Question-76277

Question Number 76277 by TawaTawa last updated on 25/Dec/19

Answered by MJS last updated on 25/Dec/19

(1) ⇒ y=4−2x ⇒ (2) 4(4−x)^(4−x) =x(4−2x)^2 3^(4−x) let t=4−x ⇔ x=4−t 4t^t =−4(t−4)(t−2)^2 3^t ((t/3))^t =−(t−4)(t−2)^2 I think we can only try y_l =((t/3))^t is defined for −t∈N ∨ t∈R∧t>0 t=−2n−1∧n∈N ⇒ y_l <0 t=−2n∧n∈N ∨ t∈R∧t>0 ⇒ y_l >0 y_r =−(t−4)(t−2)^2 t=2∨t=4 ⇒ y_r =0 t<4 ⇒ y_r >0 t>4 ⇒ y_r <0 ⇒ we have to search for solutions with 0≤t<2 ∨ 2<t<4 ⇒ t_1 ≈1.60833; t_2 =3 ⇒ x_1 ≈2.39167; y_1 ≈−.783347 x_2 =1; y_2 =2

$$\left(\mathrm{1}\right)\:\Rightarrow\:{y}=\mathrm{4}−\mathrm{2}{x} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{4}\left(\mathrm{4}−{x}\right)^{\mathrm{4}−{x}} ={x}\left(\mathrm{4}−\mathrm{2}{x}\right)^{\mathrm{2}} \mathrm{3}^{\mathrm{4}−{x}} \\ $$$$\mathrm{let}\:{t}=\mathrm{4}−{x}\:\Leftrightarrow\:{x}=\mathrm{4}−{t} \\ $$$$\mathrm{4}{t}^{{t}} =−\mathrm{4}\left({t}−\mathrm{4}\right)\left({t}−\mathrm{2}\right)^{\mathrm{2}} \mathrm{3}^{{t}} \\ $$$$\left(\frac{{t}}{\mathrm{3}}\right)^{{t}} =−\left({t}−\mathrm{4}\right)\left({t}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{try} \\ $$$${y}_{{l}} =\left(\frac{{t}}{\mathrm{3}}\right)^{{t}} \:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:−{t}\in\mathbb{N}\:\vee\:{t}\in\mathbb{R}\wedge{t}>\mathrm{0} \\ $$$${t}=−\mathrm{2}{n}−\mathrm{1}\wedge{n}\in\mathbb{N}\:\Rightarrow\:{y}_{{l}} <\mathrm{0} \\ $$$${t}=−\mathrm{2}{n}\wedge{n}\in\mathbb{N}\:\vee\:{t}\in\mathbb{R}\wedge{t}>\mathrm{0}\:\Rightarrow\:{y}_{{l}} >\mathrm{0} \\ $$$${y}_{{r}} =−\left({t}−\mathrm{4}\right)\left({t}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$${t}=\mathrm{2}\vee{t}=\mathrm{4}\:\Rightarrow\:{y}_{{r}} =\mathrm{0} \\ $$$${t}<\mathrm{4}\:\Rightarrow\:{y}_{{r}} >\mathrm{0} \\ $$$${t}>\mathrm{4}\:\Rightarrow\:{y}_{{r}} <\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{search}\:\mathrm{for}\:\mathrm{solutions}\:\mathrm{with} \\ $$$$\mathrm{0}\leqslant{t}<\mathrm{2}\:\vee\:\mathrm{2}<{t}<\mathrm{4} \\ $$$$\Rightarrow\:{t}_{\mathrm{1}} \approx\mathrm{1}.\mathrm{60833};\:{t}_{\mathrm{2}} =\mathrm{3} \\ $$$$\Rightarrow\:{x}_{\mathrm{1}} \approx\mathrm{2}.\mathrm{39167};\:{y}_{\mathrm{1}} \approx−.\mathrm{783347} \\ $$$$\:\:\:\:\:\:{x}_{\mathrm{2}} =\mathrm{1};\:{y}_{\mathrm{2}} =\mathrm{2} \\ $$

Commented by TawaTawa last updated on 25/Dec/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time} \\ $$

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