## Calculus 10th Edition

The function has a jump discontinuity at $x=2.$ This discontinuity is irremovable.
For this piece-wise function, the only possible point where a discontinuity could occur is $x=2:$ $\lim\limits_{x\to2^+}f(x)=3-(2^+)=1; \lim\limits_{x\to2^-}f(x)=\frac{1}{2}(2^-)+1=2.$ Since $\lim\limits_{x\to2^-}f(x)\ne\lim\limits_{x\to2^+}f(x)\to\lim\limits_{x\to2}f(x)$ does not exist. Since the limit does not exist, the function is not continuous at the point $x=2.$