Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.8 - One-to-One Functions and Their Inverses - 2.8 Exercises - Page 228: 96

Answer

(a)$g^{-1}=\sqrt{\frac{4625-r}{18500}}$ It represents distance from central axis ($g^{-1}$) for given velocity ($r$). (b) $g^{-1}(30)\approx 0.4984$ It means, that for $30$ $cm/s$ (blood speed) its distance from central axis is $\approx0.4984cm$

Work Step by Step

$g(r)=18500(0.25-r^2)$ (a) To find $g^{-1}$ we will go through the following steps: First write the function in terms of $y$ and $r$ (Originally $x$): $y=18500(0.25-r^2)$ Now replace $y$ by $r$ and vice versa: $r=18500(0.25-y^2)$ And solve it for $y$: $r=4625-18500y^2$ $18500y^2=4625-r$ $y=\sqrt{\frac{4625-r}{18500}}$ $g^{-1}(r)=\sqrt{\frac{4625-r}{18500}}$ It represents distance (the radius) from central axis for given velocity. (b) $g^{-1}(30)=\sqrt{\frac{4625-30}{18500}}\approx\sqrt{0.2484}\approx0.4984$ It means, that for $30$ $cm/s$ (blood speed) its distance from central axis is $\approx0.4984cm$
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