Question

How do you find the domain and range of `h(x) = e^(-x^2)`?

Answer 1

The domain is the set of all real number hence `D_f=(-oo,+oo)`

The range is `R_f=(0,1]`

Because `e^(x^2)>=x^2+1` hence `1/e^(x^2)<=1/(x^2+1)<=1`

but always `1/e^(x^2)>0`