Find the domain of the vector functions

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The domain of the vector function r(t) = <ln(6t), sqrt(t+14), 1/sqrt(16-t)> requires that all components yield real numbers. For ln(6t), t must be greater than 0, while the square root component requires t to be greater than or equal to -14. The term 1/sqrt(16-t) imposes the condition that t must be less than 16. Therefore, the correct domain for t is (0, 16). The final domain is expressed as (0, 16).
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Find the domain of the vector functions, r(t), listed below

a.) r(t) = <ln(6t), sqrt(t+14), 1/sqrt(16-t)>

i don't extactly know how to approach this, can someone give me a hint or two
 
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The domain of a function f(x) basically means the set of all x for which the function yields valid results. In your case, I presume all vector components must be real numbers. What does that mean for the values that t is allowed to have?
 
t \geq -14
t \leq 16

right? so the domain should be from [-14,16]
 
You're on the right track, but what about the ln(6t) function?
 
t cannot equal to 0 for ln(6t). so should it be [-14,0) U (0, 16]
 
If ln(6t) is to yield a real number, t must be greater than 0. A negative value for t won't work! So one of the domains you posted is almost right - the one holding positive values. The reason that the interval isn't exactly correct is because t = 16 isn't allowed (take a look at the third term and see what happens when t=16). In other words: t is in (0,16) instead of (0,16].
 
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