1. **State the problem:** Find the domain of the function $$f(x) = \frac{1}{\sqrt{[x]^2 - 3[x] - 10}}$$ where $[x]$ is the greatest integer less than or equal to $x$ (floor function). 2. **Understand the domain restrictions:** - The expression inside the square root must be positive because the denominator cannot be zero and the square root of a negative number is not real. - So, we require $$[x]^2 - 3[x] - 10 > 0$$ 3. **Solve the inequality:** Let $n = [x]$, an integer. Then solve: $$n^2 - 3n - 10 > 0$$ 4. **Factor the quadratic:** $$n^2 - 3n - 10 = (n - 5)(n + 2)$$ 5. **Analyze the inequality:** $$(n - 5)(n + 2) > 0$$ This product is positive when both factors are positive or both are negative. - Case 1: $$n - 5 > 0 \Rightarrow n > 5$$ - Case 2: $$n + 2 < 0 \Rightarrow n < -2$$ 6. **Domain in terms of $x$:** Since $n = [x]$, the domain is all real $x$ such that: $$[x] > 5 \quad \text{or} \quad [x] < -2$$ This means: - For $$[x] > 5$$, $x$ must be at least 6 (since $[5.999] = 5$ but $[6] = 6$). - For $$[x] < -2$$, $x$ must be less than -2 (since $[-2] = -2$ but $[-2.1] = -3$). 7. **Final domain:** $$(-\infty, -3] \cup [6, \infty)$$ --- **Similar questions:** 1. Find the domain of $$g(x) = \frac{1}{\sqrt{[x]^2 - 4[x] - 5}}$$. 2. Find the domain of $$h(x) = \frac{1}{\sqrt{[x]^2 + 2[x] - 8}}$$. 3. Find the domain of $$k(x) = \frac{1}{\sqrt{[x]^2 - 5[x] + 6}}$$. 4. Find the domain of $$m(x) = \frac{1}{\sqrt{[x]^2 - [x] - 12}}$$. 5. Find the domain of $$p(x) = \frac{1}{\sqrt{[x]^2 - 6[x] + 9}}$$.