University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.1 - Functions and Their Graphs - Exercises - Page 12: 31

Answer

$\mathrm{See\:the\:attachment\:below.}$
1532433254

Work Step by Step

$a).$ Find the equation of line that goes through the points $\ (0,2)\ \ \mathrm{and}\ \ (2,0)\ $ and the line that goes through $\ (2,1)\ \ \mathrm{and}\ \ (5,0).$ Secondly, write down the domain restrictions. $\mathrm{First\:part:}$ $y-1=\frac{0-1}{0+1}(x+1)$ $y=-x\ \ $ on $\ \ [-1,0)$ $\mathrm{Second\:part:}$ Equation of the horizontal line is $\ y=1\ $ on $\ (0,1].$ $\mathrm{Third\:part:}$ $y-1=\frac{0-1}{3-1}(x-1)$ $y=-\frac{1}{2}x+\frac{3}{2}\ \ $ on $\ \ [1,3)$ $b).$ Find the equation of line that goes through the points $\ (-2,-1)\ \ \mathrm{and}\ \ (0,0)\ $ and the line that goes through $\ (0,2)\ \ \mathrm{and}\ \ (1,0).$ Secondly, write down the domain restrictions. $\mathrm{First\:part:}$ $y+1=\frac{0+1}{0+2}(x+2)$ $y=\frac{1}{2}x\ \ $ on $\ \ [-2,0]$ $\mathrm{Second\:part:}$ $y-2=\frac{0-2}{1-0}(x-0)$ $y=-2x+2\ \ $ on $\ \ (0,1]$ $\mathrm{Third\:part:}$ Equation of the horizontal line is $\ y=-1\ $ on $\ (1,3].$
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.