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Problem 32 Hard Difficulty

Find (a) $ f + g $, (b) $ f - g $, (c) $ fg $ and (d) $ f/g $ and state their domains.

$ f(x) = \sqrt {3 - x} $ , $ g(x) = \sqrt {x^2 -1} $

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Video Transcript

here we have functions. F N g. Let's start by adding them so F plus G would be the square root of three minus X plus the square root of X squared minus one. And these were not like terms. There's really nothing we can do to combine those together. F minus G will be the square root of three minus X minus the square root of X squared minus one. Again, there's no simplifying to do here. Now let's talk about the domains. So what's the domain of F? Well, we know that we have to have the square root of a non negative number, so it has to be the square root of zero or positive. So that means that X could be three or anything less than three, but not more so. The domain of F is X is less than or equal to three. What about the domain of G? Same idea. Inside the radical. We have to have zero or positive. That will happen if X is one or greater or effects is negative one or smaller. So the domain of G is X is greater than or equal to one or X is less than or equal to negative one. Now to find the domain of F plus G F minus G and F times G, we need the intersection of these two domains, so the intersection would be the over that what they have in common. So for both sets, it would be OK for X to be less than or equal to negative one. And if I write that in interval notation, that would be negative. Infinity to negative one. But then, if we combine these two ideas here less than or equal to three and greater than or equal to one, we get the interval from 1 to 3 square brackets to include the endpoints, so we'll get the same answer for the domain of F minus G and the same answer for the domain of F times G. We can just rewrite those domains up here for F. It was X is less than or equal to three m. Fergie. It was X is greater than or equal to one or less than or equal to negative one. Okay, now let's find the product F times G. So we have the square root of three minus x times the square root of X squared minus one. Now we could combine these into one single square root sign so we have the square root of three minus X times the quantity X squared minus one, and we could go ahead and multiply that out if we want to simplify it. So using the foil method, we would have three X squared minus X cubed minus three plus X. And if we want to put those terms into descending powers of X, we have the square root of the opposite of X cubed plus three X squared plus X minus three. Okay, let's go to another slide for F over G. So going to take function F, which was the square root of X of three minus X and divided by function G, which was the square root of X squared minus one. We could write those inside a single square root if we want to. And I suppose if we're really, really interested, we could rationalize the denominator. But typically and calculus, you're not required to do that. So we'll call this F over G. Now, how about the domain? So remember, in the previous problem, all we had to do to get the domain of F plus g f minus g n f. Times G was too. Find the intersection of the two separate domains. But that's not the case for the quotient for the quotient. You also have to consider that G cannot equal zero because you can't have zero in the denominator. That means we can't have X equals one. We can't have X equals negative one. So then, when we write that intersection, we can write it as so This is the domain of F over G. We can write it as negative infinity to negative one. But we have to have an open bracket. Just show we're not including the endpoint union, and then we can have an open bracket on the one as well. We can still have a closed bracket on the three because it's still okay to have a three on the top and have the numerator B zero is just not okay for the denominator to be zero