Let $f:\mathbb{R}^{\leq 5}\to\mathbb{R}$, $f(x) = \begin{cases} x & x\leq 1 \\ 4-2x & 1< x\leq 5 \end{cases}$
Use the $\epsilon$, $\delta$ definition of continuity to prove $f$ is continuous on $[-\infty,1)\cup (1,5]$ and that it is not continuous at $1$
I am not sure how to go about this problem, should I prove each interval is continuous separately and then prove it is discontinuous at 1? Any help is appreciated
For $x\le1 $, let $\epsilon \gt 0$ be given. Choose $\delta=\epsilon $. Then $\vert x-y\vert \lt \delta \implies \vert f (y)-f (x)\vert =\vert y-x \vert \lt \delta = \epsilon $. Thus $f$ is continuous for $x\lt 1$.
Similarly for $1\lt x \le 5$ if $\epsilon \gt 0$ is given, let $\delta \lt \frac\epsilon2 $. Then $\vert y-x \vert \lt \delta \implies \vert f (y)-f (x) \vert =\vert (4-2y)-(4-2x)\vert=\vert 2 (x-y )\vert\lt 2\cdot\frac \epsilon 2=\epsilon $. Thus f is continuous.
Finally, for $x=1$, given $\epsilon \lt 1$, there are $y $ with $\vert y-x \vert\lt \delta $, for any $\delta\gt 0 $, such that $\vert f (y)-f (1)\vert \gt \epsilon $. Namely, let $y=1+\eta$, where $0\lt \eta\lt \delta $. Then $\vert f(y) -f (1)\vert=\vert (4-2y)-1\vert=3-2(1+\eta)\vert=\vert 1-2\eta\vert\ge 1-\vert 2\eta\vert \gt \epsilon $ for sufficiently small $\eta $... In particular,