Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 1 - Section 1.1 - Four Ways to Represent a Function - 1.1 Exercises - Page 19: 34

Answer

$g(0)=0$ $g(3)=\frac{3}{2}$ $5g(a)=\frac{5a}{\sqrt{a+1}}$ $\frac{1}{2}g(4a)=\frac{2a}{\sqrt{4a+1}}$ $g(a^2)=\frac{a^2}{\sqrt{a^2+1}}$ $[g(a)]^2=\frac{a^2}{a+1}$ $g(a+h)=\frac{a+h}{\sqrt{a+h+1}}$ $g(x-a)=\frac{x-a}{\sqrt{x-a+1}}$

Work Step by Step

$g(x)=\frac{x}{\sqrt{x+1}}$ We can replace the given value for $x$ to calculate the value of the function. $g(0)=\frac{0}{\sqrt{0+1}}=0$ $g(3)=\frac{3}{\sqrt{3+1}}=\frac{3}{2}$ $5g(a)=5\times\frac{a}{\sqrt{a+1}}=\frac{5a}{\sqrt{a+1}}$ $\frac{1}{2}g(4a)=\frac{1}{2}\times\frac{4a}{\sqrt{4a+1}}=\frac{2a}{\sqrt{4a+1}}$ $g(a^2)=\frac{a^2}{\sqrt{a^2+1}}$ $[g(a)]^2=\left[\frac{a}{\sqrt{a+1}}\right]^2=\frac{a^2}{a+1}$ $g(a+h)=\frac{a+h}{\sqrt{a+h+1}}$ $g(x-a)=\frac{x-a}{\sqrt{x-a+1}}$
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