f(x) function pair and (−∞;0] decreases in the range f(x)≥f(−3) solve the inequality


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Question Number 127322 by MathSh last updated on 28/Dec/20

$f (x) f u n c t i o n p a i r a n d (- \infty; 0]$ $d e c r e a s e s i n t h e r a n g e$ $f (x) ⩾ f (- 3) s o l v e t h e i n e q u a l i t y$
	Answered by mindispower last updated on 28/Dec/20

	$] - \infty, - 3] \cup [3, + \infty [$
	Commented by MathSh last updated on 28/Dec/20

	$S o l u t i o n s i r p l i z$
	Commented by mindispower last updated on 28/Dec/20
	$since f decrease in ]−∞,0]⇒∀x≤−3,f(x)≥f(−3) ⇒x∈]−∞,−3] is solution f(x)=f(−x) f est une fonction paire ⇒[3,∞] est aussi solution ]−∞,−3]∪[3,∞[ sont solution mais pas que car f peut etre constante sur [−3,3] dans ce cas S=R exemple f(x)= { ((x^2 ,x∈R−[−3,3])),((9 si x∈[−3,3])) :} f(x)≥f(−3), les solution sont R tous entier$
	$s i n c e f d e c r e a s e$ $i n] - \infty, 0] \Rightarrow \forall x ⩽ - 3, f (x) ⩾ f (- 3)$ $\Rightarrow x \in] - \infty, - 3] i s s o l u t i o n$ $f (x) = f (- x) f e s t u n e f o n c t i o n p a i r e$ $\Rightarrow [3, \infty] e s t a u s s i s o l u t i o n$ $] - \infty, - 3] \cup [3, \infty [s o n t s o l u t i o n$ $m a i s p a s q u e c a r f p e u t e t r e c o n s t a n t e s u r$ $[- 3, 3] d a n s c e c a s S = R$ $e x e m p l e f (x) = {\begin{cases} x^{2}, x \in R - [- 3, 3] \\ 9 s i x \in [- 3, 3] \end{cases}$ $f (x) ⩾ f (- 3), l e s s o l u t i o n s o n t R t o u s e n t i e r$
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