## University Calculus: Early Transcendentals (3rd Edition)

$f$ is continuous on $[-1,0)\cup(0,1)\cup(1,2)\cup(2,3)$
From the graph, the following values of $x$ are where there are no breaks in the graph: $[-1,0), (0,1), (1,2)$ and $(2,3)$. - For $x\in[-1,0)$: $f(x)=x^2-1$ For all $c\in[-1,0)$, we have $\lim_{x\to c}f(x)=\lim_{x\to c}(x^2-1)=c^2-1=f(c)$ So $f$ is continuous on $[-1,0)$. - Similarly, for $x\in(0,1)$: $f(x)=2x$ For all $c\in(0,1)$, we have $\lim_{x\to c}f(x)=\lim_{x\to c}(2x)=2c=f(c)$ So $f$ is continuous on $(0,1)$. - For $x\in(1,2)$: $f(x)=-2x+4$ For all $c\in(1,2)$, we have $\lim_{x\to c}f(x)=\lim_{x\to c}(-2x+4)=-2c+4=f(c)$ So $f$ is continuous on $(1,2)$. - For $x\in(2,3)$: $f(x)=0$ For all $c\in(2,3)$, we have $\lim_{x\to c}f(x)=\lim_{x\to c}0=0=f(c)$ So $f$ is continuous on $(2,3)$. Therefore, overall, $f$ is continuous on $[-1,0)\cup(0,1)\cup(1,2)\cup(2,3)$