Answer
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 $x=2$.
Work Step by Step
*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 $x=2$.
