5.
[tex]f(x)=\frac{4x+4}{7x-5}[/tex]
1st step
Replace f(x) with y:
[tex]y=\frac{4x+4}{7x-5}[/tex]
2nd step:
Replace every x with a y and replace every y with an x:
[tex]x=\frac{4y+4}{7y-5}[/tex]
3rd step:
Solve for y:
[tex]\begin{gathered} x(7y-5)=4y+4 \\ 7xy-5x-4y-4=0 \\ y(7x-4)=5x+4 \\ y=\frac{5x+4}{7x-4} \end{gathered}[/tex]
4th step:
Replace y with f^-1(x)
[tex]f^{-1}(x)=\frac{5x+4}{7x-4}[/tex]
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The domain of f^-1(x) will be given by:
[tex]\begin{gathered} 7x-4\ne0 \\ \end{gathered}[/tex]
Since we can't divide by zero, so:
[tex]\begin{gathered} 7x\ne4 \\ x\ne\frac{4}{7} \\ so\colon \\ D\colon\mleft\lbrace x\in\R\colon x\ne\frac{4}{7}\mright\rbrace \\ or \\ D\colon(-\infty,\frac{4}{7})\cup(\frac{4}{7},\infty) \end{gathered}[/tex]
The range of f^-1(x) will be the domain of f(x), the domain of f(x) is given by:
[tex]\begin{gathered} 7x-5\ne0 \\ \end{gathered}[/tex]
Because we can't divide by zero, so:
[tex]\begin{gathered} 7x\ne5 \\ x\ne\frac{5}{7} \end{gathered}[/tex]
Therefore, the range of f^-1(x) is:
[tex]\begin{gathered} R\colon\mleft\lbrace y\in\R\colon y\ne\frac{5}{7}\mright\rbrace_{} \\ R\colon(-\infty,\frac{5}{7})\cup(\frac{5}{7},\infty) \end{gathered}[/tex]
