Question

What is the domain and range of `f(t) = sqrt(9-t^2)`?

Answer 1

Domain `{t:RR, -3<=t<=3}`

Range `{f(t):RR, 0<=f(t)<=3}`

Explanation:

For real f(t), `9>=t^2` or `3>= +-t`. That is `-3<=t<=3` would be the domain of f(t).

For range, f(t) would be 0 for t =3 or -3, it would be maximum at t=0

hence range would be`{f(t) : RR, 0<=f(t)<=3}`