Question

What is the domain and range of `f(x)=sqrt(x^2+4)`?

Answer 1

The domain is easy, as the square makes everything under the root-sign non-negative, so there are no restrictions on `x`.

Explanation:

In other words domain `-oo < x<+oo`

Since `x^2>=0->x^2+4>=4->sqrt(x^2+4)>=2`

In other words range `2<=f(x)<+oo`