If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:7:11
AP.CALC:
CHA‑3 (EU)
,
CHA‑3.F (LO)
,
CHA‑3.F.1 (EK)

Video transcript

so there are situations where you have some type of a function this is a clearly a nonlinear function f of X is equal to one over X minus one 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 and so what we're going to do is we want to find an approximation let me write this down I want to find an approximation for and actually let me be 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 approximation of f 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 is negative 1/2 which sticks us right over there let me just in a better color so it's right over there and what we want to do is approximate it with a line around that and what we're essentially going to do is we're going to approximate it with the equation of the tangent line the tangent line is going to look something something like that and as we can see as we get further and further from from x equals negative 1 the approximation gets worse and worse but if we stay around x equals negative 1 well it's a decent it is a as good as you can get for a linear approximation or at least in this example is a very good linear approximation so when people say hey find a linear approximation of f around x equals negative 1 or if they say what is the following is the best approximation and all of your choices are our 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 so 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's 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 and actually I'd like to write this point-slope form like this sometimes y minus y1 over X minus x1 is equal to B because this comes as straight out of the idea 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 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 so 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 derivative of negative 1 with respect X is 0 so we could say x 1 here if we like or we could just not write that because it changed 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 can just write this as negative all right look this way negative 1 over negative 1 minus 1 squared and so this is going to be negative 2 down here so this is equal to negative negative 1/4 so the slope of our tangent line is so I could write this M is equal to negative 1 negative 1/4 and so now we just have to write its equation down so we already know an X 1 and a y 1 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 and put it right over here F of negative 1 is negative 1/2 1 over negative 1 minus 1 negative 1/2 so we know that this negative 1 comma negative 1/2 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 and so 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 1/2 is going to be equal to is going to be equal to our slope negative 1/4 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 is 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 1/2 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 1/2 from both sides so I'm going to get Y is equal to negative 1/4 X and then I already am subtracting 1/4 and I subtract another 1/2 that's going to be negative 3/4 so -3 minus 3/4 so and that's actually pretty close to what I drew up here this should be intersecting the y axis at negative 3/4 so there you have it this this line or you could even say this this equation is going to be a very good linear approximation about as good as you can get for linear approximation for that nonlinear function around x equals negative one you might say well why didn't they just ask me to find the equation of the tangent line at x equals negative one 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 at to approximate this function around x equals negative one
AP® is a registered trademark of the College Board, which has not reviewed this resource.