## Calculus 10th Edition

Published by Brooks Cole

# Chapter 1 - Limits and Their Properties - 1.4 Exercises - Page 80: 90

#### Answer

Check work for explanation.

#### Work Step by Step

$f(x)=g(x)+h(x)\to g(x)=-\dfrac{5}{x}$ and $h(x)=\tan{\dfrac{\pi x}{10}}.$ $g(x)$ is continuous for all values of $x$ such that $x\ne0.$ $h(x)=\tan{\dfrac{\pi x}{10}}$ is continuous for all values of $x$ such that $\dfrac{\pi x}{10}\ne\dfrac{\pi}{2}+k\pi\to x=5+10k$ where k is any integer. By plugging in some values for $k, h(x)$ is continuous over the following interval: ...$(-15, -5)$ U $(-5, 5)$ U $(5, 15)$... Since $f(x)$ is the sum of both, then it is continuous over the intersection of both intervals$\to ...(-15, 5)$ U $(-5, 0)$ U $(0, 5)$ U $(5, 15)...$ Since $[1, 4]$ is a subset, $f(x)$ is continuous over $[1, 4].$ $f(1)=-\dfrac{5}{1}+\tan{\dfrac{\pi(1)}{10}}\to f(1)\lt 0.$ $f(4)=-\dfrac{5}{4}+\tan{\dfrac{\pi(4)}{10}}\to f(4)\gt0.$ Since $f(x)$ is continuous over the interval $[1, 4]$ and the sign of $f(x)$ changes over the interval $[1, 4]$, then the Intermediate Value Theorem guarantees at least one root in the interval.

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.