Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.8 - Inequalities - 1.8 Exercises - Page 90: 104

Answer

The domain of definition is: $(-2, 1]$

Work Step by Step

$\sqrt[4] {\frac{1-x}{2+x}}$ Our restriction here is, that the whole expression under $4^{th}$ root must be non-negative and denominator must not be $0$. $\left\{ \begin{array}{ll} \frac{1-x}{2+x} \geq0\\ 2+x\ne0 \end{array} \right. $ For the first restriction to be satisfied we have $2$ possible intervals: $\left\{ \begin{array}{ll} 1-x <0\\ 2+x<0 \end{array} \right. $ $\left\{ \begin{array}{ll} x >1\\ x0\\ 2+x>0 \end{array} \right. $ $\left\{ \begin{array}{ll} x <1\\ x>-2 \end{array} \right. $ In this case, we have domain $[-2, 1]$ But note, our $2^{nd}$ restriction doesn't permit us to include $-2$ in the interval: $2+x\ne0$ $x\ne-2$ So, at last the domain of definition is: $(-2, 1]$
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