Discontinuity Points of Greatest Integer Function in Interval

To find the number of points in the interval $[2, 4]$ at which the function $f(x) = \left[x^2 - x - \frac{1}{2}\right]$ is discontinuous, we analyze the behavior of the inner function $g(x) = x^2 - x - \frac{1}{2}$.

Step 1: Analyze the continuity and monotonicity of $g(x)$

The function $g(x) = x^2 - x - \frac{1}{2}$ is continuous and differentiable everywhere.

Its derivative is given by: $g'(x) = 2x - 1$

For all $x \in [2, 4]$, we have $g'(x) \ge 2(2) - 1 = 3 > 0$. Thus, $g(x)$ is strictly increasing on the interval $[2, 4]$.

Step 2: Determine the range of $g(x)$ on $[2, 4]$

We evaluate $g(x)$ at the endpoints of the interval $[2, 4]$:

• At $x = 2$: $g(2) = 2^2 - 2 - \frac{1}{2} = 4 - 2 - 0.5 = 1.5$
• At $x = 4$: $g(4) = 4^2 - 4 - \frac{1}{2} = 16 - 4 - 0.5 = 11.5$

Since $g(x)$ is strictly increasing and continuous on $[2, 4]$, the range of $g(x)$ for $x \in [2, 4]$ is: $\text{Range}(g) = [1.5, 11.5]$

Step 3: Identify the integer values of $g(x)$

The greatest integer function $f(x) = [g(x)]$ is discontinuous at points where $g(x)$ takes integer values, provided those points lie within the domain and are not continuous from the restricted side at the boundaries.

The integer values present in the range $[1.5, 11.5]$ are: $k \in \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$

Since $g(x)$ is strictly increasing, for each integer $k \in \{2, 3, \dots, 11\}$, there exists a unique value $x_k \in (2, 4)$ such that $g(x_k) = k$.

Step 4: Verify discontinuity at $x_k$ and endpoints

1. At interior points $x_k \in (2, 4)$: For any $x_k$ where $g(x_k) = k$ (where $k \in \{2, 3, \dots, 11\}$):

   • For $x < x_k$ sufficiently close to $x_k$, $k-1 \le g(x) < k \implies f(x) = [g(x)] = k - 1$.
   • For $x > x_k$ sufficiently close to $x_k$, $k \le g(x) < k+1 \implies f(x) = [g(x)] = k$.

   Since $\lim_{x \to x_k^-} f(x) = k - 1 \neq \lim_{x \to x_k^+} f(x) = k$, the function $f(x)$ is discontinuous at each of these 10 interior points $x_k$.

2. At boundary points $x = 2$ and $x = 4$:

   • At $x = 2$: $g(2) = 1.5$, so $f(2) = [1.5] = 1$. For $x \in [2, 2+\delta)$, $g(x) \in [1.5, 2)$, which means $f(x) = 1$. Thus, $f(x)$ is continuous from the right at $x = 2$.
   • At $x = 4$: $g(4) = 11.5$, so $f(4) = [11.5] = 11$. For $x \in (4-\delta, 4]$, $g(x) \in (11, 11.5]$, which means $f(x) = 11$. Thus, $f(x)$ is continuous from the left at $x = 4$.

Conclusion

The function $f(x)$ is discontinuous only at the 10 interior points corresponding to $g(x) = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11$.

Therefore, the total number of points of discontinuity is $10$.