# Question Video: Finding the Domain of a Vector-Valued Function Mathematics • Higher Education

Find the domain of the vector-valued function π«(π‘) = (1/β(2π‘ β 4))π’ + (2 csc π‘)π£ + (ln (π‘ β 1))π€.

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### Video Transcript

Find the domain of the vector-valued function π« of π‘ is equal to one divided by the square root of two π‘ minus four π’ plus two times the csc of π‘ π£ plus the natural logarithm of π‘ minus one π€.

Weβre given a vector-valued function π« of π‘. And we need to find the domain of this function. Before we begin, letβs talk about vector notation. In this video, we represent vectors and vector-valued functions with a half-arrow notation. And you can also see weβre representing our unit directional vectors with a hat notation. Thereβs a lot of different notation for vectors. For example, you might see them represented with bold notation, underline notation, or full-arrow notation. It doesnβt matter which of these you use. Itβs all personal preference.

Remember, the domain of a function is the set of all possible inputs for that function. In our case, we have a vector-valued function in the variable π‘. So we need to see which inputs of π‘ will give us a defined output. But remember, we evaluate vector-valued functions component-wise. So for our function to be defined at a specific value of π‘, all of our component functions must be defined at π‘.

To see this, think what would happen if one of our component functions was not defined at a specific value of π‘. Then, we canβt possibly evaluate π« at this value of π‘. So it canβt be in the domain of our vector-valued function. So we want to find the domain of each of our component functions.

Letβs start with the domain of one divided by the square root of two π‘ minus four. Thereβs a few different parts to finding the domain of this function. First, we notice weβre dividing one by a number. We know we can divide one by any number except for zero. So because we canβt divide by zero, we know for π‘ to be in the domain of our function, the square root of two π‘ minus four is not allowed to be equal to zero. Of course, we know when this will be equal to zero. This will be equal to zero when two π‘ minus four is equal to zero. But two π‘ minus four is only equal to zero when π‘ is equal to two. So when π‘ is equal to two, weβre dividing by zero. Therefore, π‘ is equal to two canβt be in the domain of our first component function. And in fact, this tells us itβs not in the domain of our function π« of π‘.

But this is not what we need to find the domain of our component function. Remember, weβre not allowed to take the square root of a negative number. So for the square root of two π‘ minus four to be defined, two π‘ minus four must be greater than zero. But remember, weβve already shown that π‘ is not allowed to be equal to zero. So in fact, we can just simplify this to be two π‘ minus four must be greater than zero.

Now, we can just rearrange this inequality for π‘. We add four to both sides of the inequality and divide through by two. This gives us that π‘ must be greater than two. So we were able to show that our first component function is defined for all values of π‘ greater than two. And remember, this also tells us that π« of π‘ will not be defined when π‘ is less than or equal to two.

Letβs now move on to finding the domain of our second component function, two times the csc of π‘. To do this, recall that the csc of π‘ is actually equivalent to one divided by the sin of π‘. We can then multiply both sides of this equivalent through by two. So now, instead, weβll find the domain of the function two divided by the sin of π‘.

Of course, we know the sin of π‘ is defined for all real values of π‘. And we know we can divide two by any real number except for zero. So two times the csc of π‘ will be defined for all values of π‘ except when the sin of π‘ is equal to zero. And we know when the sin of π‘ is equal to zero. We know the sin of π‘ will be equal to zero whenever π‘ is an integer multiple of π.

So weβve now found the domain of our second component function. Itβs all values of π‘ except where π‘ is an integer multiple of π. And of course this means that our function π« of π‘ will also not be defined whenever π‘ is an integer multiple of π.

Letβs now move on to our third and final component function. We need to find the domain of the natural logarithm of π‘ minus one. To do this, we need to recall the domain of the natural logarithm of π₯ is all positive values of π₯. So the natural logarithm of π‘ minus one will be defined whenever π‘ minus one is positive. And of course we can solve this inequality. We just add one to both sides. We see that π‘ must be greater than one. And thatβs it. The natural logarithm of π‘ minus one is not defined when π‘ is less than or equal to one. So its domain is π‘ is greater than one. And of course this also tells us that our function π« of π‘ will not be defined when π‘ is less than or equal to one.

So what weβve shown is, for π‘ to be in the domain of π« of π‘, π‘ must be greater than two, π‘ must not be an integer multiple of π, and π‘ must be greater than one. And we can also see that if π‘ is greater than two, then π‘ is also greater than one. So we donβt actually need to include this in our domain.

Therefore, we were able to show the vector-valued function π« of π‘ is equal to one divided by the square root of two π‘ minus four π’ plus two times the csc of π‘ π£ plus the natural logarithm of π‘ minus one π€ will have a domain of π‘ being greater than two and π‘ not being equal to an integer multiple of π.