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

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Question

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

Vedclass · Accepted Answer

$(-\infty, -1) \cup [3, \infty)$

Vedclass · Answer

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

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

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