Calculus 10th Edition

The function has an irremovable jump discontinuity at $x=2.$
For this piece-wise function, the only possible point for a discontinuity is $x=2:$ $\lim\limits_{x\to2^-}f(x)=-2(2^-)=-4; \lim\limits_{x\to2^+}f(x)=(2^+)^2-4(2^+)+1=-3.$ Since $\lim\limits_{x\to2^-}f(x)\ne\lim\limits_{x\to2^+}f(x)\to\lim\limits_{x\to2}f(x)$ does not exist, the function is not continuous at the point $x=2$. This is non-removable.