91Ó°ÊÓ

We have to find the domain and the range of $fand{f}^{-1}$for the function $f\left(x\right)=\frac{2}{3x-5}$

To find the inverse of the function $f\left(x\right)=\frac{2}{3x-5}$we will first replace f(x) with y:

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

After that, interchange the variables x and y :

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

Express the variable y ( then $y=f^{-1}\left(x\right)$):

$$\begin{gathered} x=\frac{2}{3y-5} \\ (3y-5)x=2 \\ 3xy-5x=2 \\ 3xy=5x+2;3x≠0 \\ y=\frac{5x+2}{3x};x≠0 \\ {f}^{-1}\left(x\right)=\frac{5x+2}{3x};x≠0 \end{gathered}$$

In order to check whether we have determined the inverse well we need to see if the expressions are satisfied:

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

$$\begin{gathered} f^{-1}\left(f\right(x\left)\right)=f^{-1}\frac{2}{3x-5} \\ {f}^{-1}\left(f\right(x\left)\right)=\frac{5·{\displaystyle \frac{2}{3x-5}}+2}{3·{\displaystyle \frac{2}{3x-5}}} \\ {f}^{-1}\left(f\right(x\left)\right)=\frac{\frac{10+2(3x-5)}{3x-5}}{{\displaystyle \frac{6}{3x-5}}} \\ {f}^{-1}\left(f\right(x\left)\right)=\frac{\frac{6x}{3x-5}}{{\displaystyle \frac{6}{3x-5}}} \\ {f}^{-1}\left(f\right(x\left)\right)=x \end{gathered}$$

Now

$$\begin{gathered} f\left(f^{-1}\right(x\left)\right)=f\frac{5x+2}{3x} \\ f\left(f^{-1}\right(x\left)\right)=\frac{2}{3·{\displaystyle \frac{5x+2}{3x}}-5} \\ f\left(f^{-1}\right(x\left)\right)=\frac{2}{{\displaystyle \frac{15x+6-15x}{3x}}} \\ f\left(f^{-1}\right(x\left)\right)=\frac{2}{{\displaystyle \frac{2}{x}}} \\ f\left(f^{-1}\right(x\left)\right)=x \end{gathered}$$

Since the inverse functions are bijective (one to one and surjective) then it holds

• The domain of the function f is the range of the function f^-1
• The domain of the function f^-1 is the range of the function f

Therefore if $f:D→R_f$then $f^{-1}:R_f→D$

Therefore, we will determine the domain of the function $fand{f}^{-1}$

• Domain of$f\left(x\right)=\frac{2}{3x-5}$. Since the denominator of the fraction must not be 0 then it is
  $3x-5≠0⇒x≠\frac{5}{3}$
  
• Domain of$f^{-1}\left(x\right)=\frac{5x+1}{3x}$. Since the denominator of the fraction must not be 0 then it is
  $3x≠0⇒x≠0$

Conclusion: $$\begin{gathered} Domainoff=Rangeof{f}^{-1}=R\backslash \left\{\frac{5}{3}\right\}(allrealnumbersexcept1) \\ Rangeoff=Domainof{f}^{-1}=R\backslash \left\{0\right\}(allrealnumbersexcept0) \end{gathered}$$