## Thomas' Calculus 13th Edition

$b=-\displaystyle \frac{1}{2}$
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 $4b=-2$ $b=-\displaystyle \frac{1}{2}$