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: 30

Answer

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.
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