Question

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

Hint:

For piecewise-defined functions, points of non-differentiability typically occur at (a) the boundaries between the pieces (check if left and right derivatives match) and (b) any points where an individual piece is not differentiable (like the corner in $|x|$ at $x=0$).

Answer 1

Correct Answer: 5

Given: \[ f(x) = \begin{cases} \min\{|x|,\,2-x^2\}, & -2 \le x \le 2 \\ |x|, & 2<|x| \le 3 \end{cases} \] We first determine where \(|x| = 2-x^2\). \[ |x| = 2-x^2 \] For \(x \ge 0\): \[ x = 2 - x^2 \Rightarrow x^2 + x - 2 = 0 \Rightarrow x = 1 \] For \(x<0\): \[ -x = 2 - x^2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow x = -1 \] Hence, \[ f(x) = \begin{cases} -x, & -3<x<-2 \\ 2 - x^2, & -2 \le x \le -1 \\ |x|, & -1<x<1 \\ 2 - x^2, & 1 \le x \le 2 \\ x, & 2<x<3 \end{cases} \] Points of possible non-differentiability: \[ x = -2, -1, 0, 1, 2 \] Thus, total points of non-differentiability = 5. \[ \boxed{5} \]