Analyzing Discontinuities of $ f(x) = \left[\frac{x^2}{2}\right] - [\sqrt{x}] $

We need to find the number of points of discontinuity for the function $ f(x) = \left[\frac{x^2}{2}\right] - [\sqrt{x}] $ in the interval $ [0,4] $. The greatest integer function $ [y] $ introduces jumps (discontinuities) whenever its argument $ y $ takes an integer value.

Potential Discontinuity Points Identification

Potential discontinuities arise where $ \frac{x^2}{2} $ or $ \sqrt{x} $ equals an integer within the specified interval $ [0,4] $.

• Condition 1: $ \frac{x^2}{2} $ is an integer ($ k $). This occurs when $ x^2 = 2k $. For $ x \in [0,4] $, $ x^2 $ ranges from $ 0 $ to $ 16 $. The possible integer values for $ \frac{x^2}{2} $ are $ 0, 1, 2, 3, 4, 5, 6, 7, 8 $. Solving $ x = \sqrt{2k} $ gives the points: $ x \in \{0, \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4\} $. Let this set be $ A $. Note that $ \sqrt{8} = 2\sqrt{2} $ and $ \sqrt{12} = 2\sqrt{3} $.
• Condition 2: $ \sqrt{x} $ is an integer ($ k $). This occurs when $ x = k^2 $. For $ x \in [0,4] $, the possible integer values for $ \sqrt{x} $ are $ 0, 1, 2 $. Solving $ x = k^2 $ gives the points: $ x \in \{0, 1, 4\} $. Let this set be $ B $.

The set of all points where at least one of the functions might be discontinuous is the union $ P = A \cup B $. $ P = \{0, 1, \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4\} $. This set contains 10 unique points.

Continuity Analysis at Potential Points

The function $ f(x) $ is discontinuous at a point $ c $ if the limit $ \lim_{x \to c} f(x) $ does not exist or does not equal $ f(c) $. Discontinuities can be cancelled if the jumps of $ \left[\frac{x^2}{2}\right] $ and $ [\sqrt{x}] $ offset each other.

• Case 1: Both $ \frac{x^2}{2} $ and $ \sqrt{x} $ are integers. This happens at the endpoints $ x=0 $ and $ x=4 $. - At $ x=0 $: $ f(0) = [0] - [0] = 0 $. The limit from the right is $ \lim_{x \to 0^+} f(x) = [0^+] - [0^+] = 0 $. Since $ f(0) = \lim_{x \to 0^+} f(x) $, the function is continuous at $ x=0 $. - At $ x=4 $: $ f(4) = [8] - [2] = 6 $. The limit from the left is $ \lim_{x \to 4^-} f(x) = [8^-] - [2^-] = 7 - 1 = 6 $. Since $ f(4) = \lim_{x \to 4^-} f(x) $, the function is continuous at $ x=4 $.
• Case 2: Only $ \frac{x^2}{2} $ is an integer. These are the points in $ A $ but not in $ B $: $ A \setminus B = \{\sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}\} $. There are 7 such points. At these points, $ \left[\frac{x^2}{2}\right] $ has a jump discontinuity, while $ [\sqrt{x}] $ does not. This results in a net jump discontinuity for $ f(x) $. For instance, at $ x=\sqrt{2} $, $ f(\sqrt{2}) = [1] - [\sqrt{\sqrt{2}}] = 1 - 1 = 0 $. However, the limit from the left is $ \lim_{x \to \sqrt{2}^-} f(x) = [1^-] - [\sqrt{\sqrt{2}}^-] = 0 - 1 = -1 $. Since $ -1 \neq 0 $, $ f(x) $ is discontinuous at $ x=\sqrt{2} $. This logic applies to all 7 points.
• Case 3: Only $ \sqrt{x} $ is an integer. This occurs at the point in $ B $ but not in $ A $: $ B \setminus A = \{1\} $. There is 1 such point. At $ x=1 $, $ f(1) = [\frac{1^2}{2}] - [\sqrt{1}] = [0.5] - [1] = 0 - 1 = -1 $. The limit from the left is $ \lim_{x \to 1^-} f(x) = [0.5^-] - [1^-] = 0 - 0 = 0 $. Since $ 0 \neq -1 $, $ f(x) $ is discontinuous at $ x=1 $.

Final Count of Discontinuities

The total number of points of discontinuity is the sum of points from Case 2 and Case 3, as the points in Case 1 are continuous. Total discontinuities = $ |A \setminus B| + |B \setminus A| = 7 + 1 = 8 $.