Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.6 Continuity - 2.6 Exercises - Page 109: 42

Answer

$(-\infty,-1]\cup[1,\infty)$

Work Step by Step

We are given the function: $f(x)=\sqrt{x^4-1}$. First determine the domain on which the function is defined: $x^4-1\geq 0$ $(x^2-1)(x^2+1)\geq 0$ $(x-1)(x+1)(x^2+1)\geq 0$ $D=(-\infty,-1]\cup[1,\infty)$ $f(x)$ is a composed function of the functions $x^4-1$ and $\sqrt x$. Both functions are continuous, therefore their composition is also continuous on the domain, so also left-continuous and right-continuous in $x=-1$ and $x=1$. The interval of continuity is: $(-\infty,-1]\cup[1,\infty)$
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