 Welcome back everyone to our lecture series Math 1210 Calculus 1 for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Misalign. In part four of our lecture series, we're going to talk about some approximation theory. That is, how can we use the derivative to help us to solve for approximate solutions to equations or to approximate things like the square root of 17, what have you. This lecture you're going to see, it's a pretty short lecture and it's based upon two sections in the textbook actually, James Stewart textbook. One of the sections is 3.10 entitled linear approximations and differentials. We'll talk about the other one in a second. Before we talk about Newton's method, which is the main topic we're going to get to today, I want to talk about a linear approximation, which is the precursor to Newton's method. This semester we have, I should say in this course, we've spent a lot of time on derivative and particularly the tangent line. If f is differential at some point x equals a, the equation of the tangent line you can see in front of you. Typically, we solve for y here to get the slope intercept form. We might get something like y equals f prime of a times x minus a plus f of a. Now, typically, one would distribute the slope here f prime of a in combined like terms if you want slope intercept form, but that's not what I'm too interested right now. Since we solve for y here, we could, this itself becomes a function in its own right. So we could come up with a new function say L of x, which equals this value f prime of a times x minus a plus f of a. This is going to be a linear function, and this is often referred to as the linearization of the function because this is the linearization at x equals a. We have a function and we're making a new function that approximates the original function using some type of tangent line. So if we have some nice differential function, maybe like this, and we take x equals a, maybe we'll just take one right here, x equals one. Well, we could take a, we could look at the point of tangency right there. And if we form the tangent line to the function in this region, we would get something like this. The tangent line is what we mean by L, this linearization, and the blue curve is the original function f for which we are constructing that. And the thing is, when you're really close to f, if you're really close to x equals a, you get this point right here, a comma f of a. And so the thing here we're trying to say here is when x, when x is approximately equal to a, we then get that f of a will be approximately equal to L of a. So the linearization, the line and the function are almost the same thing when you're close to the function. And this depends a lot about the function in question, right? Because again, when we say things like approximate, what do we mean? Like, we just just mean close, how close do we mean? Like we talked about in previous lectures about epsilon delta, we can make very rigorous what we mean by close. So if x is delta close, then we can guarantee that will be epsilon close over here. But we're not going to worry about that too much in this lecture, we just want to talk about how one can approximate using the linearization. Alright, so as an example, let's take the function f of x equals the square root of x plus three, and using the point a, we can compute the linearization. So first of all, to do the linearization, we need to know the derivative. So f prime of x will by the usual power rule and chain rule, the derivative here is going to be one over two times the square root of x plus three. The linearization L of x, like we saw before, it's just the tangent line f prime at one times x minus one plus f of one. So there's a couple things to compute there. So to get the linearization, the derivative evaluated one, well, let's take a look at that real quick f prime at one, we get one over two times the square root of one plus three, we're going to get one over two times the square root of four, square root of four, of course, is two. So you get one fourth as the slope of the tangent line at this point. Then we're going to get x minus one. And then the last thing that to do is evaluate the function, like we saw before, at one, this function is going to be the square root of four, which is a two. I am going to try to simplify this and put in slope intercept form, we get one fourth x minus a fourth plus two, which instead of two, we'll think of it more as eight fourths. And so our linearization is then equal to one fourth x plus seven fourths. And so our claim is that this function will be approximately the same thing as f of x, which as we saw above is the square root of x plus three when x is close to one. All right, so let's see how we can use this to approximate something like the square root of 3.98. All right. So if we were trying to compute the square root of 3.98, well, we notice that 3.98, this is the same thing as 0.98 plus three. That is to say, this is just the function evaluated at 0.98. And I think we could argue that 9.8 is close to one. Again, we could be more precise about that. Those things are approximately the same number, though. And so this would tell us that our function, we approximately the same thing as 9.8 of the linearization for which we get 0.98 over four plus seven fourths. And it might make more sense to use decimals in this context here. But the point is 9.8, although a decimal, we could write as a fraction or we could work with decimals. This is something we actually could do on like a napkin, the back of a napkin or the back of an envelope or whatever, if we had to do the calculation, we don't need a calculator to help us with the square root in this situation. Anyways, if we continue on here, 0.98, since there is a common denominator four, right, we'll just add the seven and the 0.98 there. So we get 7.98 divided by four. And so if we divide that by four, you'll end up with 1.995 as your estimate. And feel free to pause this video and pull out your scientific calculator or it's just a square root, oftentimes that shows up on a four function calculator, which should be called a five function calculator if that's the case. But nonetheless, you could check this real quickly. What is the square root of 3.98 on your calculator? And you'll see that, you know, for the first couple decimal places, it's pretty close, our estimate of 0.1.995 there, right? I'll double check on my calculator right here. If I were to do this, I should have opened this ahead of time, but if we take the square root of 3.98, I got an estimate of 1.99499. I'm going to stop there. If you round to three decimal places, that's spot on to what we got right there. And if we were to try this again, because the original prompt had a second one in mind, if we do the square root of what was again, 4.05, you'll notice this is the same thing as 1.05 plus three. So we're really just looking at f of 1.05, which like before, I'd say 1.5 is pretty close to one. And so that means we'll get somewhat of an estimate using L of 1.05. And so like we did on the other one, we'll get 1.05 over 4 plus 7 fourths. We can make that 8.05 over 4. And that will then become 2.0125. And that's the exact value right there. And so that gives us an estimate of what the square root of 4.05 equals. And I would make the second recommendation pull out your scientific calculator. Most likely your phone has an app that you can use here. So put that in there. And if you do this on a calculator, you might get something like 2.012461. I'm going to stop there rounding that to four decimal places. That's accurate. And so we can get a pretty quick estimate of values. We can approximate functions like square roots, exponentials, logarithms using a straight line. A straight line can be always computed with four, the four arithmetic operations, addition, multiplication, sorry, addition, subtraction, multiplication, division. You don't need more complicated algebra like square roots, exponentials, logarithms, even sine and cosine you could approximate using this technique right here. It's fairly elementary, but we're able to get sort of reasonable calculations, right? We are accurate to three and four decimal places.