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

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Question

(C) For the function $f(x) = \frac{1}{\sqrt{[x]^2 - 3[x] - 10}}$ to be defined,the expression inside the square root must be strictly positive: $[x]^2

Vedclass · Accepted Answer

$(-\infty, -2) \cup [6, \infty)$

Vedclass · Answer

(C) For the function $f(x) = \frac{1}{\sqrt{[x]^2 - 3[x] - 10}}$ to be defined,the expression inside the square root must be strictly positive: $[x]^2 - 3[x] - 10 > 0$ Let $t = [x]$. Then $t^2 - 3t - 10 > 0$. Factoring the quadratic: $(t - 5)(t + 2) > 0$. This inequality holds when $t  5$. Since $t = [x]$,we have $[x]  5$. If $[x] If $[x] > 5$,then $[x] \geq 6$,which implies $x \geq 6$. Thus,the domain is $(-\infty, -2) \cup [6, \infty)$.

The domain of the function $f(x) = \frac{1}{\sqrt{[x]^2 - 3[x] - 10}}$ is (where $[x]$ denotes the greatest integer less than or equal to $x$).

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