University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.1 - Functions and Their Graphs - Exercises - Page 11: 6

Answer

$1.\ \ \mathrm{Domain:}\ \ (-\infty ,-4)\cup (-4,4)\cup (4,\infty )$ $2.\ \ \mathrm{Range:}\ \ (-\infty ,-\frac{1}{8}]\cup (0,\infty )$

Work Step by Step

$1.$ Domain of the rational number must not be zero. So take the denominator and compare it to zero to find the undefined points. $t^2-16=0$, $\ \ $ solve for $x$. $t^2=16$ For $\ \ x^2=f(a)\ \ $ the solutions are $\ \ x=\sqrt{f(a)}\ and-\sqrt{f(a)}$ $t=\sqrt{16}\ and \ t=-\sqrt{16}$ $t=4\ and \ t=-4$ These are the undefined points. So, the function domain is, $-4< t <4$, which, in interval notation, can be represented as, $\ \ (-\infty ,-4)\cup (-4,4)\cup (4,\infty )$ $2.$ The smallest value the denominator can take is -16. So, when the denominator is negative, the function can only take on values between $\ \ -\infty\ \ $ and $\ \ \frac{-2}{16}\ \ $, not between $\ \ -\infty\ \ $ and $\ \ 0$. So the range of the function is, $(-\infty ,-\frac{1}{8}]\cup (0,\infty )$
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