Show that if $f: R - \{\frac{7}{5}\} \rightarrow R - \{\frac{3}{5}\}$ is defined by $f(x) = \frac{3x+4}{5x-7}$ and $g: R - \{\frac{3}{5}\} \rightarrow R - \{\frac{7}{5}\}$ is defined by $g(x) = \frac{7x+4}{5x-3}$,then $f \circ g = I_{A}$ and $g \circ f = I_{B}$,where $A = R - \{\frac{3}{5}\}$,$B = R - \{\frac{7}{5}\}$; $I_{A}(x) = x, \forall x \in A$,$I_{B}(x) = x, \forall x \in B$ are called identity functions on sets $A$ and $B$,respectively.
(A) To show $g \circ f = I_{B}$,we calculate:
$g(f(x)) = g\left(\frac{3x+4}{5x-7}\right) = \frac{7\left(\frac{3x+4}{5x-7}\right) + 4}{5\left(\frac{3x+4}{5x-7}\right) - 3}$
$= \frac{\frac{21x + 28 + 20x - 28}{5x - 7}}{\frac{15x + 20 - 15x + 21}{5x - 7}} = \frac{41x}{41} = x$
Thus,$g \circ f(x) = x = I_{B}(x)$ for all $x \in B$.
To show $f \circ g = I_{A}$,we calculate:
$f(g(x)) = f\left(\frac{7x+4}{5x-3}\right) = \frac{3\left(\frac{7x+4}{5x-3}\right) + 4}{5\left(\frac{7x+4}{5x-3}\right) - 7}$
$= \frac{\frac{21x + 12 + 20x - 12}{5x - 3}}{\frac{35x + 20 - 35x + 21}{5x - 3}} = \frac{41x}{41} = x$
Thus,$f \circ g(x) = x = I_{A}(x)$ for all $x \in A$.
Therefore,$f \circ g = I_{A}$ and $g \circ f = I_{B}$.
$g(f(x)) = g\left(\frac{3x+4}{5x-7}\right) = \frac{7\left(\frac{3x+4}{5x-7}\right) + 4}{5\left(\frac{3x+4}{5x-7}\right) - 3}$
$= \frac{\frac{21x + 28 + 20x - 28}{5x - 7}}{\frac{15x + 20 - 15x + 21}{5x - 7}} = \frac{41x}{41} = x$
Thus,$g \circ f(x) = x = I_{B}(x)$ for all $x \in B$.
To show $f \circ g = I_{A}$,we calculate:
$f(g(x)) = f\left(\frac{7x+4}{5x-3}\right) = \frac{3\left(\frac{7x+4}{5x-3}\right) + 4}{5\left(\frac{7x+4}{5x-3}\right) - 7}$
$= \frac{\frac{21x + 12 + 20x - 12}{5x - 3}}{\frac{35x + 20 - 35x + 21}{5x - 3}} = \frac{41x}{41} = x$
Thus,$f \circ g(x) = x = I_{A}(x)$ for all $x \in A$.
Therefore,$f \circ g = I_{A}$ and $g \circ f = I_{B}$.
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