Find the values of the constant "$A$" so that the function $f(x)$ will be continuous for all "$x$" Find the values of the constant "$A$" so that the function $f(x)$ will be continuous for all "$x$"
$f(x) =\begin{cases}A x-3 &x < 2 \\ 3-x+2x^2 &x \geq 2 \end{cases}\tag*{}$
it says the answer is $A = 6$ but I'm not sure how they reached that answer. Am I supposed to plug in values for the two functions? 
Thanks in advance.
 A: It looks like you are saying that
$f(x) =\begin{cases}A x-3 &x < 2 \\ 3-x+2x^2 &x \geq 2 \end{cases}\tag*{}$
the only possible place of discontinuity is at $x=2$
plugging that in and insisting that the two expressions for $f(x)$ are equal at this point we get
$\begin{equation}\begin{split}A(2)-3& = 3 - 2 + 2(2^2)\\2A-3 &= 9\\2A &= 12\\A&= 6\end{split}\end{equation}\tag*{}$
A: A preliminary observation shows that $f$ is continuous on the open interval $]-\infty, 2[$ and on the half-open interval $[2, +\infty[$ (Try proving this.). Note that $f$ is continuous at $2$ if and only if $\lim_{x \to 2-}f(x) = \lim_{x \to 2+}f(x)$. Recall the definition of the one-sided limits. We observed that either $f$ restricted on $]-\infty, 2[$ or on $[2, +\infty[$ is continuous, so the equality of one-sided limits is equivalent to
$$
A\cdot 2 - 3 = 3 - 2 + 2(2)^{2}.
$$
Now we have one equation and one unknown, so we have exactly one $A$ for which the equality holds ("exactly one" answers the "all" part of the question). 
