Discontinuity Analysis of Function $g(x)$

We need to find the points $x$ in the interval $(-2, 2)$ where the function $g(x) = |x|[x^2]$ is discontinuous. The function involves the greatest integer function $[x^2]$ and the absolute value function $|x|$.

Discontinuities in $[x^2]$ occur when $x^2$ is an integer. For $x \in (-2, 2)$, we have $x^2 \in [0, 4)$. The integers in this range are $0, 1, 2, 3$. Therefore, potential points of discontinuity for $[x^2]$ (and possibly $g(x)$) are the values of $x$ where $x^2 \in \{0, 1, 2, 3\}$.

• $x^2 = 0 \implies x = 0$.
• $x^2 = 1 \implies x = \pm 1$.
• $x^2 = 2 \implies x = \pm \sqrt{2}$.
• $x^2 = 3 \implies x = \pm \sqrt{3}$.

The set of candidate points for discontinuity is $\{0, \pm 1, \pm \sqrt{2}, \pm \sqrt{3}\}$.

Continuity Check at Points of Interest

We check the continuity of $g(x) = |x|[x^2]$ at these points. The function $|x|$ is continuous everywhere. Continuity issues arise from $[x^2]$.

• $x=0$: $\lim_{x \to 0} g(x) = |0|[0^2] = 0$. $g(0)=0$. $g(x)$ is continuous at $x=0$.
• $x=1$: $\lim_{x \to 1^-} g(x) = |1|[1^-] = 1 \times 0 = 0$. $\lim_{x \to 1^+} g(x) = |1|[1^+] = 1 \times 1 = 1$. Since the limits differ, $g(x)$ is discontinuous at $x=1$.
• $x=-1$: $\lim_{x \to -1^-} g(x) = |-1|[(-1)^2] = 1 \times [1^-] = 1 \times 0 = 0$. Wait, let's re-evaluate: $\lim_{x \to -1^-} g(x) = |-1||[(-1)^2] = 1 \times 1 = 1$. $\lim_{x \to -1^+} g(x) = |-1||[(-1)^2] = 1 \times [0^+] = 1 \times 0 = 0$. The limits $1$ and $0$ differ. $g(x)$ is discontinuous at $x=-1$.
• $x=\sqrt{2}$: $\lim_{x \to \sqrt{2}^-} g(x) = |\sqrt{2}|[(\sqrt{2})^-] = \sqrt{2} \times 1 = \sqrt{2}$. $\lim_{x \to \sqrt{2}^+} g(x) = |\sqrt{2}|[(\sqrt{2})^+] = \sqrt{2} \times 2 = 2\sqrt{2}$. The limits differ. $g(x)$ is discontinuous at $x=\sqrt{2}$.
• $x=-\sqrt{2}$: $\lim_{x \to -\sqrt{2}^-} g(x) = |-\sqrt{2}|[(-\sqrt{2})^2] = \sqrt{2} \times [2^-] = \sqrt{2} \times 1 = \sqrt{2}$. $\lim_{x \to -\sqrt{2}^+} g(x) = |-\sqrt{2}|[(-\sqrt{2})^2] = \sqrt{2} \times [2^+] = \sqrt{2} \times 2 = 2\sqrt{2}$. The limits differ. $g(x)$ is discontinuous at $x=-\sqrt{2}$.
• $x=\sqrt{3}$: $\lim_{x \to \sqrt{3}^-} g(x) = |\sqrt{3}|[(\sqrt{3})^-] = \sqrt{3} \times 2 = 2\sqrt{3}$. $\lim_{x \to \sqrt{3}^+} g(x) = |\sqrt{3}|[(\sqrt{3})^+] = \sqrt{3} \times 3 = 3\sqrt{3}$. The limits differ. $g(x)$ is discontinuous at $x=\sqrt{3}$.
• $x=-\sqrt{3}$: $\lim_{x \to -\sqrt{3}^-} g(x) = |-\sqrt{3}|[(-\sqrt{3})^2] = \sqrt{3} \times [3^-] = \sqrt{3} \times 2 = 2\sqrt{3}$. $\lim_{x \to -\sqrt{3}^+} g(x) = |-\sqrt{3}|[(-\sqrt{3})^2] = \sqrt{3} \times [3^+] = \sqrt{3} \times 3 = 3\sqrt{3}$. The limits differ. $g(x)$ is discontinuous at $x=-\sqrt{3}$.

The set of points where $g(x)$ is discontinuous in $(-2, 2)$ is $S = \{1, -1, \sqrt{2}, -\sqrt{2}, \sqrt{3}, -\sqrt{3}\}$.

Calculating $f(x)$ at Discontinuity Points

We need to calculate $f(x) = \min \{\sqrt{2}x, x^2\}$ for each $x$ in the set $S$.

• $f(1) = \min \{\sqrt{2}(1), 1^2\} = \min \{\sqrt{2}, 1\} = 1$
• $f(-1) = \min \{\sqrt{2}(-1), (-1)^2\} = \min \{-\sqrt{2}, 1\} = -\sqrt{2}$
• $f(\sqrt{2}) = \min \{\sqrt{2}(\sqrt{2}), (\sqrt{2})^2\} = \min \{2, 2\} = 2$
• $f(-\sqrt{2}) = \min \{\sqrt{2}(-\sqrt{2}), (-\sqrt{2})^2\} = \min \{-2, 2\} = -2$
• $f(\sqrt{3}) = \min \{\sqrt{2}(\sqrt{3}), (\sqrt{3})^2\} = \min \{\sqrt{6}, 3\}$. Since $\sqrt{6} \approx 2.45 < 3$, $f(\sqrt{3}) = \sqrt{6}$.
• $f(-\sqrt{3}) = \min \{\sqrt{2}(-\sqrt{3}), (-\sqrt{3})^2\} = \min \{-\sqrt{6}, 3\}$. Since $-\sqrt{6} \approx -2.45 < 3$, $f(-\sqrt{3}) = -\sqrt{6}$.

Summing $f(x)$ Values

The final step is to sum the calculated $f(x)$ values for $x \in S$.

$\sum_{x \in S} f(x) = f(1) + f(-1) + f(\sqrt{2}) + f(-\sqrt{2}) + f(\sqrt{3}) + f(-\sqrt{3})$

$\sum_{x \in S} f(x) = 1 + (-\sqrt{2}) + 2 + (-2) + \sqrt{6} + (-\sqrt{6})$

$\sum_{x \in S} f(x) = 1 - \sqrt{2} + 2 - 2 + \sqrt{6} - \sqrt{6}$

$\sum_{x \in S} f(x) = 1 - \sqrt{2}$