 Hello, everyone. We're going to take a look at three examples of how to find a tangent line approximation, also known as a linearization of a function at a specified point. One thing to keep in mind as we talk about linearization, what you're really doing is just writing the equation of a tangent line to the curve at that specified point. That's it. So when we go through these problems, and you'll see how they're all phrased, like this first one that you see here, you would be obtaining the exact same answer. If the directions had read, write the equation of the tangent line to the curve at this point. So that's something to keep in mind as we do these. The first one we're going to work with is the function 1 plus sine of x, and we are looking for the linearization at the point 0 comma 1. Recall what our linearization formula states. It states that l of x, that's the notation for our linearization equation, is equal to f of a plus f prime of a times the quantity x minus a. Now let's talk about how that is that this is simply a rearrangement of point slope form of an equation of a line. L of x is simply y. f of a is comparable to our y1 that you would have in the point slope form. Remember point slope form typically states on the left that y minus y1. So if you simply add that y1 value over to the right, and now is y1 on the right, plus remember slope is really your derivative value times the quantity x minus x sub 1. So really all that linearization formula is, is simply a restatement of the point slope form of a line. In order to find then the linearization of this function 1 plus sine of x, we need to calculate a few things. The first thing we need to do is find our derivative because we'll need that in the calculation of f prime of a. So our derivative is simply going to be cosine of x. The 0 here, that's the a value. So let's go ahead and find f prime of 0. That simply is going to be 1 cosine of 0, of 0 is 1. We also need f of a, well that's simply 1, so that's easy, we have that. So we should be ready to substitute into our linearization formula. Therefore we would have l of x is equal to f of 0, remember that's 1 from the ordered pair, plus our slope, that's our derivative value which happens to also be 1, times the quantity x minus n r a value is 0. If you rearrange that, you have x plus 1 as our linearization. Again, remember that you would have the same exact answer had I asked you instead to write the equation of the line tangent to the curve 1 plus sine of x at the point 0 comma 1. Let's look at this from a graphical point of view. We have here the blue curve, which is our original function 1 plus sine of x, and the red line is our linear equation 1 plus x that we just obtained on the previous screen. Near x equals 0, and here's the point right here, it's the ordered pair 0 comma 1, the function 1 plus sine of x is very well approximated by the linear equation y equals 1 plus x. You can easily see that from the graph. See how for very, very close values of x to 0, there's almost no distinction in there between the location of the tangent line, the red tangent line, and the blue curve pretty much all right all in here. The tangent and the curve are almost one in the same. There is such a little difference between the two. As you get farther and farther from 0 though, notice how that difference between the tangent line and the curve becomes a little bit more pronounced. Look for example up here. So for x values, I believe this is pi over 2. Notice how for values of x close to pi over 2, there's more of a difference in between the location of the tangent line and the curve. Same thing over here close to negative pi over 2. So the closer you are located to 0, think the delta neighborhoods, back from when we talked about the formal definition of a limit, the closer you are to 0, the better defined the curve, the original curve is by the tangent line equation. Let's look at another type of function, 1 over x at the a value of 1. Once again, we have our linearization formula. a is 1. Let's go ahead and find f of 1, since we'll be needing that, that simply is 1. We'll need our derivative. If you think of the original function as x to the negative 1, our derivative becomes negative 1 over x squared, and f prime of 1 simply is going to be negative 1. Now we have everything that we need to substitute into our linearization formula. Therefore we have l of x is equal to f of 1. Remember that was 1 plus our slope f prime of 1, that's negative 1 times the quantity x minus 1. You can go ahead and distribute the negative and simplify and you'd have negative x plus 2. And that would be our linearization of the function of 1 over x at the x value of 1. Let's take a look at this one graphically as well. Here the blue curve is our original function, 1 over x. The red linear function is our linearization negative x plus 2. You can see here the point of tangency. And again notice right in here for values of x really really close to an x value of 1, there is almost no distinction between the curve itself and the tangent line. As you get farther and farther away from 1, so for instance at x values maybe around 0.5, notice how there's a big difference in here between the placement of the original curve and the tangent line. And as you get over on the right away from 1 again maybe closer to 2, once again there becomes more of a difference between the original curve and the tangent line. But so long as you're staying really close to 1, you're going to really be able to approximate values on the curve with the simple linear equation. In this last example we're going to find the linearization but then use it to estimate the value of the function at 2.01. Our a value in this case is the 2 from our ordered pair. f of a then is simply going to be 3. We do need our derivative. Let's go ahead and find that. We'll have to use the chain rule in order to do that. We can rewrite it. The value of that derivative at 2. If you substitute in you should obtain 12 over 6 which of course simplifies to 2. So now we have everything that we need to substitute into our linearization. We then would have l of x is equal to f of 2. Remember that's just the function value so that's 3 plus f prime of 2 that was 2 times the quantity x minus 2. Simplifying we get 2x minus 1 and that's the linearization for this function the square root function close to the point 2 comma 3. Now we wanted to use that to estimate the value of f of 2.01. A little history for you as to why this used to be perhaps a little bit more useful than it is today. Back in the old days before our nifty graph and calculators and even before scientific calculators to do something like square root of 1 plus 2.01 cubed by hand was incredibly tedious. Where linearization really had its use back then was instead of substituting 2.01 into that original function and having to deal with all that tedious hand calculations it's a lot easier to simply substitute it into a linear function. And that's really where this had benefited. If we wanted to then estimate and notice the keyword estimate there the value of 2.01 in this original function you do need to use an approximately equal to symbol. All you'll do is substitute 2.01 in place of the x in our linearization equation that we obtained. You can do that in your head and you get 3.02. Now just for kicks on your calculator go ahead and do square root of 1 plus 2.01 cubed. See what you get. You should get something about 3.02003. Notice how close it is to the answer that we obtained. And that's really a great example of how it is we can work with more complicated functions by pairing it down to a simple linear equation provided we are very close to the targeted point. In this case notice how close you're to 2. You're at 2.01 that's like a delta value of 0.01. Again think back to the delta neighborhood values we talked about when we did the formal definition of a limit. The closer you are to the targeted point the better this is going to work and the better of an estimation you will receive in the end.