Question
Show that the function $f: R \rightarrow\{3\} \rightarrow R-\{2\}$ given by $f(x)=\frac{x-2}{x-3}$ is a bijection.

Answer

We have, $f: R^{+} \rightarrow R^{+}$given by
$f(x)=x^2$
$g: R^{+} \rightarrow R^{+} \text {given by }$
$g(x)=\sqrt{x}$
$\therefore f(x)=f(g(x))$
$=f(\sqrt{x})=(\sqrt{x})^2=x$
Also, $g \circ f(x)=g(f(x))$
$=g\left(x^2\right)=\sqrt{x^2}=x$
Thus,
$f \circ g(x)=\operatorname{gof}(x)$

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