$a.\quad $no $b.\quad $no
Work Step by Step
The graph of f(x) contains points $(x,f(x))$, where x belongs to the domain of f. $a.$ Observing the graph, no point of the graph has an x-coordinate of 2 (the empty circles indicate that the points do not belong to the graph). So, x=2 is not in the domain of f; f(2) is not defined. Observing the piecewise definition of f, none of the intervals include the value x=2, (the inequalities are strict, none contain "$\leq 2$" or "$\geq 2$"), so f(2) is not defined. Either approach results in the same answer: No, f(2) is not defined. $b.$ Apply "Continuity test", pg.95 The first condition ... 1. $f(c)$ exists ( $c$ lies in the domain of $f$). is not satisfied for $c=2$, so, no, f is not continuous at $x=2$.