1. Given function

We have

and we need the number of points in $(-3,3)$ where $f$ is not differentiable.

So we examine differentiability in the interval $(-3,3)$.

2. For $-2\le x\le 2$: analyze $\min\{|x|,2-x^2\}$

Let $g(x)=|x|,\qquad h(x)=2-x^2.$ We need to know where the minimum switches between these two.

So solve $|x|=2-x^2.$

Since the equation is even, solve separately.

For $x\ge 0$

$x=2-x^2 \implies x^2+x-2=0 \implies (x+2)(x-1)=0.$ Relevant root: $x=1$.

For $x\le 0$

$-x=2-x^2 \implies x^2-x-2=0 \implies (x-2)(x+1)=0.$ Relevant root: $x=-1$.

Thus the two curves meet at $x=-1,\;1.$

Now compare values:

• At $x=0$: $|0|=0$, $2-0=2$, so minimum is $|x|$.
• At $x=\frac32$: $|x|=\frac32$, $2-\frac94=-\frac14$, so minimum is $2-x^2$.

Hence,

More explicitly, using $|x|$:

3. Possible non-differentiable points in $[-2,2]$

Inside these pieces, the function is smooth except possibly at the junctions $x=-1,\;0,\;1,$ and also maybe at $x=\pm 2$ when compared with the outer definition.

(i) At $x=0$

For $x<0$, $f(x)=-x$, so $f'_-(0)=-1.$ For $x>0$, $f(x)=x$, so $f'_+(0)=1.$ Since left and right derivatives are unequal, $f$ is not differentiable at $x=0$.

(ii) At $x=1$

From left, $f(x)=x$, so $f'_-(1)=1.$ From right, $f(x)=2-x^2$, so $f'_+(1)=-2.$ Not equal, so not differentiable at $x=1$.

(iii) At $x=-1$

From left, $f(x)=2-x^2$, so $f'_-( -1)= -2(-1)=2.$ From right, $f(x)=-x$, so $f'_+( -1)=-1.$ Not equal, so not differentiable at $x=-1$.

So far: $3$ points.

4. For $2<|x|\le 3$: analyze $[|x|]$

In $(-3,3)$ this means:

• for $-3<x<-2$, we have $|x|\in(2,3)$, so $[|x|]=2$,
• for $2<x<3$, we have $|x|\in(2,3)$, so $[|x|]=2$.

Thus on both intervals $(-3,-2)$ and $(2,3)$, $f(x)=2,$ a constant function, hence differentiable everywhere there.

The only possible issues are at the joining points $x=-2$ and $x=2$.

5. Check at $x=2$

From the inner definition, $f(2)=\min\{2,2-4\}=\min\{2,-2\}=-2.$ For $x>2$, from outer definition, $f(x)=[|x|]=2.$ So the function jumps from $-2$ to $2$ at $x=2$. Hence it is discontinuous, therefore not differentiable at $x=2$.

6. Check at $x=-2$

From the inner definition, $f(-2)=\min\{2,2-4\}=-2.$ For $x<-2$, from outer definition, $f(x)=[|x|]=2.$ Again there is a jump discontinuity at $x=-2$. Hence not differentiable at $x=-2$.

7. Total count

The points in $(-3,3)$ where $f$ is not differentiable are $x=-2,-1,0,1,2.$ So the total number is $5.$

8. Comparison with stored answer

Derived answer: $5$. Stored correct answer: $5$. They agree.