Key Concepts and Formulas

• Domain of $\frac{1}{\sqrt{g(x)}}$: For the function $f(x) = \frac{1}{\sqrt{g(x)}}$ to be defined, the expression inside the square root must be strictly positive, i.e., $g(x) > 0$.
• Greatest Integer Function ($[x]$): $[x]$ denotes the greatest integer less than or equal to $x$.
  • If $[x] = k$ (where $k$ is an integer), then $k \le x < k+1$.
  • If $[x] < k$ (where $k$ is an integer), then $x < k$.
  • If $[x] > k$ (where $k$ is an integer), then $x \ge k+1$.

Step-by-Step Solution

1. Identify the condition for the domain: The given function is $f(x)=\frac{1}{\sqrt{[x]^{2}-3[x]-10}}$. For $f(x)$ to be defined, the expression under the square root in the denominator must be strictly positive. $[x]^{2}-3[x]-10 > 0$

2. Simplify the inequality using a substitution: Let $y = [x]$. Substituting $y$ into the inequality, we get a standard quadratic inequality: $y^2 - 3y - 10 > 0$

3. Solve the quadratic inequality for $y$: We factor the quadratic expression $y^2 - 3y - 10$. The roots of $y^2 - 3y - 10 = 0$ are found by factoring: $(y-5)(y+2) = 0$ The roots are $y=5$ and $y=-2$. Since the quadratic has a positive leading coefficient, the expression $(y-5)(y+2)$ is positive when $y$ is outside the interval of the roots. Therefore, the solution for $y$ is: $y < -2 \quad \text{or} \quad y > 5$

4. Substitute back $[x]$ and translate into inequalities for $x$: Now, we replace $y$ with $[x]$: $[x] < -2 \quad \text{or} \quad [x] > 5$

Case 1: $[x] < -2$ According to the properties of the greatest integer function, if $[x]$ is strictly less than an integer $k$, then $x$ must be strictly less than $k$. So, $[x] < -2$ implies: $x < -2$ This corresponds to the interval $(-\infty, -2)$.

Case 2: $[x] > 5$ According to the properties of the greatest integer function, if $[x]$ is strictly greater than an integer $k$, then $x$ must be greater than or equal to $k+1$. So, $[x] > 5$ implies: $x \ge 5+1$ $x \ge 6$ This corresponds to the interval $[6, \infty)$.

5. Combine the intervals to form the domain: The domain of $f(x)$ is the union of the intervals found in Case 1 and Case 2. $\text{Domain} = (-\infty, -2) \cup [6, \infty)$

Common Mistakes & Tips

• Strict Inequality: Remember that for $\frac{1}{\sqrt{g(x)}}$, the condition is $g(x) > 0$, not $g(x) \ge 0$. If the expression were in the numerator, $\frac{1}{\sqrt{g(x)}}$ would imply $g(x) \ge 0$ and $g(x) \ne 0$, leading to $g(x) > 0$.
• Translating $[x]$ Inequalities: Be precise when converting inequalities involving $[x]$ to inequalities for $x$.
  • $[x] < k \implies x < k$
  • $[x] > k \implies x \ge k+1$ A common error is to write $x > 5$ for $[x] > 5$. If $x=5.5$, $[x]=5$, which does not satisfy $[x]>5$. However, if $x=6$, $[x]=6$, which satisfies $[x]>5$. Thus, $x \ge 6$ is the correct deduction.

Summary

The domain of the function $f(x)=\frac{1}{\sqrt{[x]^{2}-3[x]-10}}$ requires the expression under the square root to be strictly positive. This leads to the inequality $[x]^{2}-3[x]-10 > 0$. By substituting $y=[x]$, we solved the quadratic inequality $(y-5)(y+2) > 0$ to get $y < -2$ or $y > 5$. Translating back to $[x]$, we have $[x] < -2$ or $[x] > 5$. These inequalities are then correctly interpreted as $x < -2$ and $x \ge 6$, respectively, leading to the domain $(-\infty, -2) \cup [6, \infty)$.

The final answer is $\boxed{(-\infty,-2) \cup[6, \infty)}$.