Question-145879

Question Number 145879 by mathdanisur last updated on 09/Jul/21

Answered by mr W last updated on 09/Jul/21

let u=f(f(x)) f(u)≤0 ⇒((1−u)/(1+u^2 ))≤0 ⇒1−u≤0 ⇒u≥1 let v=f(x) u=f(v)≥1 ⇒((1−v)/(1+v^2 ))≥1 ⇒1−v≥1+v^2 ⇒v^2 +v≤0 ⇒v(v+1)≤0 ⇒−1≤v≤0 ⇒−1≤f(x)≤0 ⇒−1≤((1−x)/(1+x^2 ))≤0 ⇒−1−x^2 ≤1−x≤0 ⇒x^2 −x+2≥0 ∧ x≥1 x^2 −x+2≥0 is always true. ⇒x≥1 ⇐ solution

$${let}\:{u}={f}\left({f}\left({x}\right)\right) \\ $$$${f}\left({u}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}−{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\leqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}−{u}\leqslant\mathrm{0} \\ $$$$\Rightarrow{u}\geqslant\mathrm{1} \\ $$$${let}\:{v}={f}\left({x}\right) \\ $$$${u}={f}\left({v}\right)\geqslant\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}−{v}}{\mathrm{1}+{v}^{\mathrm{2}} }\geqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{1}−{v}\geqslant\mathrm{1}+{v}^{\mathrm{2}} \\ $$$$\Rightarrow{v}^{\mathrm{2}} +{v}\leqslant\mathrm{0} \\ $$$$\Rightarrow{v}\left({v}+\mathrm{1}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow−\mathrm{1}\leqslant{v}\leqslant\mathrm{0} \\ $$$$\Rightarrow−\mathrm{1}\leqslant{f}\left({x}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow−\mathrm{1}\leqslant\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\leqslant\mathrm{0} \\ $$$$\Rightarrow−\mathrm{1}−{x}^{\mathrm{2}} \leqslant\mathrm{1}−{x}\leqslant\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}+\mathrm{2}\geqslant\mathrm{0}\:\wedge\:{x}\geqslant\mathrm{1} \\ $$$${x}^{\mathrm{2}} −{x}+\mathrm{2}\geqslant\mathrm{0}\:{is}\:{always}\:{true}. \\ $$$$\Rightarrow{x}\geqslant\mathrm{1}\:\Leftarrow\:{solution} \\ $$

Commented by iloveisrael last updated on 09/Jul/21

$$\mathrm{v}\left(\mathrm{v}+\mathrm{1}\right)\geqslant\mathrm{0}\Rightarrow\mathrm{v}\leqslant−\mathrm{1}\:\cup\:\mathrm{v}\geqslant\:\mathrm{0} \\ $$$$ \\ $$

Commented by iloveisrael last updated on 09/Jul/21

v ≤−1 ∪ v≥0 f(x)≤−1 ∪ f(x)≥ 0 ((1−x)/(1+x^2 )) ≤−1 ∪ ((1−x)/(1+x^2 )) ≥0 ⇒1−x≤−x^2 −1 ∪ 1−x≥0 ⇒x^2 −x+2≤0_(x=∅) ∪ x≤1

$$\mathrm{v}\:\leqslant−\mathrm{1}\:\cup\:\mathrm{v}\geqslant\mathrm{0} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\leqslant−\mathrm{1}\:\cup\:\mathrm{f}\left(\mathrm{x}\right)\geqslant\:\mathrm{0} \\ $$$$\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\leqslant−\mathrm{1}\:\cup\:\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{x}\leqslant−\mathrm{x}^{\mathrm{2}} −\mathrm{1}\:\cup\:\mathrm{1}−\mathrm{x}\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} \underset{\mathrm{x}=\varnothing} {\underbrace{−\mathrm{x}+\mathrm{2}\leqslant\mathrm{0}}}\:\cup\:\mathrm{x}\leqslant\mathrm{1} \\ $$$$ \\ $$