Transcription
So, one of the YouTube viewers asked me to do this problem. And it looked like a pretty good problem. So, I thought I would record a quick video on it.
So, the problem's this: f of g of x -- hope I get this right -- is equal to 2 times the square root of x squared plus 1 minus 1, all of that over the square root of x squared plus 1 plus 1. And f of x is equal to 2x minus 1 over x plus 1. And the question is, if we know that f of g of x is equal to this fairly complex-looking expression, and that f of x is equal to this, then what is g of x?
Well, we could actually do this problem by kind of just looking at it. Because f of g of x, what we do is, we took -- everywhere we see an x, we replace it with g of x. So, f of g of x would look like this. And I'll do it in another color. So, let's see. f of g of x. And let's say we don't know what g of x is. Would be like this. Everywhere where we saw an x, we replace it with g of x. Because -- so it would be 2 times g of x minus 1 over g of x plus 1. And we also know that f of g of x is equal to this thing. Right? Here we just took the g of x and put it in of f of x, and we wrote the g of x in the expression. But we know that that is equal to this expression up here. So, that is also equal to 2 times the square root of x squared plus 1, minus 1, all of that over the square root of x squared plus 1 plus 1.
And I think, now, at this point, you can kind of pattern match. And you can see what g of x is. 2 times something minus 1, right? 2 times something minus 1. 2 times something minus 1 in the numerator, and that's something plus 1 in the denominator. And that's something plus 1 in the denominator. So, in either case, we have g of x is equal to the square root of x squared plus 1.
And, actually, in the example he sent me, there was a bunch of choices. And if you were given the choices, then all you have to really do is take each of the choices for g of x and replace them in for x in this expression. And then see which of those choices for g of x ends up with this expression. Actually, we can do it. I mean, he gave the choice -- well, let me see. Let me delete some of this. I want to do it, let's see if I can do it in -- oh, there we go. Better. All right. So, I wanted you to see that you could do the problem even if you weren't given any choices. But if you are given choices, the problem actually even becomes a little bit more straightforward.
So, the choices were -- actually, I erased some of it. It was given that f of x is 2x minus 1 over x plus 1. And then they said, which of the following is g of x. So, g of x is equal to. And they give these choices: a), square root of x; b), square root of x squared plus 1, which we just figured out was actually the answer; c), x; d), x squared; and e), x squared plus 1. And the way you would do this problem if, you, just by looking at it you didn't -- the method we just saw, saw that if you just replaced x with square root of x squared plus 1 everywhere, then you'd get f of g of x. The other option is, you just take each of these and say, well, what is f of -- if you replaced x with this term everywhere, what do you get? Well, then you'd get 2 times square root of x minus 1 over the square root of x plus 1. And that's not what we have up here. Then if you took this and you replaced for x everywhere -- so if you took this expression, you replaced it for x everywhere, then you would get this expression. So, you would know that this would be the answer. If you just replaced x with x, you would just get -- you would just get this over again, which does not equal this. So, that doesn't work. If g of x was this, then f of g of x would be -- let's see, everywhere we see an x, you'd put an x squared, so it would be 2x squared minus 1 over x squared plus 1. Which does not equal this. And similarly, if g of x was this term right here, then f of g of x would be would be 2 times x squared plus 1 minus 1 over x squared plus 1 plus 1. Which you can simplify a bit, but that still doesn't equal this.
And so this, once again, if you're given the choices, you just try this out. And if you replace this expression everywhere, where you see an x, right? Everywhere where you see an x, and you replace it with this expression, you get these and that's what the question, essentially, was asking us. So, hopefully, I didn't confuse everyone too much. And hopefully, this was helpful for the viewer who asked me to do this. I'll see you all later. Bye.