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

a)$f(x)$ is differentiable on $[-3,2)\cup(-2,2)\cup(2,3]$. b) $y=f(x)$ is continuous but not differentiable at $x=-2$ and $x=2$. c) There are no domain points in $[-3,3]$ in which $f(x)$ is neither continuous nor differentiable.
*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 $[-3,3]$. But not at every point in the interval is $f(x)$ all differentiable. - On $[-3,-2)$, the graph of $f(x)$ is a rising curve 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 in this interval. - At $x=-2$, however, there is a corner in the graph, so no tangent lines can be drawn. $f(x)$ is not differentiable here. - On $(-2,2)$, $f(x)$ has a nice smooth curve where tangent lines can be drawn at all points (no vertical ones). $f(x)$ is all differentiable in this domain. - At $x=2$, we have another corner, so $f(x)$ is not differentiable at $x=2$ even though $f(x)$ is continuous here. - On $(2,3]$, the graph of $f(x)$ is a falling curve with no corners or cusps or any points having vertical tangents. $f(x)$ is differentiable on $(2,3]$ as a result. Overall, $f(x)$ is differentiable on $[-3,2)\cup(-2,2)\cup(2,3]$. b) $y=f(x)$ is continuous but not differentiable at $x=-2$ and $x=2$. c) There are no domain points in $[-3,3]$ in which $f(x)$ is neither continuous nor differentiable.