jjamieson42 wrote:

I get putting in f(x) for x in to the equation for f(y). This results in:

f(y) = [((x + 1) / (x-1)) + 1] / [((x+1) / (x-1))-1]

can someone illustrate the simplification steps to get to f(y) = x ? It's not apparent to me from these explanations.

Hi

jjamieson42Let me try to explain.

Given \(f(x) = \frac{x+1}{x-1}, x \neq 1\) and \(y = f(x)\)

\(f(y) =\frac{ \frac{x+1}{x-1} + 1}{\frac{x+1}{x-1} - 1} = \frac{\frac{x + 1 + x - 1}{x-1}}{\frac{x+1 - x + 1}{x-1}} = \frac{2x}{2} = x\)Hope it helps.

We can solve this question by plug in the value.

Take x =2 , \(y = f(x) = \frac{2+1}{2-1} = 3\)

\(f(y) = f(3) = \frac{3+1}{3-1} = 2 = x\) => x=f(y) => x:f(y) = 1:1 . Answer (C).

Thanks.