Question

What would the inverse be of the function f? What would the inverse domain and range be?

Answer 1

Given the function

$f(x)=\frac{4x+4}{7x-5}$

The inverse is given by

Replace f(x) with y:

$y=\frac{4x+4}{7x-5}$

Swap x with y:

$x=\frac{4y+4}{7y-5}$

Solve for y:

$\begin{gathered} x(7y-5)=\frac{4y+4}{7y-5}(7y-5) \\ x(7y-5)=4y+4 \\ 7xy-5x=4y+4 \\ 7xy-5x+5x=4y+4+5x \\ 7xy=4y+4+5x \\ 7xy-4y=4y+4+5x-4y \\ 7xy-4y=4+5x \end{gathered}$

Factor

$\begin{gathered} y(7x-4)=4+5x \\ y\frac{(7x-4)}{(7x-4)}=\frac{4+5x}{7x-4} \\ y=\frac{4+5x}{7x-4} \end{gathered}$

Answer:

the inverse of function f is, replace y with f^-1(x)

$f^{-1}(x)=\frac{4+5x}{7x-4}$

Domain of f^-1

These are the x-values of the function. First, Find undefined points

$\begin{gathered} 7x-4\ne0 \\ 7x-4+4\ne0+4 \\ 7x\ne4 \\ \frac{7x}{7}\ne\frac{4}{7} \\ x\ne\frac{4}{7} \end{gathered}$

We have that x must be different from 4/7 therefore the domain is:

$\begin{gathered} x<\frac{4}{7} \\ or\text{ } \\ x>\frac{4}{7} \end{gathered}$

Answer:

in interval notation

$(-\infty,\frac{4}{7})\cup(\frac{4}{7},\infty)$

Range of f^-1

These are the values and for which the function is defined. In this case, we have that given a function f and its inverse the domain of f becomes the range of f^-1. Therefore, we find the domain of the function f:

$\begin{gathered} Restrictions \\ 7x-5\ne0 \\ 7x-5+5\ne0+5 \\ 7x\ne5 \\ \frac{7x}{7}\ne\frac{5}{7} \\ x\ne\frac{5}{7} \end{gathered}$

We have that x must be different from 5/7, so:

$\begin{gathered} x<\frac{5}{7} \\ or \\ x>\frac{5}{7} \end{gathered}$

Combining the ranges

$\begin{gathered} f(x)<\frac{5}{7} \\ or \\ f(x)>\frac{5}{7} \end{gathered}$

Answer:

the range in interval notation

$(-\infty,\frac{5}{7})\cup(\frac{5}{7},\infty)$