Step 1: Analyze the points where discontinuity might occur.
The given piecewise function transitions at \( x = -3 \) and \( x = 3 \). These are the potential points where discontinuities could arise.
Step 2: Verify continuity at \( x = -3 \).
At \( x = -3 \), calculate the left-hand and right-hand limits: \[ \text{Left-hand limit (LHL)} = |x| + 3 = |-3| + 3 = 3 + 3 = 6, \] \[ \text{Right-hand limit (RHL)} = -2x = -2(-3) = 6. \] The function value at \( x = -3 \) is: \[ f(-3) = |x| + 3 = |-3| + 3 = 6. \] Since the left-hand limit, right-hand limit, and the function's value at \( x = -3 \) are all equal, the function is continuous at \( x = -3 \).
Step 3: Verify continuity at \( x = 3 \).
At \( x = 3 \), calculate the left-hand and right-hand limits: \[ \text{Left-hand limit (LHL)} = -2x = -2(3) = -6, \] \[ \text{Right-hand limit (RHL)} = 6x + 2 = 6(3) + 2 = 18 + 2 = 20. \] Since the left-hand limit and right-hand limit are not equal, the function is discontinuous at \( x = 3 \).
Step 4: Final Conclusion.
There is exactly one point of discontinuity, which occurs at \( x = 3 \). Therefore, the total number of points of discontinuity is: \[ \boxed{1}. \]
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: