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 :
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Official solution
\[f:R \to R\] defined as
f(x)=x−1 and \[g:R \to \{ 1, - 1\} \to R,\,g(x) = {{{x^2}} \over {{x^2} - 1}}\]
Now \[fog(x) = {{{x^2}} \over {{x^2} - 1}} - 1 = {1 \over {{x^2} - 1}}\]
\[\therefore\] Domain of \[fog(x) = R - \{ - 1,1\} \]
And range of \[fog(x) = ( - \infty , - 1] \cup (0,\infty )\]
Now, \[{d \over {dx}}(fog(x)) = {{ - 1} \over {{x^2} - 1}}\,.\,2x = {{2x} \over {1 - {x^2}}}\]
\[\therefore\] \[{d \over {dx}}(fog(x)) > 0\] for \[{{2x} \over {(1 - x)(1 + x)}} > 0\]