91Ó°ÊÓ

First, consider the expression \(|x|-1\). We can break it into two cases: 1. For \(x\geq 1\), the expression can be simplified to \(x-1\), which is differentiable for all \(x\geq 1\). 2. For \(x< 0\), the expression can be simplified to \(-x-1\), which is differentiable for all \(x< 1\). Thus, \(|x|-1\) is not differentiable only at \(x=1\). Now, consider the expression \(|x-2|-1\). We can also break it into two cases: 1. For \(x\geq 3\), the expression can be simplified to \(x-3\), which is differentiable for all \(x\geq 3\). 2. For \(x< 2\), the expression can be simplified to \(2-x-1\), which is differentiable for all \(x< 2\). Thus, \(|x-2|-1\) is not differentiable only at \(x=2\).

Now, we will analyze the differentiability of the min function based on these intervals and values: 1. If \(x<1\), both \(|x|-1\) and \(|x-2|-1\) are differentiable, so the min function is also differentiable. 2. The function switches from \(|x|-1\) to \(|x-2|-1\) at the intersection of the two expressions, which occurs when \(|x|-1 = |x-2|-1\). Solving for \(x\), we get \(x = \frac{3}{2}\). 3. When \(1\leq x<\frac{3}{2}\), \(f(x)=|x|-1\) is differentiable as it is a simple linear function \(x-1\). However, \(f(x)\) is not differentiable at \(x=1\) as \(|x|-1\) is not differentiable at this point. 4. When \(\frac{3}{2}\leq x<2\), \(f(x)=|x-2| - 1\) is differentiable as it is a simple linear function \(3 - x\). However, \(f(x)\) is not differentiable at \(x=2\) as \(|x-2|-1\) is not differentiable at this point. 5. If \(x\geq 2\), both \(|x|-1\) and \(|x-2|-1\) are differentiable, so the min function is also differentiable.

We found that the function is not differentiable at: 1. \(x = 1\) (transition point for the first absolute value expression) 2. \(x = \frac{3}{2}\) (intersection point of the two expressions) 3. \(x = 2\) (transition point for the second absolute value expression) That gives us a total of 3 points where the function is not differentiable. Therefore, the correct answer is (A) 3 points.