## Calculus 10th Edition

The function is continuous over the interval $(-\infty, 2)$ U $(2, \infty).$
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 $(2, \infty).$