In order for a function [tex]f(x)[/tex] to be continuous at a point [tex]x=c[/tex], it has to satisfy the criterion,
[tex]\displaystyle\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c)[/tex]
First, notice that [tex]x<2[/tex] means that [tex]x\neq2[/tex], so the "leftmost" piece can be simplified to
[tex]\dfrac{x^2-4}{x-2}=\dfrac{(x-2)(x+2)}{x-2}=x+2[/tex]
Then we have
[tex]f(x)=\begin{cases}x+2&\text{for }x<2\\ax^2-bx+3&\text{for }2\le x<3\\4x-a+b&\text{for }x\ge3\end{cases}[/tex]
Each of the pieces are polynomials, so they are continuous everywhere. This means we only need to worry about the endpoints of the pieces, so we need to find [tex]a,b[/tex] such that [tex]f(x)[/tex] is continuous at both [tex]x=2[/tex] and [tex]x=3[/tex].
- At [tex]x=2[/tex]: by definition of [tex]f[/tex], we have [tex]f(2)=4a-2b+3[/tex]. We also need
[tex]\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2}(x+2)=4[/tex]
[tex]\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2}(ax^2-bx+3)=4a-2b+3[/tex]
[tex]\implies4a-2b+3=4[/tex]
- At [tex]x=3[/tex]: by definition, [tex]f(3)=12-a+b[/tex], we need
[tex]\displaystyle\lim_{x\to3^-}f(x)=\lim_{x\to3}(ax^2-bx+3)=9a-3b+3[/tex]
[tex]\displaystyle\lim_{x\to3^+}f(x)=\lim_{x\to3}(4x-a+b)=12-a+b[/tex]
[tex]\implies9a-3b+3=12-a+b[/tex]
Now solve for [tex]a,b[/tex]:
[tex]\begin{cases}4a-2b+3=4\\9a-3b+3=12-a+b\end{cases}\implies\begin{cases}4a-2b=1\\10a-4b=9\end{cases}\implies\boxed{a=\frac72,b=\frac{13}2}[/tex]
