#### Answer

a) $y=f(x)$ is differentiable at all points in its domain $[-2,3]$.
b) There are no domain points in $[-2,3]$ where $y=f(x)$ is continuous but not differentiable.
c) There are no domain points in $[-2,3]$where $y=f(x)$ 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 on the closed interval $[-2,3]$. Looking at the curve, we see that it is smooth at all points in the interval; there are no corners or cusps or any points having vertical tangents.
Therefore, we conclude $y=f(x)$ is differentiable at all points in its domain $[-2,3]$.
b) There are no domain points in $[-2,3]$ where $y=f(x)$ is continuous but not differentiable.
c) There are no domain points in $[-2,3]$where $y=f(x)$ is neither continuous nor differentiable, as the graph is a continuous curve in the interval.