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

a) $f(x)$ is differentiable on $[-2,-1)\cup(-1,0)\cup(0,2)\cup(2,3]$ b) $y=f(x)$ is continuous but not differentiable at $x=-1$. c) $y=f(x)$ is neither continuous nor differentiable at $x=0$ and $x=2$.
*Some things to remember about differentiability: - If $f(x)$ is differentiable at $x=c$, then $f(x)$ is continuous at $x=c$. (Theorem 1) - $f(x)$ is not differentiable at $x=c$ if the secant lines passing $x=c$ fail to take up a limiting position or can only take up a vertical tangent. In other words, we can look at differentiability as the ability to draw a tangent line at a point, or the smoothness of the graph. a) In this exercise, we have a rather complicated discontinuous curve $y=f(x)$ on the closed interval $[-2,3]$. We would examine each domain point: - On $[-2,0)$, $f(x)$ is continuous. But at $x=-1$, the graph makes a corner, so $f(x)$ is not differentiable at $x=-1$. At all other points in the domain, we can always draw a tangent line to the graph, so $f(x)$ is all differentiable there. - At $x=0$, $f(x)$ is discontinuous, so it is not differentiable as well. - On $(0,2)$, the graph is continuous. And it is also smooth with no corners or cusps or any points having vertical tangents. So $f(x)$ is differentiable on $(0,2)$. - At $x=2$, $f(x)$ is discontinuous, so it is not differentiable as well. - On $(2,3]$, the graph is continuous. And it is also smooth with no corners or cusps or any points having vertical tangents. So $f(x)$ is differentiable on $(2,3]$. Overall, $f(x)$ is differentiable on $[-2,-1)\cup(-1,0)\cup(0,2)\cup(2,3]$ b) $y=f(x)$ is continuous but not differentiable at $x=-1$. c) $y=f(x)$ is neither continuous nor differentiable at $x=0$ and $x=2$.