Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Section 1.8 - Inequalities - 1.8 Exercises - Page 90: 118

Answer

During the interval: $(30, +\infty)$ $x>30$

Work Step by Step

We can find the range of distances by at first calculating the distance at key point of $500°C$ : $500=\frac{600,000}{x^2+300}$ $x^2+300=\frac{600,000}{500}$ $x^2=1200-300$ $x^2=900$ $x=30 $ $meters$ (In general, $x$ would be $±30$ but in this case $x$ represents distance, which cannot be negative. So we omit it) It is a decreasing function, the temperature decreases as the distance increases, so the temperature would decrease when $x$ is greater than $30$. That means interval $(30, +\infty)$ $x>30$ Using almost the same idea, we could also write: $T<500$ $\frac{600,000}{x^2+300}<500$ $x^2+300>\frac{600,000}{500}$ $x^2>1200-300$ $x^2>900$ $x>30 $
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