Question

What is the range of `1/sqrt(9-x^2-y^2)`?

Answer 1

Range is `(1/3,oo)`

Explanation:

Here we cannot have `9-x^2-y^2<0`, as we cannot have square root of a negative number. In fact we cannot even have `9-x^2-y^2=0`, as it makes denominator `0`.

In other words `x^2+y^2<9` i.e. `x` and `y` lie within the boundary of circle `x^2+y^2=9` - not even on circle.

As maximum value of `x^2+y^2->9`, `f(x,y)` can have maximum value of `oo` as `x^2+y^2->9`. Further its minimum value could be zero and then `f(x)=1/3` and hence range is `[1/3,oo)`

graph{1/sqrt(9-x) [-10, 10, -5, 5]}