over {{x^2} - 1}}\,.\,2x = {{2x} \over {1 - {x^2}}}$$</p> <p>$\therefore$ ${d \over {dx}}(fog(x)) > 0$ for ${{2x} \over {(1 - x)(1 + x)}} > 0$</p> <p>$\Rightarrow {x \over {(x - 1)(x + 1)}} < 0$</p> <p>$\therefore$ $x \in ( - \infty , - 1) \cup (0,1)$</p> <p>and ${d \over {dx}}(fog(x)) < 0$ for $x \in ( - 1,0) \cup (1,\infty )$</p> <p>$\therefore$ $fog(x)$ is neither one-one nor onto.</p>