find-the-domain-of-thefunction-f-x-1-x-2-x-2-where-is-the-fractional-part-function-

Question Number 192852 by York12 last updated on 29/May/23

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 \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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