### Instruction: If $f(a) = \frac{1}{1-a}$, find the product $f^{-1}(a) \times a \times f(a)$.  (Assume $a \neq 0$ and $a \neq 1$.).### Response: To find the inverse function $f^{-1}(a)$, I need to switch the roles of $a$ and $f(a)$ in the original equation and then solve for $f(a)$.
That is, I want to find $f(a)$ such that $a = \frac{1}{1-f(a)}$.
To solve for $f(a)$, I can cross-multiply and get $a - af(a) = 1$.
Then I can isolate $f(a)$ by adding $af(a)$ to both sides and dividing by $a$.
I get $f(a) = \frac{a-1}{a}$.
This means that $f^{-1}(a) = \frac{a-1}{a}$, which is the inverse function of $f(a)$.
Now I can plug this into the product I want to find: $f^{-1}(a) \times a \times f(a)$.
This becomes $\frac{a-1}{a} \times a \times \frac{1}{1-a}$.
I can simplify this by canceling out one $a$ in the numerator and denominator.
I get $\frac{a-1}{1-a}$.
This is the final answer, unless I want to rewrite it in a different form.
For example, I could multiply the numerator and denominator by -1 to get $\frac{1-a}{a-1}$.