Asked by John
f(t)=3sqrt(2t-1)
why domain is all real #'s ?
3sqrt(2t-1) >= 0
( 3sqrt(2t-1) )^3 >= (0)^3
t>= 1/2
?
why domain is all real #'s ?
3sqrt(2t-1) >= 0
( 3sqrt(2t-1) )^3 >= (0)^3
t>= 1/2
?
Answers
Answered by
oobleck
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
2t-1 >= 0
t >= 1/2
so the domain is [1/2,∞)
You are correct
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