 Hi there and welcome to the screencast where we're going to look at an example of both constructing and using the local linearization of a function at a point. Here I have on the screen here three data points about temperature. These are actual temperatures in Grand Rapids, Michigan over the course of a morning on a single day. The temperatures here are in degrees Fahrenheit and we have temperatures at 7am, 10am and 1pm. Only these three data points were measured and suppose I want to estimate the temperature at a point where it wasn't measured. This is kind of like a forecast in some ways. We're going to use a local linearization to estimate the temperature right at 10.15am. Now again, the reason we need a local linearization or some kind of estimation technique in this problem is that I don't have a formula for my temperature. I can't just plug in x equals 10.15am and get the temperature out. So we have to use a pretty good estimation technique and local linearization is one such technique. Let's first of all kind of normalize some of the scaling here especially in the time. Let's arbitrarily set 7am to be time equals zero. That would make 10am time equals three hours later and then three hours later than that would be 1pm and that's time equals six. And so that would make 10.15am. That's three hours plus another quarter an hour. So that would make that x equals 3.25. So now that we've rescaled some of the time values. Let's first of all think our way through the problem. So first of all it says use a local linearization. So the first thing we should do is go back and just recall the formula for a local linearization. I'm going to write that down here in green. So the local linearization at a point x equals a is L of x equals in the definition it says f for my function. But my function is actually called t here. So this should be t prime at a times x minus a plus t of a. Again this is just the formula for the equation of the tangent line to the graph of in this case t at x equals a. So let's first of all interpret what is the a here. I've copied down the formula and that's a good strategy but now I have to sort of interpret what all the pieces mean. Again I'm using t because that's the name of my function even though it says f in the definition. So what is the a value? Well a is the point where we are attaching the tangent line. A has to be a known point and which known point should I use? I have three of them to choose from. This one, this one, and this one. Should I let a equal zero? Should I let a equal three? Should I let a equal six? Well that's kind of a function of where I want the estimate. I'm going to use the local linearization to estimate the temperature at x equals 3.25. By a I'm going to want to choose the closest value to 3.25 where I actually know the function value. So that would be time equals three in this case. Again I could conceivably use any of these three time values to create a local linearization but the farther I get away from the point where I'm going to eventually estimate the worse the estimate is going to be. So we're going to let a equals three here in my formula. So that will make the local linearization t prime of three times x minus three plus t of three. t prime of three we'll have to think about in just a second. x is, we'll talk about that in a second, but t of three is right up here in the table. That is the temperature value at time equals three. So that's just going to be 64. So I'll make that quick substitution. Now what remains to be done here in this problem is first of all I need to figure out what t prime of three is. And then second of all I have to deal with the x. Now I can, let's talk about that second point first. The x is going to be the place where I make my estimate. So eventually once I figure out what t prime of three is I'm going to have an equation that has an x in it and this problem I'm going to use x equals 3.25. That's the value that I actually want to estimate. Okay, so what really needs to be done the hard work here is figuring out what t prime of three is and we're going to go on a separate slide and do that. Okay, so here's just a copy of the data table that we have. I'm trying to calculate t prime of three. I can't do this exactly because I don't have a full specification for t. I don't have a formula that I can use to calculate a limit for example. So what I will do here is kind of fall back on a previous technique of using a central difference approximation. I happen to be in a position where I have two data points, one before x equals three and one after x equals three at the same distance. Both of those data points are three hours away from ten o'clock. And so t prime of three, I really shouldn't say equals here. That's not totally right. I should say approximately would be t of six minus t of zero divided by six minus zero. Again, this is a central difference approximation that we saw in an earlier section. Okay, so reading off the table, t of six is 75, t of zero is 54, and of course, six minus zero is six. This gives me 21 over six, and that is exactly 3.5. Okay, that's the value of t prime of three. Now let's copy over the formula for the linearization at x equals three. That was t prime of three times x minus three plus t of three. And now we can specify every part of this function except the x. The t prime of three is 3.5. The x is a variable, we leave it alone. And then the t of three sets the temperature at time equals three and that's 64. So now I have the linearization at x equals three, and what I want to do is use it to estimate the value of the temperature at x equals 3.25. That's 10, 15 a.m. That's to simply calculating l of 3.25. This formula I have here in the previous line is just the equation of the tangent line to the graph of t at a certain point. So I'm just going to evaluate it at the point that I want. So this is 3.5 times 3.25 minus three, all that plus 64. I'll let you crunch the numbers on this and you will come up with 64.875. There's a decimal point there and that's in degrees Fahrenheit. So that will be an estimate for the temperature value at 10.15 a.m. This is a believable temperature value because the temperature right at 10 a.m. is 64. This is just a little bit warmer and I can see that there's an increase in temperature going on at this point. So I believe that it's only 15 minutes later so the temperature can't go up by that much but it should be going up by a little and this does so I kind of believe the answer. So that's an example of calculating a linearization at a point and then using it to estimate unknown values of the function. Thanks for watching.