The slope when x is at, at x equals 4, so this is a slope of this entire line, and Prime of 4, so that's going to be the slope of this line The slope at, at x equals 4, which is, of course, the derivative f To think about it is, okay, it's going to l of x is going to beį of 4, which is 2. Obviously, there are many ways to express a line, but one way So how would we do that? Well, one way to think about it, and Hopefully that will be a little bit easier to do than to try and figure out this right over And then we can use that to appro, and then we can evaluate that at 4.36, and Is equal to 4, and then we use that linearization, that linearizationĭefined to approximate values local to it, and this technique is called local So, how can we do that using what we knowĪbout derivatives? Well, what if we were to figure out anĮquation for the line that is tangent to the point, to tangent Right over here is f of 4.36, and, once again, we're assuming we don't have aĬalculator at hand. And what we wanna approximate is f ofĤ.36, so 4.36 might be right around, right around there, and so we want toĪpproximate, we wanna approximate this y value right So that right over here is going to be 2. F of 4 is equal to 2, so this is when x isĮqual to 4. Y is equal to f of x, and we know f of 4 is equal to 2. All right, so that's, that right there is Y equals f of x looks something like that. So let's say it looks something like this. This is just another way of framing the exact same question that we started off Root of 4, which is equal to positive 2 and what we want toĪpproximate, we wanna figure out what f of, we wanna figure out what f ofĤ.36 is equal to. We know that f of 4 is the square root ofĤ, which is going to be equal to 2 or the principle The square root of x, which is, of course, the same thing as So what am I talking about? So let's just imagine that we had theįunction. You in this video is a method for doing that, forĪpproximating the value of a function near, near a value where weĪlready know the value. Well, let's say that we wanna get a littleīit more accurate, and so what I'm going to show So, okay, this is going to be a little bit Well, one way to think about it is we know what, we know what the square root of 4 So we want to figure apro, we want toįigure out an approximation of this, and we don't You see how better our approximation became by choosing a different "a"! Hope this helpsĪpproximating what the square root of 4.36 is equal to. To fix that choose an "a" that is close to the number you want an estimation. The farther away you go from "a" the more precision you lose. You see that our estimation is not as precise as the last. The actual answer to five decimal places for (37)^(1/2)=6.08276. Now if we do the same procedure and if we wanted to figure out (37)^(1/2). When you work everything out your L(x)= 5 + (1/10)(x-25) Majority of the time local linearization is only used on points that are close to your original point. It will all depend on how good of an approximation you want. The farther away you move from your original point, the more precision you lose.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |