let f(x)=((λ-x)/(1+x^2 )) and λ≥((-3)/4) solve in R f(f(f(x))) ≤ 0


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Question Number 151747 by mathdanisur last updated on 22/Aug/21

$let f (x) = \frac{λ - x}{1 + x^{2}} and λ ⩾ \frac{- 3}{4}$ $solve in R f (f (f (x))) ⩽ 0$
	Answered by mr W last updated on 22/Aug/21
	$say g(x)=f(f(x)) f(f(f(x)))≤0 ⇒f(g(x))≤0 ⇒((λ−g(x))/(g^2 (x)+1))≤0 ⇒g(x)≥λ ⇒((λ−f(x))/(f^2 (x)+1))≥λ ⇒λf^2 (x)+f(x)≤0 ⇒f(x)(λf(x)+1)≤0 ⇒ { ((f(x)≥0, λf(x)+1≤0 ...(i))),((f(x)≤0, λf(x)+1≥0 ...(ii))) :} (i): ((λ−x)/(x^2 +1))≥0 ⇒x≤λ ((λ(λ−x))/(x^2 +1))+1≤0 x^2 −x+λ^2 +1≤0 (x−(λ/2))^2 +((3λ^2 )/4)+1≤0 ⇒no solution! (ii): ((λ−x)/(x^2 +1))≤0 ⇒x≥λ ((λ(λ−x))/(x^2 +1))+1≥0 x^2 −λx+λ^2 +1≥0 (x−(λ/2))^2 +((3λ^2 )/4)+1≥0 ⇒true ⇒solution is x≥λ$
	$s a y g (x) = f (f (x))$ $f (f (f (x))) ⩽ 0 \Rightarrow f (g (x)) ⩽ 0$ $\Rightarrow \frac{λ - g (x)}{g^{2} (x) + 1} ⩽ 0$ $\Rightarrow g (x) ⩾ λ$ $\Rightarrow \frac{λ - f (x)}{f^{2} (x) + 1} ⩾ λ$ $\Rightarrow λ f^{2} (x) + f (x) ⩽ 0$ $\Rightarrow f (x) (λ f (x) + 1) ⩽ 0$ $\Rightarrow {\begin{cases} f (x) ⩾ 0, λ f (x) + 1 ⩽ 0 . . . (i) \\ f (x) ⩽ 0, λ f (x) + 1 ⩾ 0 . . . (i i) \end{cases}$ $(i) :$ $\frac{λ - x}{x^{2} + 1} ⩾ 0 \Rightarrow x ⩽ λ$ $\frac{λ (λ - x)}{x^{2} + 1} + 1 ⩽ 0$ $x^{2} - x + λ^{2} + 1 ⩽ 0$ ${(x - \frac{λ}{2})}^{2} + \frac{3 λ^{2}}{4} + 1 ⩽ 0 \Rightarrow n o s o l u t i o n!$ $(i i) :$ $\frac{λ - x}{x^{2} + 1} ⩽ 0 \Rightarrow x ⩾ λ$ $\frac{λ (λ - x)}{x^{2} + 1} + 1 ⩾ 0$ $x^{2} - λ x + λ^{2} + 1 ⩾ 0$ ${(x - \frac{λ}{2})}^{2} + \frac{3 λ^{2}}{4} + 1 ⩾ 0 \Rightarrow t r u e$ $\Rightarrow s o l u t i o n i s x ⩾ λ$
	Commented by mr W last updated on 22/Aug/21

	$c o n d i t i o n λ ⩾ - \frac{3}{4} i s n o t n e c c e s s a r y .$ $t h e q u e s t i o n i s t r u e f o r a n y λ \in R .$
	Commented by mathdanisur last updated on 23/Aug/21

	$Thank you Ser$
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