Question

A function $f$ is defined on $[–3,3]$ as $f\left(x\right)=\left\{\begin{array}{l}min\left\{\left|x\right|,2-x^2\right\},-2\le x\le 2\\ \left[\left|x\right|\right],2<\left|x\right|\le 3\end{array}\right.$, where $\left[x\right]$ denotes the greatest integer $\le x$.

The number of points, where $f$ is not differentiable in $(–3,3)$ is __________ .

Answer 1

The explanation for the answer:

Step 1: Checking the points of the function

The function is $f\left(x\right)=\left\{\begin{array}{l}min\left\{\left|x\right|,2-x^2\right\},-2\le x\le 2\\ \left[\left|x\right|\right],2<\left|x\right|\le 3\end{array}\right.$

The limit is $\left[-3,3\right]$.

So we need to check the points $\left(-2,-1,0,1,2\right)$

We know that, if a function is continuous in the point, then it will be differentiable at that point.

Now we are going to check the continuity of the function at these points.

Step 2: Finding continuity of the function at $x=-2,-1,0$

(a) Take $x=-2,$

$$\begin{gathered} \Rightarrow f(-2)=\left|2-{\left(-2\right)}^2\right| \\ =\left|-2\right| \\ =2\ne 0 \end{gathered}$$

(b) Take $x=-1$,

$$\begin{gathered} \Rightarrow f(-1)=\left|2-{\left(-1\right)}^2\right| \\ =\left|1\right| \\ =1\ne 0 \end{gathered}$$

(c) Take $x=0$,

$$\begin{gathered} \Rightarrow f\left(0\right)=\left|2-{\left(0\right)}^2\right| \\ =\left|2\right| \\ =2\ne 0 \end{gathered}$$

Step 3: Finding continuity of the function at $x=1,2$

(d) Take $x=1,$

$$\begin{gathered} \Rightarrow f\left(1\right)=\left|2-{\left(1\right)}^2\right| \\ =\left|1\right| \\ =1\ne 0 \end{gathered}$$

(e) Take $x=2,$

$$\begin{gathered} \Rightarrow f\left(2\right)=\left|2-{\left(2\right)}^2\right| \\ =\left|-2\right| \\ =2\ne 0 \end{gathered}$$

Thus the function $f\left(x\right)$ is not continuous at $\left(-2,-1,0,1,2\right)$.

Hence, the number of points, where f is not differentiable in $(–3,3)$ is $5$ points.