find-domine-and-range-of-function-f-x-y-x-y-2-x-2-y-2-4-

Question Number 195520 by moh777 last updated on 04/Aug/23

f i n d d o m i n e a n d r a n g e o f f u n c t i o n

f (x, y) = \sqrt{\frac{x + y^{2}}{x^{2} + y^{2} - 4}}

Answered by MM42 last updated on 04/Aug/23

D_{f} \subseteq R^{2}

i f A = {(x, y) ∣ x + y^{2} < 0} & B = {(x, y) ∣ x^{2} + y^{2} < 4}

\Rightarrow D_{f} = R^{2} - (A - B) \cup (B - A)

Answered by witcher3 last updated on 04/Aug/23

(x, y) \to \frac{x + y^{2}}{x^{2} + y^{2} - 4} is continus

(x, y) = (rcos (a), rsin (a))

r \in [0, \infty [, a \in [0, 2 π] r

f (x, y) = f (rcos (a), rsin (a)) = g (r, a)

= \sqrt{\frac{rcos (a) + r^{2} \sin^{2} (a)}{r^{2} - 4}}

for r > 2, a \in [0, \frac{π}{2}] g is well defind

g (0, 0

\underset{r \to \infty}{\lim g} (r, a) = 0

\underset{r \to 2^{+}}{\lim g} (r, a) = + \infty

\Rightarrow range f is] 0, + \infty [

0 = g (0, a) range is [0, \infty [

Answered by Frix last updated on 04/Aug/23

f (x, y) defined for x, y \in R with

(y < - \sqrt{4 - x^{2}} \lor y > \sqrt{4 - x^{2}}) \land (x ⩾ 0 \lor (x < 0 \land (y ⩽ - \sqrt{- x} \lor y ⩾ \sqrt{- x}))

0 ⩽ f (x, y) < + \infty

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