Answer
The function is continuous over the interval $(-\infty, 2)$ U $(2, \infty).$
Work Step by Step
The only possible point of discontinuity is $x=2.$
$\lim\limits_{x\to2^-}f(x)=\lim\limits_{x\to2^-}(5-x)=5-2^-=3.$
$\lim\limits_{x\to2^+}f(x)=\lim\limits_{x\to2^+}(2x-3)=2(2^+)-3=1.$
Since $\lim\limits_{x\to2^-}f(x)\ne\lim\limits_{x\to2^+}f(x)\to\lim\limits_{x\to2}f(x)$ does not exist then the function has a discontinuity at $x=2.$
The function is continuous over the interval $(-\infty, 2)$ U $(2, \infty).$
