Answer:
Answer C: [tex]f(x)=\frac{x+1}{x-1}[/tex]
Step-by-step explanation:
Notice that all the functions look like x plus a constant in the numerator, and x minus the same constant in the denominator.
So we can do a general analysis finding the inverse of a function of the form:
[tex]f(x)=\frac{x+a}{x-a}[/tex]
and finding which value of "a" will make [tex]f(x)=f^{-1}(x)[/tex]
To find the inverse we need to solve for "x" in the equivalent equation:
[tex]y=\frac{x+a}{x-a} \\y\,(x-a)=x+a\\yx-ya=x+a\\yx-x=a+ya\\x(y-1)=a(1+y)\\x=\frac{a(1+y)}{y-1}[/tex]
Then, we see that if [tex]a=1[/tex], then the expression for [tex]f(x)=f^{-1}(x)[/tex]
Then the function that verifies this condition is:
[tex]f(x)=\frac{x+1}{x-1}[/tex]
