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# Local linearity

AP.CALC:
CHA‑3 (EU)
,
CHA‑3.F (LO)
,
CHA‑3.F.1 (EK)

## Video transcript

Let's say that we're interested in approximating what the square root of 4.36 is equal to. So we want to figure apro, we want to figure out an approximation of this, and we don't have a calculator at hand. Well, one way to think about it is we know what, we know what the square root of 4 is. We know that this is positive 2. The principle root of 4 is positive 2. So, okay, this is going to be a little bit more than two. Well, let's say that we wanna get a little bit more accurate, and so what I'm going to show you in this video is a method for doing that, for approximating the value of a function near, near a value where we already know the value. So what am I talking about? So let's just imagine that we had the function. We have the function f of x is equal to the square root of x, which is, of course, the same thing as x to the one-half power. So we know what f of 2 is. We know, we know that f of 2, I'm sorry, we know that f of 4 is. We know that f of 4 is the square root of 4, which is going to be equal to 2 or the principle root of 4, which is equal to positive 2 and what we want to approximate, we wanna figure out what f of, we wanna figure out what f of 4.36 is equal to. This is just another way of framing the exact same question that we started off this video. So let's just imagine our function. Let's just imagine it for a second. So, let me draw some axes. This is my y-axis. This is my, this is my x-axis, and let's graph y is equal to f of x. So let's say it looks something like this. Y equals f of x looks something like that. That's pretty decent. All right, so that's, that right there is y is equal to f of x, and we know f of 4 is equal to 2. F of 4 is equal to 2, so this is when x is equal to 4. I haven't drawn it really to scale, but hopefully, this is clear enough. So that right over here is going to be 2. That's f of 4. And what we wanna approximate is f of 4.36, so 4.36 might be right around, right around there, and so we want to approximate, we wanna approximate this y value right over here. We want to approximate that. Right over here is f of 4.36, and, once again, we're assuming we don't have a calculator at hand. So, how can we do that using what we know about derivatives? Well, what if we were to figure out an equation for the line that is tangent to the point, to tangent to this point right over here. So the equation of the tangent line at x is equal to 4, and then we use that linearization, that linearization defined to approximate values local to it, and this technique is called local linearization. So what I'm saying is, let's figure out what this, the equation of this line is. Let's call that l of x. And then we can use that to appro, and then we can evaluate that at 4.36, and hopefully that will be a little bit easier to do than to try and figure out this right over here. So how would we do that? Well, one way to think about it, and obviously, there are many ways to express a line, but one way to think about it is, okay, it's going to l of x is going to be f of 4, which is 2. It's going to be f of 4 plus the slope, the slope at, at x equals 4, which is, of course, the derivative f prime of 4, so that's going to be the slope of this line of l of x is f prime of 4. Let me make that clear. So this right over here is the slope. The slope when x is at, at x equals 4, so this is a slope of this entire line, and so any other point on this is gonna be f' of 4 plus the slope times how far you are away from x equals 4. So it's going to be times x minus 4. Let's just, let's just validate that this makes sense. When we put 4.36 here, when we put 4.36 here, actually let me zoom in on this graph just to make things a little bit clearer. So, if this is, so I'm gonna do a zoom in. I'm gonna do a zoom in. I'm gonna try to zoom in into this region right over here. So, this is the point. This is the point (4, f of 4), and we are going to graph l of x. So let me do that. So this right over here is l of x. That's l of x. And let's say this, right over here, this right over here, is the point (4.36, f of 4.36), and the way that we're, we're gonna approximate this value is to figure out what, to figure out, what this value is right over here. And what is this one going to be? This right over here is going to be, this is going to be (4.36, L of 4.36). This line evaluated when x is equal to 4.36, and what is that going to be? What is that going to be equal to? Well, let's see. Let's just evaluate it. L of 4.36 is going to be f of 4. So, it's going to be 2 plus the derivative, so the slope of this line plus f prime of 4 times x minus 4. So, 4.36 minus 4 is going to be times 0.36, and that makes sense. You're starting at 2 and you're saying, okay, my change in x, my change in x is 4.36. So, my change in y is going to be my slope times that change in x to get me that value, to get me that value right over there. So, let's figure out, let's figure out what this, let's figure out what this thing, what this thing actually is. So to do that, we need to figure out f prime of 4, so let's go back up here. I'll try to leave actually, I'll leave this little visualization here. So let's see, f prime, f prime of x is going to be one-half x to the negative one-half just using the power rule over here. So f prime of 4, f prime of 4 is equal to one-half times 4 to the negative one-half, which is, of course, equal to one-half times one-half. 4 to the one-half would be 2. 4 to the negative one-half is going to be one-half. So, this is equal to 1 4th. So L of, we deserve a little bit of a drum roll now, L of 4.36 is equal to f of 4, is equal to f of 4, which is eq, let me just rewrite it. It's f of 4 plus f prime of 4 plus, gee, why am I switching to that color. Let me do the yellow. Plus f prime of 4, times, times, times, 4.36. 4.36. Let me make this actually a new color, just so we see it. So, 4.36, so times 4.36 minus 4, minus 4, and actually let me make all the 4s one color, too, so you see it's the same, so just like that. So what is this going to be? Well, this we already established is positive 2. This we already established, and we do this in a yellow color. This, we already established is 1 4th, and this part right over here is 0.36. So, this is going to be equal to 2 plus 1 4th times 0.36, is 0.9 or is 0.09, so this is going to be equal to, this is going to be equal to 2.09. So that is our approximation and should be at least the way, based on how I graphed it, a little higher than the actual value of the square root of 4.36, but we could write that up here. This is going to be approximately, let me just write it this way. The square root, I'll just write it down here. So we could say the square root of 4.36, which is the same thing as f of 4.36. This is approximately equal to 2.09. Now let's just say we happen to find a calculator, and just out of curiosity let's see how good of an approximation that is. Let's get a calculator out. And so, we wanna do the square root of 4.36, and we get 2.088. So, we actually, if we round to the nearest hundredths, we got a pretty good approximation just like we saw. In this, in this indicative graph right over here, it is, our approximation was indeed a little bit higher than the actual value.
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