### 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€.

Weβre given a vector-valued function π« of π‘. And we need to find the domain of this function. Before we begin, letβs talk about vector notation. In this video, we represent vectors and vector-valued functions with a half-arrow notation. And you can also see weβre representing our unit directional vectors with a hat notation. Thereβs a lot of different notation for vectors. For example, you might see them represented with bold notation, underline notation, or full-arrow notation. It doesnβt matter which of these you use. Itβs 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 canβt possibly evaluate π« at this value of π‘. So it canβt be in the domain of our vector-valued function. So we want to find the domain of each of our component functions.

Letβs start with the domain of one divided by the square root of two π‘ minus four. Thereβs a few different parts to finding the domain of this function. First, we notice weβre dividing one by a number. We know we can divide one by any number except for zero. So because we canβt 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, weβre dividing by zero. Therefore, π‘ is equal to two canβt be in the domain of our first component function. And in fact, this tells us itβs 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, weβre 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, weβve 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.

Letβs 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, weβll 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 weβve now found the domain of our second component function. Itβs 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.

Letβs 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 thatβs 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 weβve 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 donβt 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.