The domain of the function f(x) = frac{1}{sqr | vedclass

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Question

(A) For the function $f(x) = \frac{1}{\sqrt{[x]^2 - [x] - 6}}$ to be defined,the expression inside the square root must be strictly positive: $[x]^2 -

Vedclass · Accepted Answer

$(-\infty, -2) \cup [4, \infty)$

Vedclass · Answer

(A) For the function $f(x) = \frac{1}{\sqrt{[x]^2 - [x] - 6}}$ to be defined,the expression inside the square root must be strictly positive: $[x]^2 - [x] - 6 > 0$ Factoring the quadratic expression: $([x] - 3)([x] + 2) > 0$ This inequality holds when: $[x] > 3$ or $[x] If $[x] > 3$,then the smallest integer value for $[x]$ is $4$,which implies $x \geq 4$,or $x \in [4, \infty)$. If $[x] Combining these,the domain is $x \in (-\infty, -2) \cup [4, \infty)$.

The domain of the function $f(x) = \frac{1}{\sqrt{[x]^2 - [x] - 6}}$,where $[x]$ is the greatest integer function $\leq x$,is:

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