## University Calculus: Early Transcendentals (3rd Edition)

For $g(x)$ to be continuous at every $x$, $b=-1/2$
$g(x)=x$ for $x\lt-2$ and $g(x)=bx^2$ for $x\ge-2$ For $g(x)$ to be continuous at every $x$, $g(x)$ must be continuous at every point on $(-\infty,\infty)$ We would examine 3 intervals: 1) $x\in(-\infty,-2)$ Here $g(x)=x$. And for all $x\in(-\infty,-2)$, we have $$\lim_{x\to c}g(x)=\lim_{x\to c}x=c=f(c)$$ So $g(x)$ is continuous on $(-\infty,-2)$ 2) $x\in(-2,\infty)$ Here $g(x)=bx^2$. And for all $x\in(-2,\infty)$, we have $$\lim_{x\to c}g(x)=\lim_{x\to c}(bx^2)=bc^2=f(c)$$ So $g(x)$ is continuous on $(-2,\infty)$ 3) $x=-2$ Here $g(x)=bx^2$, so $g(-2)=b(-2)^2=4b$. However, as $x\to-2^-$, $g(x)=x$, so $$\lim_{x\to-2^-}g(x)=\lim_{x\to-2^-}(x)=-2$$ While at the same time, as $x\to-2^+$, $g(x)=bx^2$, so $$\lim_{x\to-2^+}g(x)=\lim_{x\to-2^+}(bx^2)=b\times(-2)^2=4b=g(-2)$$ For $g(x)$ to be continuous at $x=-2$, $\lim_{x\to-2}g(x)$ must exist, which requires that $$\lim_{x\to-2^+}g(x)=\lim_{x\to-2^-}g(x)$$ $$4b=-2$$ $$b=-\frac{1}{2}$$ $\lim_{x\to-2}g(x)=g(-2)$ is already satisfied, since $\lim_{x\to-2^+}g(x)=g(-2)$. Therefore, for $g(x)$ to be continuous at every $x$, $b=-1/2$