f(t)=3sqrt(2t-1)

why domain is all real #'s ?

3sqrt(2t-1) >= 0

( 3sqrt(2t-1) )^3 >= (0)^3
t>= 1/2

?

1 answer

if f(t) = 3√(2t-1) the domain is not all real numbers. You need
2t-1 >= 0
t >= 1/2
so the domain is [1/2,∞)
You are correct