Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.5 - Continuity - 2.5 Exercises: 58

Answer

a. The equation has at least one real root. b. The root is $1.35$.
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Work Step by Step

$\ln x = 3 -2x$ Equal this function to $f(x)$. $f(x) = \ln x +2x - 3$ From the definition of limits involving $\ln$ we know that: $\lim\limits_{x \to \infty} \ln x = \infty$ and $\lim\limits_{x \to 0} ln x = -\infty$. So we note that in this problem: $\lim\limits_{x \to 0^{+}} \ln x +2x - 3 = -\infty$ $\lim\limits_{x \to \infty} \ln x +2x - 3 = \infty$ We know from the intermediate value theorem that $f(x)$ takes all values between $-\infty$ and $+\infty$. So there is at least one root for $f(x)$. b. To find the interval make one graph using $(\ln x)$ and $(3-2x)$ From the graph we can see that the root lies in 1.35
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