Functions Modeling Change: A Preparation for Calculus, 5th Edition

Published by Wiley
ISBN 10: 1118583191
ISBN 13: 978-1-11858-319-7

Chapter 2 - Functions - 2.5 Preview of Composite and Inverse Functions - Exercises and Problems for Section 2.5 - Exercises and Problems - Page 103: 37

Answer

$f^{-1}(f(A))$ gives the area which can be covered by $f(A)$ gallons; that is, $A$ square feet. Similarly, $f\left(f^{-1}(n)\right)$ gives the number of gallons needed for an area of $f^{-1}(n)$; that is, $n$ gallons.

Work Step by Step

Since $f(A)=A / 250$ and $f^{-1}(n)=250 n$, we have $$ \begin{aligned} f^{-1}(f(A)) & =f^{-1}\left(\frac{A}{250}\right)=250 \frac{A}{250}=A . \\ f\left(f^{-1}(n)\right) & =f(250 n)=\frac{250 n}{250}=n . \end{aligned} $$ We know that $f(A)$ gives the number of gallons of paint needed to cover an area $A$, and $f^{-1}(n)$ gives the area covered by $n$ gallons. Thus $f^{-1}(f(A))$ gives the area which can be covered by $f(A)$ gallons; that is, $A$ square feet. Similarly, $f\left(f^{-1}(n)\right)$ gives the number of gallons needed for an area of $f^{-1}(n)$; that is, $n$ gallons.
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