Linear approximation of a rational function | Derivative rules | AP Calculus AB | Khan Academy

So there are situations where you have some type of a function. This is clearly a non-linear function. f of x is equal to 1 over x minus 1. This is its graph, or at least part of its graph, right over here. But where you want to approximate it with a linear function, especially around a certain value.

What we're going to do is find an approximation. Let me write this down. I want to find an approximation for—actually, let me clear—I want to find a linear approximation. So I'm going to approximate it with a line. I want to find a linear approximation of f around, and you need to know where you're going to be approximating it around x equals negative 1.

So what do we mean by that? Well, let's look at this graph over here on this curve. When x is equal to negative 1, f of negative 1 is negative one-half, which sticks us right over there. Let me do this in a better color, so it's right over there. What we want to do is approximate it with a line around that. Essentially, we're going to approximate it with the equation of the tangent line.

The tangent line is going to look something like that. As we can see, as we get further and further from x equals negative one, the approximation gets worse and worse. But if we stay around x equals negative one, well, it's a decent—it's as good as you can get for a linear approximation. At least in this case example, it is a very good linear approximation.

So when people say, "Hey, find a linear approximation of f around x equals negative one," or if they say, "What is the following is the best approximation?" and all of your choices are lines, well, essentially, they're asking you to find the equation of the tangent line at x equals negative 1. So let's do that.

In order to find the equation of the tangent line, the equation of a line is y is equal to mx plus b, where m is the slope and b is the y-intercept. There are other ways that you could think about it. You could think about it in terms of point-slope, where you could say y minus some y that sits on that line is equal to the slope times x minus the corresponding x1. So x1 comma y1 sits on that line someplace.

Actually, I like to write this point-slope form like this sometimes: y minus y1 over x minus x1 is equal to b. This comes straight out of the idea of look, if x1 and y1 are on the line, the slope between any other point on the line and that point is going to be your slope of your line. So we could think about it any of these ways.

So let's first find the slope of the tangent line, and that's where the derivative is useful. So f—well, actually, let me just write f of x again. I'm going to write it as x minus 1 to the negative 1 power because that makes it a little bit clearer that we can use the power rule and a little bit of the chain rule.

So the derivative of f with respect to x is equal to—so the derivative of x minus 1 to the negative 1 with respect to x minus 1, well that's just going to be—we're just going to use the power rule here. It's going to be negative 1 times x minus 1 to the negative 2, and then we're going to multiply that times the derivative of x minus 1 with respect to x. Well, that's just going to be 1, right? The derivative of x with respect to x is 1; the derivative of negative 1 with respect to x is 0. So we could say times 1 here if we like, or we could just not write that because it doesn't change the value.

And so let's evaluate that when x is equal to negative 1. So f prime of negative 1 is equal to—I could just write this as negative—alright, look this way—negative 1 over negative 1 minus 1 squared. This is going to be negative 2 down here. So this is equal to negative negative 1 over 4. So the slope of our tangent line is—so I could write this way, m is equal to negative 1 over 4.

And so now we just have to write its equation down. We already know an x1 and a y1 that sits on the line. In fact, we want to use the point when x equals negative 1. So we know that the point negative 1 comma—we could just put it right over here—f of negative 1 is negative one-half. One over negative one minus one is negative one-half.

So we know that this negative one comma negative one-half—that that is on our curve and it is on our line. That's the point at which the tangent and the curve actually intersect. We can use any of these to now write the equation of our line. We could say y—I’ll do it right here—y minus y1, so minus negative one-half, is going to be equal to—it's going to be equal to our slope, negative one-fourth. I'm just using the point-slope version of our equation—is equal to our slope times x minus x1. So x minus our x coordinate that we know sits on this, so minus negative 1.

And so let me now write all of this in a neutral color. This will be y plus one-half is equal to—and I can—so this is going to be plus 1 right over there. So I can distribute the negative 1/4. So it's negative 1/4 x minus 1/4 minus 1/4, and then I can subtract one-half from both sides.

So I'm going to get y is equal to negative 1/4 x. And then if I already am subtracting 1/4 and I subtract another half, that's going to be negative three-fourths. So minus three-fourths. That’s actually pretty close to what I drew up here. This should be intersecting the y-axis at negative three-fourths.

So there you have it! This line, or you could even say this equation, is going to be a very good linear approximation, about as good as you can get for a linear approximation for that non-linear function around x equals negative 1. You might say, "Well, why didn't they just ask me to find the equation of the tangent line at x equals negative 1?" Well, they could have, but there’s a little bit of extra cognitive processing here where you say, "Okay, I can actually use the equation of the tangent line to approximate this function around x equals negative 1."