We already know that $f^{-1}$ undoes the action of $f\text{:}$ $$f^{-1}(f(x)) = x$$ In this problem we will see what happens when we compose a function and its inverse in reverse order: $$f(f^{-1}(x))$$ Consider the function $f$ and its inverse: $$\begin{aligned} f(x)\amp= \frac{12}{x+2}\amp f^{-1}(x)\amp = \frac{12-2x}{x} \end{aligned}$$ and use them to answer the following.

1. Evaluate the expression by working from the inside out. In the first blank, you are evaluating $f^{-1}(4)\text{,}$ where it is being entered inside parentheses as the input for $f\text{.}$ In the second blank, you are evaluating to get the final answer.
   $f(f^{-1}(4)) = f\big($$\big) =$
2. Evaluate the expression by working from the inside out. In the first blank, you are evaluating $f^{-1}(3)\text{,}$ where it is being entered inside parentheses as the input for $f\text{.}$ In the second blank, you are evaluating to get the final answer.
   $f(f^{-1}(3)) = f\big($$\big) =$

You should notice that the function $f$ undid the action of $f^{-1}\text{.}$ In function notation, this says: $$f(f^{-1}(x)) = x$$