Question

Show that, if $f:R-\left\{\frac{7}{5}\right\}\to R-\left\{\frac{3}{5}\right\}$ is defined by f (x) = $\frac{3x+4}{5x-7}$ and $g:R-\left\{\frac{3}{5}\right\}\to R-\left\{\frac{7}{5}\right\}$ is defined by g(x) = $\frac{7x+4}{5x-3}$,
then log = $l_A$ and gof = $l_B$, where $l_A$ and $l_B$ are called identity functions on sets A and B.

Answer 1

$f\left(x\right)=\frac{3x+4}{5x-7},g\left(x\right)=\frac{7x+4}{5x-3}$

Note Heat $f\left(x\right)$ and $g\left(x\right)$ are both one one and onto in the given domain and co-domain hence both are unvertible

$fog\left(x\right)=g\left(g\right(x\left)\right)=\frac{3\left(\frac{7x+4}{5x-3}\right)+4}{5\left(\frac{7x+4}{5x-3}\right)-7}=\frac{21x+12+20x-12}{35x+20-35x+21}=\frac{41x}{41}=x=I_A$

$gof\left(x\right)=g\left(f\right(x\left)\right)=\frac{7\left(\frac{3x+4}{5x-7}\right)+4}{5\left(\frac{3x+4}{5x-7}\right)-3}=\frac{21x+28+20x-28}{15x+20-15x+21}=\frac{41x}{41}=x=I_B$

Hence proved