Question Number 224922 by fantastic last updated on 12/Oct/25
$${Domain} \\ $$$${f}\left({x}\right)=\sqrt{\mathrm{log}\:_{{e}} \left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{6}\right)} \\ $$
Answered by Raphael254 last updated on 12/Oct/25
![log_e (x^2 −6x+6) ≥ 0 and x^2 − 6x + 6 > 0 x = ((6 ± (√(36 − 24)))/2) = ((6 ± 2(√3))/2) x_1 = 3 + (√3) x_2 = 3 − (√3) x^2 − 6x + 6 > 0 ⇒ x < 3 − (√3) or x > 3 + (√3) log_e (x^2 − 6x + 6)≥ 0 ⇒ x^2 − 6x + 6 ≥ 1 x^2 − 6x + 5 ≥ 0 x = ((6 ± (√(36 − 20)))/2) = ((6 ± 4)/2) x_1 = 5 x_2 = 1 x^2 − 6x + 6 ≥ 1 ⇒ x ≤ 1 or x ≥ 5 (x ≤ 1 or x ≥ 5) and (x < 3 − (√3) or x > 3 + (√3)) 1 < 3 − (√3) and 5 > 3 + (√3) D(f) = (−∞, 1] ∪ [5, ∞)](https://www.tinkutara.com/question/Q224923.png)
$$ \\ $$$${log}_{{e}} \left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{6}\right)\:\geqslant\:\mathrm{0}\:{and}\:{x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:\mathrm{6}\:>\:\mathrm{0} \\ $$$$ \\ $$$${x}\:=\:\frac{\mathrm{6}\:\pm\:\sqrt{\mathrm{36}\:−\:\mathrm{24}}}{\mathrm{2}}\:=\:\frac{\mathrm{6}\:\pm\:\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$ \\ $$$${x}_{\mathrm{1}} \:=\:\mathrm{3}\:+\:\sqrt{\mathrm{3}} \\ $$$${x}_{\mathrm{2}} \:=\:\mathrm{3}\:−\:\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:\mathrm{6}\:>\:\mathrm{0}\:\Rightarrow\:{x}\:<\:\mathrm{3}\:−\:\sqrt{\mathrm{3}}\:{or}\:{x}\:>\:\mathrm{3}\:+\:\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${log}_{{e}} \:\left({x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:\mathrm{6}\right)\geqslant\:\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:\mathrm{6}\:\geqslant\:\mathrm{1} \\ $$$$ \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:\mathrm{5}\:\geqslant\:\mathrm{0} \\ $$$$ \\ $$$${x}\:=\:\frac{\mathrm{6}\:\pm\:\sqrt{\mathrm{36}\:−\:\mathrm{20}}}{\mathrm{2}}\:=\:\frac{\mathrm{6}\:\pm\:\mathrm{4}}{\mathrm{2}} \\ $$$$ \\ $$$${x}_{\mathrm{1}} \:=\:\mathrm{5} \\ $$$${x}_{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$ \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:\mathrm{6}\:\geqslant\:\mathrm{1}\:\Rightarrow\:{x}\:\leqslant\:\mathrm{1}\:{or}\:{x}\:\geqslant\:\mathrm{5} \\ $$$$ \\ $$$$\left({x}\:\leqslant\:\mathrm{1}\:{or}\:{x}\:\geqslant\:\mathrm{5}\right)\:{and}\:\left({x}\:<\:\mathrm{3}\:−\:\sqrt{\mathrm{3}}\:{or}\:{x}\:>\:\mathrm{3}\:+\:\sqrt{\mathrm{3}}\right) \\ $$$$ \\ $$$$\mathrm{1}\:<\:\mathrm{3}\:−\:\sqrt{\mathrm{3}}\:{and}\:\mathrm{5}\:>\:\mathrm{3}\:+\:\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${D}\left({f}\right)\:=\:\left(−\infty,\:\mathrm{1}\right]\:\cup\:\left[\mathrm{5},\:\infty\right) \\ $$