Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.8 Continuity - 1.8 Exercises - Page 92: 30


continuous on $(-2, -\pi/2)\cup(-\pi/2, \pi/2)\cup(\pi/2,2)$

Work Step by Step

$f(x)=\tan x$ is continuous on its domain, $\displaystyle \{x|x\neq\frac{\pi}{2}+\pi\pi\}$ (Theorem 7$: $polynomials, rational functions, root functions, trigonometric functions are continuous on their domains) $v(x)=4-x^{2}$, a polynomial is continuous everywhere (Th.5) $u(x)=\sqrt{x}$, is continuous for $x \geq 0.$ $g(x)=u(v(x))$, a composite of continuous functions is continuous by Th.9. The domain of $g(x)$ is $4-x^{2}\geq 0,$ $4\geq x^{2}$ $x\in[-2,2]$ $B(x)=\displaystyle \frac{f(x)}{g(x)}$ has domain: $ x\displaystyle \neq\frac{\pi}{2}+\pi\pi,$ and $ x\in(-2,2),$ (not defined for $\pm 2$ because of zero in the denominator, and the domain of f demands $\displaystyle \pm\frac{\pi}{2}$ also to be excluded.) Domain= $(-2, -\pi/2)\cup(-\pi/2, \pi/2)\cup(\pi/2,2)$ On this domain B(x) is continuous by Th.4.5.
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.