Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 7 - Functions and Graphs - 7.1 Introduction to Function - 7.1 Exercise Set: 43


g(0) = 5 g(-4 ) = -3 g(-7) = -9 g(8) = 21 g(a+2) = 2a+9 g(a)+2 = 2a+7

Work Step by Step

The problem is asking for us to find the function values. the function is $g(x)=2x+5$ In order to solve for g, we just replace x with g(x). (a) $g(0)$ Part (a) is telling us to replace x with 0. $g(0)=2(0) + 5$ $g(0)=0+5$ $g(0)=5$ Part (b) wants us to replace x with -4. (b) $g(-4)$ $g(-4)=2(-4)+5$ $g(-4)=-8+5$ $g(-4)=-3$ Problem (c) asks use to replace x with -7 (c) $g(-7)$ $g(-7)=2(-7)+5$ $g(-7)=-14+5$ $g(-7)=-9$ Problem (d) asks for us to replace x with 8 (d) $g(8)$ $g(8)=2(8)+5$ $g(8)=16+5$ $g(8)=21$ Problem (e) asks for x to equal (a+2) (e)$g(a+2)$ $g(a+2)=2(a+2)+5$ This problem is different from our previous ones. The other problems required for use to multiply two numbers. This problem, instead has us distribute 2. $g(a+2)=2a+4+5$ You now add all numbers that can be added. 2a cannot be added with the other numbers since it has a variable. $g(a+2)=2a+9$ Part (f) tells us x equals (a) +2. (f) $g(a)+2$ This problem looks similar to part (e). However, there are asking for different things. $g(a)+2) =2(a)+2+5$ In the other problem, you distributed the 2 amongst(a+2). In this problem, however, only a is in the parenthesis. This means only a is multiplied by the number 2. The two parenthesis signify that the equation replaces x, but only a is multiplies by 2, Because of this the next part should look like this. g((a)+2)= 2a+2+5 g((a)+2)= 2a+7 This is your answer for part (f).
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