Hey, everyone, and welcome back. So let's see if we can solve this problem. In this problem, we have 2 functions. We have \( f(x) = \sqrt{x+4} + 30 \), and then we have \( g(x) = \sqrt{x+4} - 2x + 35 \). We're asked to complete the following operations below and determine the domain of the new functions.
So let's see if we can solve this problem. What I'm first going to do is see if I can find \( f+g \). And to do this, recall that this is the same thing as \( f(x) + g(x) \). So I just need to add the 2 functions above together. I can see here that \( f(x) = \sqrt{x+4} + 30 \).
And I can see here that \( g(x) \), this is \( f(x) \), the \( g(x) \) is equal to \( \sqrt{x+4} - 2x + 35 \). Now since I don't have any kind of negative signs or numbers that I need to distribute into these parentheses over here, I can kinda just drop the parentheses on all of these functions that we plugged in here. So all I need to do really is combine like terms. Now I see that we have a \(\sqrt{x+4}\) and a \(\sqrt{x+4}\), which will give us \(2\sqrt{x+4}\). So we have 2 of these here, so we add them together.
And then I also see that we have a positive 30 and a positive 35. \(30+35 = 65\). So we're gonna get \(+ 65\), and then we're gonna have this \(-2x\). Although to do this in better order, I'm actually going to write this as \(-2x + 65\). Typically, we like to have the \(x\)'s come before the constants.
That's just a general preference thing. And this right here would be the functions when we've added them together. So \(f(x) + g(x)\) would look like this. Now to find the domain, what you have to do is find the combination of the domains from the functions when you initially plug them in. Now looking at these functions when I did the initial plug in, I can really see that the only thing that's going to give us restrictions for our domain is the square root function, because typically it's going to be square roots or fractions that you see that will cause restrictions.
And since we have a square root, we need to take this into account. Recall that nothing underneath the square root can be negative. So what we can do is take this whole \(x+4\) and say this has to be greater than or equal to 0. Now what I can do from here is solve this mini equation by subtracting 4 on both sides, cancelling the 4s there, giving me that \(x\) has to be greater than or equal to \(-4\). So the \(x\) values can equal \(-4\), but they just cannot be less than \(-4\).
So that means that our domain is going to go from \(-4\) to positive infinity. And since I see that it's the same quantity we have under the square root, this is the only restriction that I need to take into account, meaning this is the only domain. So this right here is going to be the domain of \(f+g\). And that's how you can solve these types of problems.
But what would happen if instead we had \(f - g\) in this situation over here? Well, to solve this situation, what I'm going to do is the same strategy I used up here, except now I'm going to subtract the 2 functions. So we're going to have \(f(x)\), which is the square root of \(x + 4\), plus 30, and this this is going to be minus the \(g(x)\) function, which is the square root of \(x + 4\) minus \(2x + 35\). Now from here, what I'm going to do is since I have a negative sign out front, I need to distribute this negative sign into each of these terms. So I can drop the parenthesis on this first thing, this will give me \(\sqrt{x + 4}\), this first quantity I should say, plus 30, and then I can take this negative sign and distribute it into each of these terms. Taking the negative sign and distributing it to here, we'll get \(-\sqrt{x + 4}\).
The two negative signs here will cancel giving me \(+2x\), and then the negative sign multiplied by 35 will give me \(-35\). So this right here is what the function is going to look like when we distribute the negative sign. Now I can see here that because we have a positive and a negative quantity for square root of \(x+4\), these are going to cancel each other out. And I can also see that we have a positive 30 and a negative 35. \(30 - 35\) or \(30 + (-35)\) is going to give you \(-5\).
So that means that what we're going to end up with is this \(2x\) by itself, and this is going to be \(-5\). So this is what the function \(f - g(x)\) looks like. And as for the domain, you may look at this total function and think there are no restrictions on the domain because we only have an \(x\) here, and it's like we got this polynomial which is a linear equation, but if you look up here notice that we have to take into account the functions before we simplify them. So when we initially plug things in we have the square root of \(x + 4\). So because of that, the inside of the square root has to stay positive, meaning the \(x + 4\) still has to be greater than or equal to 0 since we started with this square root.
And if I go ahead and solve this mini equation, well, we already did that up here. We already know \(x\) is going to be greater than or equal to \(-4\). So that means our domain for this situation is still going to be all real numbers from \(-4\) to positive infinity. So this is going to be our domain when we subtract the two functions. Notice it's the same for when we added them.
Even though this came out to a function that didn't have the square root in it anymore, we still needed the same domain because the square root was inside the function when we initially combined them together. So this is how you can do addition and subtraction of functions and find the domain. Hope you found this video helpful, and thanks for watching.