Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.1 Composite Functions - 5.1 Assess Your Understanding - Page 255: 31

Answer

a) $\dfrac{4}{x+4}$; $D=(-\infty,-4)\cup(-4,0)\cup\left(0,\infty\right)$ b) $\dfrac{4-4x}{x}$; $D=(-\infty,0)\cup(0,1)\cup(1,\infty)$ c) $x$; $D=\left(-\infty,1\right)\cup\left(1,\infty\right)$ d) $x$; $D=(-\infty,0)\cup(0,\infty)$

Work Step by Step

We are given the functions: $f(x)=\dfrac{x}{x-1}$ $g(x)=-\dfrac{4}{x}$ Let's note: $D_f$=the domain of $f$ $D_g$=the domain of $g$ We have: $D_f=(-\infty,1)\cup(1,\infty)$ $D_g=(-\infty,0)\cup(0,\infty)$ a) Determine $f\circ g$ and its domain $D_{f\circ g}$: $(f\circ g)(x)=f(g(x))=f\left(-\dfrac{4}{x}\right)=\dfrac{-\dfrac{4}{x}}{-\dfrac{4}{x}-1}=\dfrac{-4}{-4-x}=\dfrac{4}{x+4}$ As $D_g=(-\infty,0)\cup(0,\infty)$ we exclude 0 from the domain of $f\circ g$. Also the domain of $f$ doesn't include 1, so $g(x)\not=1$. Solve: $g(x)=1$ $-\dfrac{4}{x}=1$ $x=-4$ So we also exclude $-4$ from the domain of $f\circ g$. We got: $D_{f\circ g}=(-\infty,-4)\cup(-4,0)\cup\left(0,\infty\right)$ b) Determine $g\circ f$ and its domain $D_{g\circ f}$: $(g\circ f)(x)=g(f(x))=g\left(\dfrac{x}{x-1}\right)=-\dfrac{4}{\dfrac{x}{x-1}}=-\dfrac{4(x-1)}{x}=\dfrac{4-4x}{x}$ As $D_f=(-\infty,1)\cup(1,\infty)$ we exclude 1 from the domain of $g\circ f$. Also the domain of $g$ doesn't include 0, so $f(x)\not=0$. Solve: $f(x)=0$ $\dfrac{x}{x-1}=0$ $x=0$ We exclude 0 from the domain of $f\circ g$ also. We got: $D_{g\circ f}=(-\infty,0)\cup(0,1)\cup(1,\infty)$ c) Determine $f\circ f$ and its domain $D_{f\circ f}$: $(f\circ f)(x)=f(f(x))=f\left(\dfrac{x}{x-1}\right)=\dfrac{\dfrac{x}{x-1}}{\dfrac{x}{x-1}-1}=\dfrac{x}{1}=x$ As $D_f=(-\infty,1)\cup(1,\infty)$ we exclude -1 from the domain of $f\circ f$. Also the domain of $f$ doesn't include 1, so $f(x)\not=1$. Solve: $f(x)=1$ $\dfrac{x}{x-1}=1$ $x=x-1$ $0=-1$ No solution We got: $D_{f\circ f}=\left(-\infty,1\right)\cup(1,\infty)$ d) Determine $g\circ g$ and its domain $D_{g\circ g}$: $(g\circ g)(x)=g(g(x))=g\left(-\dfrac{4}{x}\right)=-\dfrac{4}{-\dfrac{4}{x}}=x$ As $D_g=(-\infty,0)\cup(0,\infty)$ we exclude 0 from the domain of $g\circ g$. Also the domain of $g$ doesn't include 0, so $g(x)\not=0$. Solve: $g(x)=0$ $-\dfrac{4}{x}=0$ No solution We got: $D_{g\circ g}=(-\infty,0)\cup(0,\infty)$
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