Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - Review Exercises: 56

Answer

For the function to be continuous $b=-3$ and $c=4.$

Work Step by Step

$|x-2|\ge1\to x-2\ge1$ or $x-2\le1\to x\ge3$ or $x\le1.$ Hence $f(x)=x^2+bx+c$ for $x\le1$ or $x\ge3.$ $\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^-}(x^2+bx+c)=(1^-)^2+b(1^-)+c=1+b+c.$ $\lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^+}(x+1)=1^++1=2.$ $\lim\limits_{x\to3^-}f(x)=\lim\limits_{x\to3^-}(x+1)=3^-+1=4.$ $\lim\limits_{x\to3^+}f(x)=\lim\limits_{x\to3^+}(x^2+bx+c)=(3^+)^2+b(3^+)+c=9+3b+c.$ For the function to be continuous, $\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^+}f(x)$ and $\lim\limits_{x\to3^-}f(x)=\lim\limits_{x\to3^+}f(x)\to$ $2=1+b+c$ and $4=9+3b+c$ Solving the system of equations gives us $(-5-1)=(3-1)b+(1-1)c\to-6=2b\to b=-3\to$ $2=1+(-3)+c\to c=4.$
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.