The function $f$is one-to-one. Find the inverse of each function and check your answer. State the

domain and the range of $f$and $f^{-1}$.

$f\left(x\right)=\frac{2x+3}{5x-2}$

At first in $y=f\left(x\right)$,interchange the variables $x$and $y$, to obtain $x=f\left(y\right)$.

Then solve the equation for $y$in terms of $x$.

After that, check the result by showing that,

$f^{-1}\left(f\left(x\right)\right)=x$and $f\left(f^{-1}\left(x\right)\right)=x$

Given that, $f\left(x\right)=\frac{2x+3}{5x-2}$

Replacing $f\left(x\right)$with $y$and interchange the variables $x$and $y$in

$y=\frac{2x+3}{5x-2}$

to obtain

$x=\frac{2y+3}{5y-2}$

$x=\frac{2y+3}{5y-2}$

Multiplying both sides by $\left(5y-2\right)$,

$x\left(5y-2\right)=2y+3$

Applying distribution property,

$5xy-2x=2y+3$

Subtract $2y$from both sides and add $2x$with both sides,

$5xy-2y=2x+3$

By factoring,

$y\left(5x-2\right)=2x+3$

Divide by $\left(5x-2\right)$,

$y=\frac{2x+3}{5x-2}$

The inverse is

role="math" localid="1646366294031" $f^{-1}\left(x\right)=\frac{2x+3}{5x-2}$

$f^{-1}\left(f\left(x\right)\right)=f^{-1}\left(\frac{2x+3}{5x-2}\right)$

role="math" localid="1646365915415" $=\frac{2\left({\displaystyle \frac{2x+3}{5x-2}}\right)+3}{5\left({\displaystyle \frac{2x+3}{5x-2}}\right)-2}$[Putting values]

role="math" localid="1646365936240" $=\frac{2\left(2x+3\right)+3\left(5x-2\right)}{5\left(2x+3\right)-2\left(5x-2\right)}$[By simplifying]

role="math" localid="1646365954941" $=\frac{4x+6+15x-6}{10x+15-10x+4}$

$=\frac{19x}{19}$

$=x$

$f\left(f^{-1}\left(x\right)\right)=f\left(\frac{2x+3}{5x-2}\right)$

$=\frac{2\left({\displaystyle \frac{2x+3}{5x-2}}\right)+3}{5\left({\displaystyle \frac{2x+3}{5x-2}}\right)-2}$[Putting values]

$=\frac{2\left(2x+3\right)+3\left(5x-2\right)}{5\left(2x+3\right)-2\left(5x-2\right)}$[Simplifying]

$=\frac{4x+6+15x-6}{10x+15-10x+4}$

$=\frac{19x}{19}$

$=x$

Thus, the result is checked.

$f^{-1}\left(f\left(x\right)\right)=x$, where $x$is in domain of $f$and

$Domainoff=Rangeof{f}^{-1}$

$f\left(f^{-1}\left(x\right)\right)=x$, Where $x$is in domain of $f^{-1}$and

$Rangeoff=Domainof{f}^{-1}$