91Ó°ÊÓ

For each case in the piecewise function, we need to check if there is any discontinuity within the specified interval of the case. Case 1: \(-\frac{1}{x}\) for \(x>2\) or \(-2<x<0\). In this case, the function is continuous on its domain as \(-\frac{1}{x}\) is continuous. Case 2: \(\frac{1}{x}\) for \(0<x<2\). This function is also continuous on its domain.

Now, we need to verify the continuity at the points where the function transitions between cases: \(x=-2, 0, 2\). At \(x=-2\): Since \(-\frac{1}{x}\) is continuous for \(-2<x<0\) and \(\frac{1}{x}\) is continuous for \(0<x<2\). The left and right limits exist for these intervals and they are equal to the value of the function. Thus, the function is discontinuous at this point. At \(x=0\): The left limit is given by the function \(-\frac{1}{x}\) as \(x\) approaches zero from the left, and the right limit is given by the function \(\frac{1}{x}\) as \(x\) approaches zero from the right. Since both left and right limits are not equal and approach infinity, the function is discontinuous at this point as well. At \(x=2\): Similar to the case at \(2\), the left and right limits exist and they are equal to the value of the function at this point. Thus, the function is discontinuous at this point. Hence, \(f(x)\) is discontinuous at \(x=0,+2,-2\).