Answer
a) $f(x)$ is differentiable on $[-1,0)\cup(0,2]$.
b) $y=f(x)$ is continuous but not differentiable at $x=0$.
c) There are no domain points in $[-1,2]$ in which $f(x)$ is neither continuous nor differentiable.
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 continuous curve $y=f(x)$ on the closed interval $[-1,2]$. But not at all points in the interval is $f(x)$ differentiable.
- On $[-1,0)\cup(0,2]$, the graph $f(x)$ is smooth with no corners or cusps or any points having vertical tangents, so tangent lines can always be drawn at all points in the domain. $f(x)$ is differentiable on $[-1,0)\cup(0,2]$ then.
- At $x=0$, however, there is a cusp. $f(x)$ is not differentiable there.
Overall, $f(x)$ is differentiable on $[-1,0)\cup(0,2]$.
b) $y=f(x)$ is continuous but not differentiable at $x=0$.
c) There are no domain points in $[-1,2]$ in which $f(x)$ is neither continuous nor differentiable.
