find the domain of thefunction f(x) = (1/( (√(x^2 −{x}^2 )))) where {.} is the fractional part function.


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Question Number 192852 by York12 last updated on 29/May/23
$find the domain of thefunction f(x) = (1/( (√(x^2 −{x}^2 )))) where {.} is the fractional part function.$
$f i n d t h e d o m a i n o f t h e f u n c t i o n$ $f (x) = \frac{1}{\sqrt{x^{2} - {x}^{2}}} w h e r e {.} i s t h e f r a c t i o n a l p a r t f u n c t i o n .$
Commented by York12 last updated on 30/May/23

$t h e a n s w e r i s$ $(- \infty - 1) \cup [- 1 - \frac{1}{2}) \cup [1, \infty)$
Commented by MM42 last updated on 30/May/23

$(- \infty, - 1) \cup [- 1, - \frac{1}{2}] \cup [1, \infty) = (- \infty, - \frac{1}{2}] \cup ⌈ 1, \infty)$
	Answered by MM42 last updated on 29/May/23

	$D = (- \infty, - \frac{1}{2}) \cup [1, \infty)$
	Commented by York12 last updated on 30/May/23

	$How do you define {x} for x < 0 ?$ ${- 1.3} = - . 3 is what I learned but I^{'} ve$ $also seen {- 1.3} = . 7$ ${x} = x - ⌊ x ⌋$ $f o r e x a m p l e x = - 3.4$ ${- 3.5} = - 3.4 - ⌊ - 3.4 ⌋ = - 3.4 + 4 = . 6$
	Commented by MM42 last updated on 30/May/23

	$I n a u t h o r i t a t i v e b o o k s a n d m a t h e m a t i c a l r e f e r e n c e s, t h e f o l l o w i n g d i f i n a t i o n i s u s e$ $c o r r e c t p a r t . [x] = x - {x}; 0 ⩽ {x} < 1$ $t h e r f o r e a l w a y s 0 ⩽ {x} < 1$
	Commented by York12 last updated on 30/May/23

	$y e a h s i r e x a c t l y$
	Commented by York12 last updated on 30/May/23

	$(- \infty, - 1) \cup [- 1, - \frac{1}{2}] \cup [1, \infty) = (- \infty, - \frac{1}{2}] \cup ⌈ 1, \infty)$ $t h a n k s s i r y e a h y o u a r e r i g h t$
	Commented by MM42 last updated on 30/May/23

	$g o o d l u c k$
	Answered by witcher3 last updated on 31/May/23
	$x^2 >{x}^2 x=[x]+{x} if ∣x∣≥1⇒x^2 −1≥x^2 −{x}^2 >0 {x}∈[0,1[ if x∈]−1,1[ x∈]−1,0[ x=−1+{x}⇒ x^2 −{x}^2 =1−2{x}>0⇒{x}<(1/2) ⇒x<−(1/2)⇒x∈]−1,−(1/2)[ if 1>x≥0 ⇒x={x}⇒x−{x}=0 ⇒D_f =]−∞,−(1/2)[∪[1,∞[$
	$x^{2} > {x}^{2}$ $x = [x] + {x}$ $if ∣ x ∣⩾ 1 \Rightarrow x^{2} - 1 ⩾ x^{2} - {x}^{2} > 0$ ${x} \in [0, 1 [$ $if x \in] - 1, 1 [$ $x \in] - 1, 0 [$ $x = - 1 + {x} \Rightarrow$ $x^{2} - {x}^{2} = 1 - 2 {x} > 0 \Rightarrow {x} < \frac{1}{2}$ $\Rightarrow x < - \frac{1}{2} \Rightarrow x \in] - 1, - \frac{1}{2} [$ $if 1 > x ⩾ 0$ $\Rightarrow x = {x} \Rightarrow x - {x} = 0$ $\Rightarrow D_{f} =] - \infty, - \frac{1}{2} [\cup [1, \infty [$
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