Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.5 - Continuity - Exercises 2.5 - Page 85: 44


$b=-\displaystyle \frac{1}{2}$

Work Step by Step

The piecewise defined functions are continuous on their intervals because they are polynomials. The only problem that could arise is at $x=-2$. Find the one-sided limits at $x=-2$ and set them to be equal, in which case there will exist a limit and g will be continuous at $x=-2$: $\displaystyle \lim_{x\rightarrow-2^{-}} g(x)=\displaystyle \lim_{x\rightarrow-2^{-}}x=-2$ $\displaystyle \lim_{x\rightarrow-2^{+}} g(x)=\displaystyle \lim_{x\rightarrow-2^{+}}bx^{2}=4b$ Set the one-sided limits to be equal $4b=-2$ $b=-\displaystyle \frac{1}{2}$
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