Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 1: Functions - Section 1.2 - Combining Functions; Shifting and Scaling Graphs - Exercises 1.2 - Page 18: 10


$(f\circ g\circ h)(x)=\dfrac{8-3x}{7-2x}.$

Work Step by Step

We are given $f(x)=\dfrac{x+2}{3-x}$, $g(x)=\dfrac{x^2}{x^2+1}$, and $h(x)=\sqrt{2-x}$. So $(f\circ g\circ h)(x)=f(g(h(x)))=f\left(g\left(\sqrt{2-x}\right)\right)=f\left(\dfrac{\left(\sqrt{2-x}\right)^2}{\left(\sqrt{2-x}\right)^2 +1}\right)$. Simplifying, we get $f\left(\dfrac{2-x}{2-x +1}\right)=f\left(\dfrac{2-x}{3-x}\right).$ Then $f\left(\dfrac{2-x}{3-x}\right)=\dfrac{\left(\dfrac{2-x}{3-x}\right)+2}{3-\left(\dfrac{2-x}{3-x}\right)}.$ Now, we multiply the numerator and denominator of our fraction by $3-x$ to get $\dfrac{2-x+2(3-x)}{3(3-x)-(2-x)}=\dfrac{2-x+6-2x}{9-3x-2+x}=\dfrac{8-3x}{7-2x}.$
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