Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.2 - Basics of Functions and Their Graphs - Exercise Set - Page 176: 54

Answer

See the full explanation below.

Work Step by Step

We have to plot the graph of the functions with the help of the points individually for $f\left( x \right)$ and $g\left( x \right)$. The co-ordinates of $f\left( x \right)$ for $x=0,1,4,9$ are as follows: Put $x=0$ $\begin{align} & f\left( x \right)=\sqrt{x} \\ & =0 \end{align}$ Put $x=1$ $\begin{align} & f\left( x \right)=\sqrt{x} \\ & =\sqrt{1} \\ & =1 \end{align}$ Put $x=4$ $\begin{align} & f\left( x \right)=\sqrt{x} \\ & =\sqrt{4} \\ & =2 \end{align}$ Put $x=9$ $\begin{align} & f\left( x \right)=\sqrt{x} \\ & =\sqrt{9} \\ & =3 \end{align}$ The co-ordinates of $g\left( x \right)=\sqrt{x+2}$ for $x=-2,-1,2,7$ are as follows: Put $x=-2$ $\begin{align} & g\left( x \right)=\sqrt{x+2} \\ & =\sqrt{-2+2} \\ & =0 \end{align}$ Put $x=-1$ $\begin{align} & g\left( x \right)=\sqrt{x+2} \\ & =\sqrt{-1+2} \\ & =\sqrt{1} \\ & =1 \end{align}$ Put $x=2$ $\begin{align} & g\left( x \right)=\sqrt{x+2} \\ & =\sqrt{2+2} \\ & =\sqrt{4} \\ & =2 \end{align}$ Put $x=7$ $\begin{align} & g\left( x \right)=\sqrt{x+2} \\ & =\sqrt{7+2} \\ & =\sqrt{9} \\ & =3 \end{align}$
Small 1567742585
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.