Let
\(f(x) = \left\{ \begin{array}{cl} ax+3, &\text{ if }x>2, \\ x-5 &\text{ if } -2 \le x \le 2, \\ 2x-b &\text{ if } x <-2. \end{array} \right.\)
Find \(a+b\) if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper).
+2666 Because the graph is continuous when x = 2, \(x- 5 = ax+ 3\). Subbing in what we know for x, we get \(a = -3\)
Now, do the same thing when x = -2.
