 In this video, we provide the solution to question 14 for the practice final exam for math 1210. We're given a function f of x, which equals x to the seventh plus four. We're supposed to use Newton's method to approximate the root of this function, that is its x-intercept. And we're gonna use the initial value x equals one. If that's the case, what would be x2? What is the second number in this sequence? Well, by Newton's method, x2 is gonna equal x1 minus f of x1 over f prime at x1. So what do we have here? So x1, of course, is just one. So if we plug one into the function, we're gonna get one to the seventh plus four, which of course, one of the seventh is just one. So we get one plus four, which is gonna be five. We'll do that in a second. We have to then also do f prime and value it at one, for which the derivative of the function is gonna equal seven x to the sixth. So we end up with seven times one to the sixth. So we get one minus, well, like I said, one to the seventh is one. One to the sixth, of course, is likewise one. So I'm just gonna write the numerator as a five. One plus four is five. And then we get one to the sixth, which is one. Seven times one is seven. So we get one minus five sevenths. So we should take seven sevenths minus five sevenths. And we see that our approximation is gonna be two sevenths. That is x2 is choice A.