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

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

Answered by MM42 last updated on 04/Aug/23

$$D_f\subseteq {\mathbb{R}}^2$$$$ifA=\{(x,y)\mid x+y^2<0\}\mathrm{\&}B=\{(x,y)\mid x^2+y^2<4\}$$$$\Rightarrow D_f={\mathbb{R}}^2-(A-B)\cup (B-A)$$

Answered by witcher3 last updated on 04/Aug/23

$$(\mathrm{x},\mathrm{y})\to \frac{\mathrm{x}+{\mathrm{y}}^2}{{\mathrm{x}}^2+{\mathrm{y}}^2-4}\mathrm{is}\mathrm{continus}$$$$(\mathrm{x},\mathrm{y})=(\mathrm{rcos}\left(\mathrm{a}\right),\mathrm{rsin}\left(\mathrm{a}\right))$$$$\mathrm{r}\in [0,\mathrm{\infty }[,\mathrm{a}\in [0,2\pi ]\mathrm{r}$$$$\mathrm{f}(\mathrm{x},\mathrm{y})=\mathrm{f}(\mathrm{rcos}\left(\mathrm{a}\right),\mathrm{rsin}\left(\mathrm{a}\right))=\mathrm{g}(\mathrm{r},\mathrm{a})$$$$=\sqrt{\frac{\mathrm{rcos}\left(\mathrm{a}\right)+{\mathrm{r}}^2{\sin }^2\left(\mathrm{a}\right)}{{\mathrm{r}}^2-4}}$$$$\mathrm{for}\mathrm{r}>2,\mathrm{a}\in [0,\frac{\pi }{2}]\mathrm{g}\mathrm{is}\mathrm{well}\mathrm{defind}$$$$\mathrm{g}(0,0$$$$\underset{\mathrm{r}\to \mathrm{\infty }}{\lim g}(\mathrm{r},\mathrm{a})=0$$$$\underset{\mathrm{r}\to 2^{+}}{\lim g}(\mathrm{r},\mathrm{a})=+\mathrm{\infty }$$$$\Rightarrow \mathrm{range}\mathrm{f}\mathrm{is}]0,+\mathrm{\infty }[$$$$0=\mathrm{g}(0,\mathrm{a})\mathrm{range}\mathrm{is}[0,\mathrm{\infty }[$$

Answered by Frix last updated on 04/Aug/23

$$f(x,y)\mathrm{defined}\mathrm{for}x,y\in \mathbb{R}\mathrm{with}$$$$(y<-\sqrt{4-x^2}\vee y>\sqrt{4-x^2})\wedge (x⩾0\vee (x<0\wedge (y⩽-\sqrt{-x}\vee y⩾\sqrt{-x}))$$$$0⩽f(x,y)<+\mathrm{\infty }$$

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