At this point in the course, we should be very familiar with a tangent line. We've discussed what a tangent line is, how to graph it, and how we can find the equation of a tangent line based on its relation to derivatives and some particular point we have on our graph. Now in this video, we're going to take all this knowledge about tangent lines we've learned and apply it to this new idea of linear approximations. This might sound complicated and abstract, but don't sweat it because the process for solving linear approximation problems that you're going to see in this course is very similar, in fact, nearly identical to the problem-solving we did when finding the equation of a tangent line. Let's just go ahead and get right into things.
The first step for solving a linear approximation problem is to find something called the linearization. Linearization is the process where we take any smooth curve or function and approximate the entire thing to be a line. This might sound like a bad idea because we've seen some pretty complicated-looking functions. The thought that we could approximate any of these functions to look like a line sounds ridiculous, and I would agree with that. But there's a trick that we use when we do these linearization cases.
What we do is zoom in because if you zoom in very closely to one point on your curve or one specific area, you're going to find that this entire section you zoom into will look very much like a line. The more and more you zoom in, the more and more it's going to resemble a straight line. That's the process we use to solve linearization problems. We know how to find the equation of a tangent line, something we've discussed before.
Notice how a tangent line is going to be at one specific point. We know the process for finding the equation of the tangent line there, and it turns out we can use the same process for finding a linearization since the linearization is just the line that represents our function. Now let's actually get into this example. Here, we're told to find the linearization of this function, \( f(x) = x^2 \) at a value of \( a = 1 \). Now notice that our function \( x^2 \) looks nothing like a line.
But remember, when we're doing linearization, what we're doing is focusing on a very specific point. In this case, we're looking at the point where the x-value is 1. Notice what happens if I zoom in very closely here on my graph. If I look closely at that point, the entire thing eventually is just going to look like a line, which is exactly what we want for our linearization problem. Now I can see here is one is this value of interest we have, which is \( a \).
This value right here would be the output when our input is \( a \). So our function is \( x^2 \). And then if \( a \) is 1, one squared is 1. So this right here would be the output \( f(a) \). Now if I want to go ahead and find an equation for this line, wouldn't this just be the tangent line at the certain point?
It turns out that is exactly what happens with linearization. \( L(x) \), this linearization case, that's just the tangent line of your function at this specific value of interest. So if I want to find my linearization \( L(x) \), what I need to do is have my output value, which we know is just going to be \( f(a) \). I need to have the slope, which we know is just the derivative at a certain point. That's the slope of a tangent line, and then it's going to be \( x - a \). So this would be our equation for \( L(x) \).
Solving this situation where we want to find \( L(x) \), well, this is just going to end up being the equation of a line. So \( L(x) \) is going to be \( f(a) \), which we figured out is 1. See, we have 1 right here as our output. That's going to be plus our derivative evaluated at \( a \). Now \( f'(x) \), the derivative of our function \( f(x) \), that's going to be \( 2x \) using the power rule. And \( f'(a) \) would be \( f'(1) \) since we have this value right here, which is 2 times 1 or 2. So in this case, we're going to have \( f'(a) \) which is 2, and that's going to be multiplied by \( x - a \) which is 1. So this is where our function \( L(x) \) becomes, and what I can do is actually simplify this, and you should get that this whole linearization is \( 2x - 1 \).
So that's going to be the equation for our line, and that would be the solution to this problem where \( L(x) = 2x - 1 \). Notice it comes out to being the exact same equation that we had for our tangent line, and that's because it's the exact same process in both situations. All we're doing is using the equation of a tangent line and just zooming in really closely. Now, of course, there is an interesting question when we do this, which is what's the point? How is this really any different from just finding the equation of a tangent line, and why would we ever need this?
Well, because with linearization, we're able to put this into application much better. Over the course of the past couple of videos, we've been spending a lot of time talking about how we can apply certain situations that we see in calculus or math in general to the real world, and that's what we're doing with linearization. So when it comes to linearization, since we found this line, notice any values that we find that are close to this region, close to this area we zoomed in on, are going to be pretty close to the points that are actually on our function. So if we have some complicated function where we can't figure out what it evaluates to, we can just look at our line, which we know is a lot simpler of an equation. Let's actually.VideoCapture(0) go ahead and try some examples of this where we need to approximate values on our line.
So here we have an example down here where we are given this function \( f(x) = x^2 \), and we're also given the linearization, which we calculated up here. Now what we're asked to do is approximate \( f(1.05) \). Notice something about the problem we just solved. Recall that our example, the value of interest we were looking at is \( a = 1 \). And notice how 1.05 is very close to 1.
So that means we have a point that is somewhere in this region up here. So since it's 1.05, it's probably going to be somewhere up here, which is about where the function value is too. Let's actually see what happens when we approximate this. Since we're in close proximity, we know that our function is approximately going to be a line at that point. So what I can do is find \( L(1.05) \), which is approximately going to be \( f(1.05) \).
So plugging these values in, I'll go to my linearization up here, which is going to be 2. It will replace \( x \) with 1.05 minus 1. Now 2 times 1.05, that's going to be 2.1. We can do that math in our heads. And then it's going to be minus 1, which means this whole thing is going to come out to 1.1.
So this right here is going to be the approximate solution for \( f(1.05) \). That's how we can approximate this answer. Now the question becomes how close is this value to our actual solution? Because if 1.05 is right about there on the x-axis, how close is this to the function itself? What we can do is we can actually take this value and type it into our calculator.
If we have 1.05, here we're going to have \( 1.05^2 \). Now if you type this in, your calculator should output 1.1025. So this would be the result that you get, and notice something right here. Notice how the result that we calculated using our calculator came out very close to what we found when we just used this approximation with the linearization that we found. And that's the idea of linear approximation.
It allows you to find values that are super close to the function that you_CURR have. So if you were a scientist and you had some sort of data with a very complicated function you were dealing with, you could approximate it all to be a line so you could solve it by hand like we did up here. Now one thing that I will also mention is that when you approximate your function at a specific x value and you plug your value into the linearization, the farther this value is from your point of interest \( a \), the less accurate the result is going to be. So notice \( 1.05 \) and \( 1 \) were very close together. Let's say that you were trying to approximate \( f(1.5) \) or \( f(1.8) \) or even something like \( f(2) \).
Notice we're getting further and further away from the original \( a \) value that we had, which means that your result is going to be less and less accurate of an approximation the further you get. So you do want to remain in this region right here because that's going to allow you to get a more accurate approximation since your function looks most like a line at your certain point of interest. So hope you found this video helpful. That's the main idea of linear approximations and linearization.
