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

What is the expression, domain, and range for the inverse of f(x)?-example-1

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5 votes

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}

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