 In this video, we present the solution to question number six for exam, practice exam number one for math 1210, in which case we're given a function capital F of X equals the fourth root of X squared plus X and little f of X equals the fourth root of X. We are supposed to find a function G of X such that capital F of X is equal to F of G of X. That is we're doing a problem of function decomposition. We want to find a function G so that when you put G inside of little F you're gonna get capital F. And so let's see the relationships here. So when you look at, when you see F of X as the fourth root of X right here, and you see there is a fourth root of X inside of capital F, we can see that, okay, this function right here, you have the fourth root of something, we're gonna call it U, but what's left inside? What's inside of the fourth root of this function? Well, the answer is gonna be X squared plus one. That is X squared plus one is the thing we put inside of the fourth root to construct the function capital F right there. And that would indicate to us that G of X then should be X squared plus one, which would be option in D. And so that's how one should approach this one, but at the very least, if that seems too insurmountable, another option you have is that if we know little f and we have these options for G, each and every one of them, you could plug into F and see what happens. If you put the fourth root inside the fourth root, you're gonna get the 16th root. If you put the square root inside of the fourth root, you're gonna get the eighth root. If you put the square root of X plus one inside the fourth root, you're gonna get the eighth root of X plus one. You can quite process elimination, answer this question.