Question Number 204853 by depressiveshrek last updated on 29/Feb/24
$${f}\left({x}\right)=\sqrt{\mathrm{1}−\mathrm{log}_{\left(\mathrm{2}{x}+\mathrm{5}\right)} \left(\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right)} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{this}\:\mathrm{function} \\ $$
Answered by Frix last updated on 29/Feb/24
$$−\frac{\mathrm{5}}{\mathrm{2}}<{x}\leqslant\mathrm{2}\wedge{x}\neq−\mathrm{2}\wedge{x}\neq−\mathrm{1} \\ $$$$ \\ $$$$\mathrm{We}\:\mathrm{could}\:\mathrm{include}\:\mathrm{the}\:\mathrm{limits} \\ $$$$\underset{{x}\rightarrow−\frac{\mathrm{5}}{\mathrm{2}}} {\mathrm{lim}}\:{f}\left({x}\right)\:=\mathrm{1} \\ $$$$\underset{{x}\rightarrow−\mathrm{2}} {\mathrm{lim}}\:{f}\left({x}\right)\:=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:−\frac{\mathrm{5}}{\mathrm{2}}\leqslant{x}\leqslant\mathrm{2}\wedge{x}\neq−\mathrm{1} \\ $$
Answered by TonyCWX08 last updated on 29/Feb/24
![Separate the function: (√(1−log_(2x+5) ((x+1)^2 ))) 1 log_(2x+5) ((x+1)^2 ) 2x+5 (x+1)^2 x+1 Domain (√(1−log_(2x+5) ((x+1)^2 ))) x∈⟨−(5/2),−2⟩∪⟨−2,2] Domain 1 x∈R Domain log_(2x+5) ((x+1)^2 ) x∈⟨−(5/2),−2⟩∪⟨−2,−1⟩∪⟨−1,∞⟩ Domain 2x+5 x∈R Domain (x+1)^2 x∈R Domain x+1 x∈R Finding Intersection Domain x∈⟨−(5/2),−2⟩∪⟨−2,2]](https://www.tinkutara.com/question/Q204865.png)
$${Separate}\:{the}\:{function}: \\ $$$$\sqrt{\mathrm{1}−{log}_{\mathrm{2}{x}+\mathrm{5}} \left(\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right)} \\ $$$$\mathrm{1} \\ $$$${log}_{\mathrm{2}{x}+\mathrm{5}} \left(\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right) \\ $$$$\mathrm{2}{x}+\mathrm{5} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}+\mathrm{1} \\ $$$$ \\ $$$${Domain}\:\sqrt{\mathrm{1}−{log}_{\mathrm{2}{x}+\mathrm{5}} \left(\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right)} \\ $$$$\left.{x}\in\langle−\frac{\mathrm{5}}{\mathrm{2}},−\mathrm{2}\rangle\cup\langle−\mathrm{2},\mathrm{2}\right] \\ $$$${Domain}\:\mathrm{1} \\ $$$${x}\in{R} \\ $$$${Domain}\:{log}_{\mathrm{2}{x}+\mathrm{5}} \left(\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right) \\ $$$${x}\in\langle−\frac{\mathrm{5}}{\mathrm{2}},−\mathrm{2}\rangle\cup\langle−\mathrm{2},−\mathrm{1}\rangle\cup\langle−\mathrm{1},\infty\rangle \\ $$$${Domain}\:\mathrm{2}{x}+\mathrm{5} \\ $$$${x}\in{R} \\ $$$${Domain}\:\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}\in{R} \\ $$$${Domain}\:{x}+\mathrm{1} \\ $$$${x}\in{R} \\ $$$$ \\ $$$${Finding}\:{Intersection} \\ $$$${Domain} \\ $$$$\left.{x}\in\langle−\frac{\mathrm{5}}{\mathrm{2}},−\mathrm{2}\rangle\cup\langle−\mathrm{2},\mathrm{2}\right] \\ $$