Question

What is the expression, domain, and range for the inverse of f(x)?

Answer 1

5.

`f(x)=(4x+4)/(7x-5)`

1st step

Replace f(x) with y:

`y=(4x+4)/(7x-5)`

2nd step:

Replace every x with a y and replace every y with an x:

`x=(4y+4)/(7y-5)`

3rd step:

Solve for y:

`\begin{gathered} x(7y-5)=4y+4 \\ 7xy-5x-4y-4=0 \\ y(7x-4)=5x+4 \\ y=(5x+4)/(7x-4) \end{gathered}`

4th step:

Replace y with f^-1(x)

`f^(-1)(x)=(5x+4)/(7x-4)`

---------------------------------

The domain of f^-1(x) will be given by:

`\begin{gathered} 7x-4\\e0 \\ \end{gathered}`

Since we can't divide by zero, so:

`\begin{gathered} 7x\\e4 \\ x\\e(4)/(7) \\ so\colon \\ D\colon\mleft\lbrace x\in\R\colon x\\e(4)/(7)\mright\rbrace \\ or \\ D\colon(-\infty,(4)/(7))\cup((4)/(7),\infty) \end{gathered}`

The range of f^-1(x) will be the domain of f(x), the domain of f(x) is given by:

`\begin{gathered} 7x-5\\e0 \\ \end{gathered}`

Because we can't divide by zero, so:

`\begin{gathered} 7x\\e5 \\ x\\e(5)/(7) \end{gathered}`

Therefore, the range of f^-1(x) is:

`\begin{gathered} R\colon\mleft\lbrace y\in\R\colon y\\e(5)/(7)\mright\rbrace_{} \\ R\colon(-\infty,(5)/(7))\cup((5)/(7),\infty) \end{gathered}`