Question

Need help ASAP please

Answer 1

Answer:

Answer C:  $f(x)=\frac{x+1}{x-1}$

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:

$f(x)=\frac{x+a}{x-a}$

and finding which value of "a" will make $f(x)=f^{-1}(x)$

To find the inverse we need to solve for "x" in the equivalent equation:

$y=\frac{x+a}{x-a} \\y\,(x-a)=x+a\\yx-ya=x+a\\yx-x=a+ya\\x(y-1)=a(1+y)\\x=\frac{a(1+y)}{y-1}$

Then, we see that if $a=1$, then the expression for $f(x)=f^{-1}(x)$

Then the function that verifies this condition is:

$f(x)=\frac{x+1}{x-1}$