91Ó°ÊÓ

We are given the function \(f(x)\) defined as: $$ f(x) = \begin{cases} x^3 - 1, & \text{if}\ sgn(-x) \geq 0 \\ -x + 1, & \text{otherwise} \end{cases} $$ We have to determine the points where the function is continuous, which means the limit of the function exists at those points and is equal to the function's value at that point. Let's find the value of x where \(sgn(-x) = 0\): $$ sgn(-x) = \begin{cases} 1, & x < 0 \\ 0, & x = 0 \\ -1, & x > 0 \\ \end{cases} $$ Since the \(sgn(-x) \geq 0\) for \(x \leq 0\), we rewrite our function to have overlapping cases: $$ f(x) = \begin{cases} x^3 - 1, & x \leq 0 \\ -x + 1, & x > 0 \end{cases} $$ For continuity, we need to check if the function is continuous at \(x = 0\): $$ \lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} (x^3 - 1) $$ $$ \lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} (-x + 1) $$ Since the left and right limits both exist, we need to check if they are equal: $$ \lim_{x \rightarrow 0^{-}} (x^3 - 1) = (0^3 - 1) = -1 $$ $$ \lim_{x \rightarrow 0^{+}} (-x + 1) = -(0) + 1 = 1 $$ Since the left and right limits are not equal, the function is discontinuous at \(x=0\).

The function is already discontinuous at \(x=0\), so we don't need to check for differentiability at \(x=0\). Moreover, the function is defined in separate regions, and in each region, the function is in a differentiable form (polynomial). There are no other points to consider for continuity and differentiability. Therefore, there are no points where the function is continuous but not differentiable. The correct answer is (C) Zero.