Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 2 - Section 2.7 - Combining Functions - 2.7 Exercises - Page 217: 52

Answer

$f\circ g(x)=x-3;\qquad $ domain: $[3,\infty)$ $ g\circ f(x)=\sqrt{x^{2}-3}; \quad$ domain: $(-\infty,-\sqrt{3})\cup(\sqrt{3},\infty)$ $ f\circ f(x)=x^{4}; \quad$ domain: $(-\infty,\infty)$ $ g\circ g(x)==\sqrt{\sqrt{x-3}-3}; \quad$ domain: all reals, $[12,\infty)$

Work Step by Step

f(x) is defined for all x, g(x) is defined on $[3,\infty)$ , because the radicand can not be negative. $ f\circ g(x)=f[g(x)]= [g(x)]^{2}\quad$on the domain of $g$(x) $=[\sqrt{x-3}]^{2}\quad$ on $[3,\infty)$ $=x-3;\qquad $ domain: $[3,\infty)$ $g\circ f(x)=g[f(x)]=\sqrt{f(x)-3} $on the domain of f(x),\ and $f(x)\geq 3$ $=\sqrt{x^{2}-3},\quad x^{2}\geq 3\Rightarrow x\leq-\sqrt{3}$ or $x\geq\sqrt{3}$ $=\sqrt{x^{2}-3}; \quad$ domain: $(-\infty,-\sqrt{3})\cup(\sqrt{3},\infty)$ $ f\circ f(x)=f[f(x)]=[f(x)]^{2},\qquad$on the domain of f(x) $=(x^{2})^{2},\quad $ on $(-\infty,\infty)$ $=x^{4}; \quad$ domain: $(-\infty,\infty)$ $g\circ g(x)=g[g(x)]=\sqrt{g(x)-3} \quad$on the domain of $g$(x), and $g(x)\geq 3$ $=\sqrt{\sqrt{x-3}-3},\ \qquad\left\{\begin{array}{ll} \sqrt{x-3}\geq 3 & /(..)^{2}\\ x-3\geq 9 & \\ x\geq 12 & \end{array}\right\}$ so the domain is $[12,\infty)$
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