Finding $a$ and $b$ so that the function is continuous
$$f(x) = \begin{cases} \displaystyle\frac{x^2-4}{x-2}&\quad x<2\\[0.4em] ax^2-bx+3&\quad 2 \leq x <3\\[0.3em] 2x-a+b&\quad x \geq 3 \end{cases}$$
I can't make the right limit of second piece equal to left limit of the first piece. And then how to solve for a and b? I get $4a-2b = 1$ but what do I do with that?
When I try to graph it I'm confused about how to get the $y$-values.
$\endgroup$ 31 Answer
$\begingroup$First note that $f$ is continuous over $\mathbb R \backslash \{2, 3\}$. Then take the limits as $x$ approaches those two points: $$\begin{align} \lim_{x \to 2^-}f(x) &= \lim_{x \to 2^-}\frac{x^2-4}{x-2} = 4\\ \lim_{x \to 2^+}f(x) &= \lim_{x \to 2^+}\left[ax^2 - bx + 3\right] = 4a - 2b + 3 = f(2)\\ \lim_{x \to 3^-}f(x) &= \lim_{x \to 3^-}\left[ax^2 - bx + 3\right] = 9a - 3b + 3\\ \lim_{x \to 3^+}f(x) &= \lim_{x \to 3^+}\left[2x - a + b\right] = 6 - a + b = f(3)\\ \end{align}$$
The first two have to be equal, and that holds for the last two as well. That's because if $x_0$ is an accumulation point for $f$, then $f$ is continuous in $x_0$ if $\lim_\limits{x \to x_0} f(x) = f(x_2)$: for this to happen the limits from the sides have to be equal. We then have a system of two equations in two unknowns, which is solvable: $$\begin{cases} 4a - 2b + 3 &= 4\\ 9a - 3b + 3 &= 6 - a + b \end{cases}$$
The solution to the system is $$\begin{cases} a = \frac12\\ b = \frac12 \end{cases}$$
And those two values are the only values for which $f$ is continuous over $\mathbb R$.
$\endgroup$ 5
