Question

19. Consider the function below.

a) Find and sketch the domain of f

b) Find the range of f​

Answer 1

(a)
`f(x,y)` is defined only as long as

`x^2 + y^2 - 9 > 0 \implies x^2 + y^2 > 3^2`

which is the region outside the closed disk with radius 3. So

`\mathrm{dom} f(x,y) = \left\{(x,y) \in \Bbb R^2 \mid x^2 + y^2 > 9\right\}`

which is easy to sketch.

(b)
`x^2+y^2>9`, so
`x^2+y^2` can only approach 9 from above. As we approach the circle
`x^2+y^2=9`, the first term of
`f(x,y)\to\infty`, since
`\frac1{\sqrt{\text{small number}}} = \text{large number}`. As
`(x,y)` goes off to some infinity in the
`x,y`-plane,
`x^2+y^2\to\infty`, so the first term
`f(x,y)\to0`. It never actually takes on the value of 0, since
`√(x^2+y^2-9) > 0`.

Meanwhile
`5\sin(x+y)` is bounded between -5 and 5. This means there is an overall lower bound of -5 in some infinite corner of the plane.

`\mathrm{ran} f(x,y) = \left\{z \in \Bbb R \mid -5 < z < \infty\right\}`