Question Number 99612 by bemath last updated on 22/Jun/20
Commented by john santu last updated on 22/Jun/20
![f(x)= (1/( (√([x]^2 −[x]−6)))) , defined if [x]^2 −[x]−6 > 0 ([x]−3)([x]+2 ) >0 ⇒[x] <−2 or [x] > 3 ⇒ x <−2 or x ≥ 4 Hence Domain x ∈ (−∞,−2) ∪ [4 ,∞)](https://www.tinkutara.com/question/Q99616.png)
$${f}\left({x}\right)=\:\frac{\mathrm{1}}{\:\sqrt{\left[{x}\right]^{\mathrm{2}} −\left[{x}\right]−\mathrm{6}}}\:,\:\mathrm{defined}\:\mathrm{if} \\ $$$$\left[{x}\right]^{\mathrm{2}} −\left[{x}\right]−\mathrm{6}\:>\:\mathrm{0} \\ $$$$\left(\left[{x}\right]−\mathrm{3}\right)\left(\left[{x}\right]+\mathrm{2}\:\right)\:>\mathrm{0} \\ $$$$\Rightarrow\left[{x}\right]\:<−\mathrm{2}\:\mathrm{or}\:\left[{x}\right]\:>\:\mathrm{3} \\ $$$$\Rightarrow\:{x}\:<−\mathrm{2}\:\mathrm{or}\:{x}\:\geqslant\:\mathrm{4}\: \\ $$$$\mathrm{Hence}\:\mathrm{Domain}\:{x}\:\in\:\left(−\infty,−\mathrm{2}\right)\:\cup\:\left[\mathrm{4}\:,\infty\right)\: \\ $$