A function f is defined on [-3,3] as f(x)=cases|x|,2-x^2, & -2 x2[|x|], & 2

Question

A function $f$ is defined on $[-3,3]$ as $f(x)=\begin{cases}\min\{|x|,2-x^2\}, & -2\le x\le2\[|x|], & 2<|x|\le3\end{cases}$, where $[x]$ denotes the greatest integer $\le x$. The number of points where $f$ is not differentiable in $(-3,3)$ is ..........

Answer: 5

Accepted Answer

Plotting the graph on $(-3,3)$: the curve $\min\{|x|,2-x^2\}$ has "breaking" points where $|x|=2-x^2$ (at $x=\pm1$) and where the two pieces $\min$ and $[|x|]$ join (at $x=\pm2$), plus the corner of $|x|$ at $x=0$. So the points of non-differentiability are $x=-2,-1,0,1,2$ --- a total of 5 points.

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