Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 8 - Section 8.3 - Polynomial and Rational Functions - Exercise Set - Page 595: 40

Answer

a) $f(20)=25000$ dollars b) $f(60)=150000$ dollars, $f(80)=400000$ dollars c)$f(90)=900000$ dollars, $f(95)=1900000$ dollars, $f(99)=9900000$ dollars As $x$ approaches 100%, the cost drastically increases.

Work Step by Step

$f(x)=\frac{100000x}{100-x}$ a) x=20 $f(x)=\frac{100000x}{100-x}$ $f(20)=\frac{100000*20}{100-20}$ $f(20)=\frac{2000000}{80}$ $f(20)=25000$ b) $x=60$ $f(x)=\frac{100000x}{100-x}$ $f(60)=\frac{100000*60}{100-60}$ $f(60)=\frac{6000000}{40}$ $f(60)=150000$ $x=80$ $f(x)=\frac{100000x}{100-x}$ $f(80)=\frac{100000*80}{100-80}$ $f(80)=\frac{8000000}{20}$ $f(80)=400000$ c) $x=90$ $f(x)=\frac{100000x}{100-x}$ $f(90)=\frac{100000*90}{100-90}$ $f(90)=\frac{9000000}{10}$ $f(90)=900000$ $x=95$ $f(x)=\frac{100000x}{100-x}$ $f(95)=\frac{100000*95}{100-95}$ $f(95)=\frac{9500000}{5}$ $f(95)=1900000$ $x=99$ $f(x)=\frac{100000x}{100-x}$ $f(99)=\frac{100000*99}{100-99}$ $f(99)=9900000/1$ $f(99)=9900000$
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