Video Transcript
Find the domain of the function 𝑓 of 𝑥 is equal to negative three divided by the square root of two 𝑥 minus one.
In this question, we’re given a function 𝑓 of 𝑥. And we’re asked to find the domain of this function. So we’re going to need to start by recalling exactly what we mean by the domain of a function. We recall that when we say the domain of a function, we mean the set of inputs for that function. But before we find the domain of our function, let’s discuss exactly what we mean by its domain. When we say the inputs of our function, this means our function has to be defined for this value of 𝑥 because otherwise we couldn’t possibly input this value of 𝑥 into our function because we would just get an undefined output.
And this gives us a good way of looking for the domain of our function 𝑓 of 𝑥. Instead of looking for the values of 𝑥 where 𝑓 of 𝑥 is defined, we’ll start by looking for the values of 𝑥 where 𝑓 of 𝑥 is not defined. And to find the values of 𝑥 where 𝑓 of 𝑥 is not defined, we’re going to need to notice two key things about our function 𝑓 of 𝑥. First, we’re taking a square root, and, remember, we’re not allowed to take the square root of a negative number. Next, we’re also dividing by 𝑥, and, remember, we can’t divide by zero.
So in both of these cases our function 𝑓 of 𝑥 won’t be defined. Let’s look for the values of 𝑥 where this occurs. First, we’ll be taking the square root of a negative number if the expression inside of our square root sign is less than zero, in other words, if two 𝑥 minus one is less than zero. And of course, we can rearrange this inequality to find the values of 𝑥. We add one to both sides of our inequality and then divide through by two; we get that 𝑥 is less than one-half. So that explicitly state what we’ve shown. We’ve shown if 𝑥 is less than one-half, then two 𝑥 minus one will be less than zero; it’s negative.
Therefore, for these values of 𝑥, if we were to try and take the square root of two 𝑥 minus one, we’ll be taking the square root of a negative number. And we know this does not exist. Hence, if we were to try and input a value of 𝑥 less than one-half into 𝑓 of 𝑥, we would take the square root of a negative number. Therefore, 𝑓 cannot be defined when 𝑥 is less than one-half. So the values of 𝑥 less than one-half are not allowed to be inputs in our function.
But this wasn’t the only possible problem we saw. We also saw we might be able to divide by zero. So let’s also try and find any possible input values of 𝑥 where we’re dividing through by zero. To be dividing through by zero, we’re going to need our denominator equal to zero. So, we must have the square root of two 𝑥 minus one is equal to zero. And once again, we can solve this equation for 𝑥. One way of doing this is to square both sides of our equation, add one to both sides of our equation, and then divide through by two. We see that 𝑥 would be equal to one-half. And we can verify this solution directly. If we substitute 𝑥 is equal to one-half into the square root of two 𝑥 minus one, we do see it’s equal to zero.
And once again, we can see if we input the value of 𝑥 is equal to one-half into our function 𝑓 of 𝑥, then we see we’re dividing through by zero. So 𝑥 is equal to one-half can’t be an input of our function, and, therefore, it’s not in the domain of this function. So so far, we’ve shown two things about the domain of our function 𝑓 of 𝑥. First, we’ve shown all values of 𝑥 less than one-half are not in the domain of 𝑓 of 𝑥. And we’ve shown the value of 𝑥 is equal to one-half is also not in the domain.
And it’s worth pointing out here we can combine these into two. Saying that 𝑥 is not allowed to be less than one-half and 𝑥 is not allowed to be equal to one-half is exactly the same as saying that 𝑥 is not allowed to be less than or equal to one-half. So we’ll write these values in this way. But now we need to ask the question, what about all the other values of 𝑥? We want to know what happens if we input a value of 𝑥 bigger than one-half into our function 𝑓 of 𝑥.
To do this, let’s take a look at our definition of 𝑓 of 𝑥. 𝑓 evaluated at 𝑥 is equal to negative three divided by the square root of two 𝑥 minus one. But now we can see something interesting. If 𝑥 is bigger than one-half, then two 𝑥 minus one must be positive. So now, in our denominator, we’re just taking the square root of a positive number. This is also a positive number. So what we’ve shown is when 𝑥 is bigger than one-half, we just have a negative number divided by a positive number. And this will always be defined. We can always divide a negative number by a positive number.
Therefore, all of the values of 𝑥 bigger than one-half are allowed to be possible inputs of our function. Now we could give our answer as 𝑥 is being greater than one-half. However, we can also write this in set notation. We can write it as the open interval from one-half to ∞. However, these both represent the same values of 𝑥. Therefore, we were able to show the domain of the function 𝑓 of 𝑥 is equal to negative three divided by the square root of two 𝑥 minus one is equal to the open interval containing one-half and ∞.
