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Question Number 208149 by mnjuly1970 last updated on 06/Jun/24

	Answered by A5T last updated on 06/Jun/24

	$⌊ 2 x^{2} ⌋ > 2 x^{2} - 1 \Rightarrow x - ⌊ 2 x^{2} ⌋ < 1 - 2 x^{2} + x$ $S u p p o s e D_{f}, R_{f} \subseteq R$ $1 - 2 x^{2} + x < 0 \Rightarrow x > 1 o r x < - \frac{1}{2}$ $I f R_{f} \subseteq R, t h e n f (x) i s u n d e f i n e d f o r x < \frac{- 1}{2} \cup x > 1$ $w h e n x = \frac{- 1}{2}, 1; f (x) i s u n d e f i n e d i n R$ $\frac{- 1}{2} < x < 0 \Rightarrow∣ x ∣< \frac{1}{2} \Rightarrow 0 < x^{2} < \frac{1}{4} \Rightarrow 0 < 2 x^{2} < \frac{1}{2}$ $\Rightarrow ⌊ 2 x^{2} ⌋ = 0 \Rightarrow f (x) = \sqrt{x - 0} = f (x) = \sqrt{x}$ $w h i c h i s u n d e f i n e d f o r \frac{- 1}{2} < x < 0$ $w h e n x = 0; f (x) = 0$ $w h e n 0 < x < \frac{1}{\sqrt{2}}; ⌊ 2 x^{2} ⌋ = 0 \Rightarrow f (x) = \sqrt{x} \in (0, \frac{\sqrt[4]{8}}{2})$ $w h e n x = \frac{1}{\sqrt{2}}; f (x) i s u n d e f i n e d i n R$ $w h e n \frac{1}{\sqrt{2}} < x < 1, ⌊ 2 x^{2} ⌋ = 1 \Rightarrow f (x) = \sqrt{x - 1}$ $w h i c h i s u n d e f i n e d i n R s i n c e \frac{1}{\sqrt{2}} < x < 1 .$ $S o, f o r D_{f}, R_{f} \subseteq R, R_{f} = [0, \frac{\sqrt[4]{8}}{2}), D_{f} = [0, \frac{1}{\sqrt{2}})$
	Commented by mnjuly1970 last updated on 06/Jun/24


	Commented by A5T last updated on 06/Jun/24

	$D o y o u h a v e t h e a n s w e r ? I s i t c o r r e c t ?$
	Commented by mnjuly1970 last updated on 07/Jun/24

	$y e s s i r A 5 T$
	Answered by MM42 last updated on 06/Jun/24

	$D = [0, \frac{1}{\sqrt{2}}) & R = [0, \frac{1}{\sqrt[4]{2}})$
	Commented by MM42 last updated on 06/Jun/24

	$i f x < 0 \Rightarrow x - ⌊ 2 x^{2} ⌋ < 0 \Rightarrow\notin D_{f}$ $i f \sqrt{\frac{n}{2}} ⩽ x < \sqrt{\frac{n + 1}{2}} \Rightarrow x < n; \forall n \in N$ $\Rightarrow n ⩽ 2 x^{2} < n + 1 \Rightarrow ⌊ 2 x^{2} ⌋ = n$ $\Rightarrow x - ⌊ 2 x^{2} ⌋ < 0 \Rightarrow\notin D_{f}$ $f (0) = 0 & {l i m}_{x \to \frac{1}{\sqrt{2}}^{}} f (x) = \frac{1}{\sqrt[4]{2}}$ $\Rightarrow R_{f} = [0, \frac{1}{\sqrt[4]{2}})$
	Answered by mnjuly1970 last updated on 07/Jun/24
	$x−⌊2x^2 ⌋≥0 ⇒ x ≥ ⌊ 2x^2 ⌋≥0 (★) −−−−−− ⌊ 2x^2 ⌋=k ⇒^(k∈Z^(≥0) ) k ≤ 2x^2 < k+1 from (★): x ≥ k ⇒^(x≥0) 2x^2 ≥ 2k^2 (★★) −−−−−− from(★),(★★)⇒^(★∩★★={}) : 2k^2 ≥ k+1 k ≥ 1 (■) complement of(■) is our purpose (★★★) k =0 ⇒ 0 ≤ 2x^2 < 1 ⇒ { ((D_f = 0≤x <(1/( (√2))))),(( ⌊2x^2 ⌋=0 ⇒ f(x)=(√x))) :} ⇒ { (( 0≤x <(1/( (√2))))),(( f(x)=(√x) ⇒ R_f = [0 , (1/( (2)^(1/4) )) ))) :}$
	$x - ⌊ 2 x^{2} ⌋ ⩾ 0 \Rightarrow x ⩾ ⌊ 2 x^{2} ⌋ ⩾ 0 (★)$ $- - - - - -$ $⌊ 2 x^{2} ⌋ = k \overset{k \in Z^{⩾ 0}}{\Rightarrow} k ⩽ 2 x^{2} < k + 1$ $f r o m (★) : x ⩾ k \overset{x ⩾ 0}{\Rightarrow} 2 x^{2} ⩾ 2 k^{2} (★ ★)$ $- - - - - -$ $f r o m (★), (★ ★) \overset{★ \cap ★ ★ = {}}{\Rightarrow} : 2 k^{2} ⩾ k + 1$ $k ⩾ 1 (◼)$ $c o m p l e m e n t o f (◼) i s o u r p u r p o s e (★ ★ ★)$ $k = 0 \Rightarrow 0 ⩽ 2 x^{2} < 1 \Rightarrow {\begin{cases} D_{f} = 0 ⩽ x < \frac{1}{\sqrt{2}} \\ ⌊ 2 x^{2} ⌋ = 0 \Rightarrow f (x) = \sqrt{x} \end{cases}$ $\Rightarrow {\begin{cases} 0 ⩽ x < \frac{1}{\sqrt{2}} \\ f (x) = \sqrt{x} \Rightarrow R_{f} = [0, \frac{1}{\sqrt[4]{2}}) \end{cases}$
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