Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 4 - Section 4.2 - The Natural Exponential Function - 4.2 Exercises - Page 343: 29

Answer

(a) $t=200$ $P(200)\approx11.7911$ Billions $t=300$ $P(200)\approx11.9713$ Billions (b) See the graph below. (c) According to this model, the world population seems to approach $12$ Billions.

Work Step by Step

According to the information provided we have the following function: $P(t)=\frac{73.2}{6.1 + 5.9e^{-0.02t}}$ Where $t$ stands for years passed after $2000$ year. (a) In the year $2200$ we have $t=200$ $P(200)=\frac{73.2}{6.1 + 5.9e^{-0.02\times200}}=\frac{73.2}{6.1 + 5.9e^{-4}}\approx11.7911$ Billions In the year $2300$ we have $t=300$ $P(200)=\frac{73.2}{6.1 + 5.9e^{-0.02\times300}}=\frac{73.2}{6.1 + 5.9e^{-6}}\approx11.9713$ Billions (b) See the graph above. Note, $t$ stands for years passed after the year $2000$ (c) According to this model, the world population seems to approach $12$ Billions. Which, according to the information, is the amount the planet can support.
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