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Step-by-Step Solution
1. Understand the Definition of f(x)
The function f is given by:
f(x) = \begin{cases}
\min \{\,|x|,\ 2 - x^2 \} & \text{for } -2 \le x \le 2,\\
[\,|x|\,] & \text{for } 2 < |x| \le 3,
\end{cases}
where [\,\cdot\,] denotes the greatest integer function. We need to find all points in the interval (-3,3) where f is not differentiable.
2. Break Down the Function in the Interval [-2, 2]
Within -2 \le x \le 2 , we have
f(x) = \min\{\,|x|,\ 2 - x^2\}.
Since |x| is -x for x < 0 and x for x \ge 0 , we analyze separately on negative and positive subintervals to see when |x| equals 2 - x^2 .
2.1 Case: x \ge 0
Solve x = 2 - x^2. This gives
x^2 + x - 2 = 0 \quad \Longrightarrow \quad (x-1)(x+2) = 0,
so x = 1 or x = -2 . Only x=1 is valid for x \ge 0 . Thus:
For 0 \le x < 1 , we have x < 2 - x^2 , so f(x) = x.
For x > 1 (up to 2), we have x > 2 - x^2 , so f(x) = 2 - x^2.
At x=1 , both expressions are equal ( 1 = 2 - 1 ), so f(1)=1.
2.2 Case: x \le 0
Here, |x| = -x. Solve -x = 2 - x^2 . That gives
x^2 - x - 2 = 0 \quad \Longrightarrow \quad (x-2)(x+1) = 0,
so x=2 or x=-1 . Only x=-1 is valid for x \le 0 . Thus:
For -2 \le x < -1 , we have -x > 2 - x^2 \Longrightarrow f(x) = 2 - x^2.
For -1 < x \le 0 , we have -x < 2 - x^2 \Longrightarrow f(x) = -x.
At x=-1 , both expressions are equal ( 1 = 2 - 1 ), so f(-1)=1.
2.3 Summary for -2 \le x \le 2
Putting it together:
-2 \le x < -1 : f(x) = 2 - x^2 .
x = -1 : f(x)=1 (common value).
-1 < x < 0 : f(x) = -x.
x = 0 : f(x)=0.
0 < x < 1 : f(x) = x.
x = 1 : f(x)=1 (common value).
1 < x \le 2 : f(x) = 2 - x^2.
3. Analyze the Function for 2 < |x| \le 3
For 2 < |x| \le 3 ,
f(x) = [\,|x|\,].
This is the greatest integer function of |x| . For 2 < |x| < 3 , [\,|x|\,] is constant and equals 2 because 2 < |x| < 3 . At |x|=3 , that is x=3 or x=-3 , we have [3]=3 , but those endpoints are outside (-3,3) or on its boundary, so they do not affect differentiability strictly inside (-3,3) except potentially at x = \pm 3 . However, the question asks for non-differentiability in (-3,3) , so x= \pm 3 are not included.
4. Identify Points of Possible Non-Differentiability
We list all boundary and “corner” points where the definition or formula for f(x) changes or has absolute value behavior:
x=-2 : boundary between f(x)= \min\{|x|, 2 - x^2\} and f(x)= [\,|x|\,].
x=-1 : corner where the formula switches from 2 - x^2 to -x.
x=0 : absolute value “cusp” for |x|.
x=1 : corner where the formula switches from x to 2 - x^2.
x=2 : boundary between f(x)= \min\{|x|, 2 - x^2\} and f(x)= [\,|x|\,].
Next, we check which of these points are inside the open interval (-3,3) and whether the derivative fails to exist there.
5. Checking Non-Differentiability at Each Candidate Point
x = -2 : There is a jump from f(-2)= \min(\,2,\, -2\,)=-2 to f(x) just left of -2 where f(x)= [\,|x|\,]=2. This jump makes f not differentiable at x=-2.
x = -1 : To the left of -1, f(x) = 2 - x^2 (derivative is -2x ), and to the right of -1, f(x) = -x (derivative is -1 ). At x=-1 , these derivatives differ ( 2 on the left vs. -1 on the right), so f is not differentiable at x=-1.
x = 0 : For x0 , f(x)= x (derivative is +1 ). The left and right derivatives do not match, so x=0 is not differentiable.
x = 1 : For x1 , f(x)= 2 - x^2 (derivative is -2x evaluated at 1 gives -2 ). Hence, f is not differentiable at x=1.
x = 2 : There is a jump from f(2)= \min\{2, 2-4\}= -2 to f(x) just right of 2 in which f(x)= [\,|x|\,]=2. Hence, f is also not differentiable at x=2.
6. Count the Non-Differentiable Points
All five points x = -2, -1, 0, 1, 2 lie in (-3,3) and are points of non-differentiability. Thus, the total number of non-differentiable points in (-3,3) is
\boxed{5}.
Reference Diagram from Provided Solution
