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 - Page 126: 58


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

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
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.