Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.8 - The Derivative as a Function - 2.8 Exercises: 29

Answer

$$G'(t)=\frac{-7}{(3+t)^2}$$ The domain of both $G(t)$ and $G'(t)$ is $(-\infty,-3)U(-3,\infty)$

Work Step by Step

$$G(t)=\frac{1-2t}{3+t}$$ According to definition, $$G'(t)=\lim\limits_{h\to0}\frac{G(t+h)-G(t)}{h}$$ $$G'(t)=\lim\limits_{h\to0}\frac{\frac{1-2(t+h)}{3+t+h}-\frac{1-2t}{3+t}}{h}$$ $$G'(t)=\lim\limits_{h\to0}\frac{[1-2(t+h)](3+t)-(1-2t)(3+t+h)}{h(3+t+h)(3+t)}$$ $$G'(t)=\lim\limits_{h\to0}\frac{(1-2t-2h)(3+t)-(3+t+h-6t-2t^2-2th)}{h(3+t+h)(3+t)}$$ $$G'(t)=\lim\limits_{h\to0}\frac{3+t-6t-2t^2-6h-2th-3-t-h+6t+2t^2+2th}{h(3+t+h)(3+t)}$$ $$G'(t)=\lim\limits_{h\to0}\frac{-7h}{h(3+t+h)(3+t)}$$ $$G'(t)=\lim\limits_{h\to0}\frac{-7}{(3+t+h)(3+t)}$$ $$G'(t)=\frac{-7}{(3+t+0)(3+t)}$$ $$G'(t)=\frac{-7}{(3+t)^2}$$ *Domain of $G(t)$ $G(t)$ is defined for $R$ except where $3+t=0$, or $t=-3$. Therefore, the domain of $G(t)$ is $(-\infty,-3)U(-3,\infty)$ *Domain of $G'(t)$ $G'(t)$ is defined for $R$ except where $(3+t)^2=0$, or $t=-3$. Therefore, the domain of $G'(t)$ is $(-\infty,-3)U(-3,\infty)$
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