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

For $f(x)$ to be continuous at every $x$, $a=4/3$
$f(x)=x^2-1$ for $x\lt3$ and $f(x)=2ax$ for $x\ge3$ For $f(x)$ to be continuous at every $x$, $f(x)$ must be continuous at every point on $(-\infty,\infty)$ We would examine 3 intervals: 1) $x\in(-\infty,3)$ Here $f(x)=x^2-1$. And for all $x\in(-\infty,3)$, we have $$\lim_{x\to c}f(x)=\lim_{x\to c}(x^2-1)=c^2-1=f(c)$$ So $f(x)$ is continuous on $(-\infty,3)$ 2) $x\in(3,\infty)$ Here $f(x)=2ax$. And for all $x\in(3,\infty)$, we have $$\lim_{x\to c}f(x)=\lim_{x\to c}(2ax)=2ac=f(c)$$ So $f(x)$ is continuous on $(3,\infty)$ 3) $x=3$ Here $f(x)=2ax$, so $f(3)=2a\times3=6a$. However, as $x\to3^-$, $f(x)=x^2-1$, so $$\lim_{x\to3^-}f(x)=\lim_{x\to3^-}(x^2-1)=3^2-1=8$$ While at the same time, as $x\to3^+$, $f(x)=2ax$, so $$\lim_{x\to3^+}f(x)=\lim_{x\to3^+}(2ax)=2a\times3=6a=f(3)$$ For $f(x)$ to be continuous at $x=3$, $\lim_{x\to3}f(x)$ must exist, which requires that $$\lim_{x\to3^+}f(x)=\lim_{x\to3^-}f(x)$$ $$6a=8$$ $$a=\frac{4}{3}$$ $\lim_{x\to3}f(x)=f(3)$ is already satisfied, since $\lim_{x\to3^+}f(x)=f(3)$. Therefore, for $f(x)$ to be continuous at every $x$, $a=4/3$