Answer
a) Check the results below.
b) the graph of $f$ must be symmetric over the line $y = x$.
Work Step by Step
To show that when $f(x) = \frac{3- x}{1 -x}$, $f^{-1}(x) = f(x)$.
We are simply showing the above result, we are not deriving it. Thus, it is sufficient to show that $f^{-1}(x) = f(x)$.
When $f^{-1}(x) = f(x)$, $f^{-1}(f(x)) = f(f(x)) =x$.
Thus, we need to show that $f(f(x)) = x$.
$f(f(x) = f(\frac{3- x}{ 1 -x})$
= $\frac{3 - \frac{3- x}{1 - x}}{1 - \frac{3- x}{1 - x}}$
= $\frac{\frac{3(1 - x) - (3 - x)}{1 - x}}{\frac{1(1- x) - (3 - x)}{1 - x}}$ ...... (We use the fact: $\frac{\frac{a}{b}}{\frac{c}{b}} = \frac{a}{c}$)
= $\frac{3 - 3x - 3 + x}{1 - x - 3 + x}$
= $\frac{-2x}{-2}$
=$x$
Hence, $f^{-1}(x) = f(x)$
When $f^{-1}(x) = f(x)$, the graph of $f$ must be symmetric over the line $y = x$.
