Answer
(a) $f(3)=4 \cdot 3=12$ is the perimeter of a square of side 3 .
b) (b) $f^{-1}(20)= 5$ is the side of a square of perimeter 20.
c) $f^{-1}(P)=\frac{P}{4}$
Work Step by Step
(a) $f(3)=4 \cdot 3=12$ is the perimeter of a square of side 3 .
(b) $f^{-1}(20)$ is the side of a square of perimeter 20 . If $20=4 s$, then $s=5$, so $f^{-1}(20)=5$.
(c) To find $f^{-1}(P)$, solve for $s$ :
$$
\begin{aligned}
P & =4 s \\
s & =\frac{P}{4} \\
f^{-1}(P) & =\frac{P}{4} .
\end{aligned}
$$
