Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 2 - Review - Exercises - Page 234: 87

Answer

$\sqrt x + 1$

Work Step by Step

f o g o h, given $f(x) = \sqrt{1-x}$ $g(x) = 1-x^2$, $h(x) = 1 + \sqrt x$ $f(g(h(x)))$ = $f(g(1+ \sqrt x))$ =$f(1-(1+\sqrt x)^{2})$ = $f(-2\sqrt x - x)$ $f(-2\sqrt x - x)$ = $\sqrt{1- (-2 \sqrt x - x)}$ = $\sqrt {(\sqrt x + 1)}^2$ = $\sqrt x + 1$
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