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## Differential Calculus (2017 edition)

### Course: Differential Calculus (2017 edition) > Unit 11

Lesson 1: Linear approximation# Linear approximation of a rational function

AP.CALC:

CHA‑3 (EU)

, CHA‑3.F (LO)

, CHA‑3.F.1 (EK)

Sal finds a linear expression that approximates y=1/(x-1) around x=-1. This is done by finding the equation of the line tangent to the graph at x=-1, a process called "linear approximation.".

## Want to join the conversation?

- What's the benefit of using linear approximation in this specific function? It's as easy/hard to calculate any point of the graph with the original function as it is with the linear approximation (which is only less accurate). So why use it here?(15 votes)
- With this particular function, it's easy enough to plug in whole numbers, but it might be more difficult to plug in decimal numbers that don't have nice fraction equivalents, especially if you don't have a calculator.(18 votes)

- Could anyone explain (or tell me the video explaining) why (y - y_1) = m(x - x_1) ?

I'm only familiar with (y = mx + b) version and I can't even come up with keywords for search.(4 votes)- those two formulas are the two basic forms of a line

y=mx+b (standard)

y-y1=m(x-x1) (point-slope)

However, those two equations are equivalent, let's see.

( y - y1 ) = m ( x - x1 )

( y - y1 ) = mx - mx1

y = mx - mx1 + y1

y = mx + ( y1 - mx1 ) = mx + b

which means b= y1 - mx1 , this is the formula calculating y- intercept of the line at any point with the slope of the line.

And that's it.(22 votes)

- is this the sames as the taylor series?(6 votes)
- You are correct -- linear approximation is equivalent to using a first degree Taylor series.(7 votes)

- How do you calculate the Error and Percentage Error after finding the Linear Approximation? (if you could add a video on it, it would be greatly appreciated)(6 votes)
- %Error = (Approximate Value - Exact Value) / Exact Value(6 votes)

- At time stamp2:50, Sal is calculating the value of the linear approximation using the point slope formula in the form, (y-y1)/(x-x1)=b, and he points to b and calls it the slope. But I always thought that b was the y intercept. So b would be equal to: (y-y1) – m(x-x1)=b, and that would be the y intercept, not the slope. Anyway, I'm sure he wanted to calculate the slope here, so it seems to me that he should have written (y-y1) / (x-x1) = m, not b. Am I missing something?(6 votes)
- What grade math is this?(2 votes)
- The beginning of calculus is usually 12th grade math and/or early college math. Some advanced math students take calculus earlier than 12th grade.

Have a blessed, wonderful day!(7 votes)

- I'm still not sure what the difference between this and finding the equation of the tangent line is. Could someone try to help me understand this?(4 votes)
- They're the same thing. A linear approximation to a function at a point is the same as its tangent line at that point.(3 votes)

- At3:32Sal after applying the power rule says "then we're going to multiply that times the derivative of x minus 1 with respect to x."

I have been following all Calculus lectures and exercises so far and this step is a little confusing for me, could someone please help me?(3 votes)- So he is just multiplying 'that' with the derivative of x-1 (with respect to x) which is 1.

Hope this helps! If you have any questions or need help, please ask! :)(4 votes)

- What does it mean when you’re asked to use the tangent line at an x point to approximate a different x value? When you solve for it and get the y value, what does that y value represent?(3 votes)
- We are working with the understanding that the tangent line lies very close to the function in the region around the point of tangency. So when we solve for the y-value on the tangent line, we have the y-value of a point very close to the desired point on the function.(3 votes)

- How would you solve for the y-intercept algebraically?(1 vote)
- You can solve for the y-intercept algebraically by evaluating the function when x = 0(5 votes)

## Video transcript

- [Voiceover] So there are situations where you have some type of a function, this is 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 wanna approximate
it with a linear function especially around a certain value, and so what we're going to do is, we wanna find an approximation,
let me write this down, I wanna find an approximation for, actually meant to be clear, I wanna find a linear approximation so I'm gonna approximate it with a line. I wanna 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 one. 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 one, f of negative one is
negative, is negative 1/2, which sticks us right over
there, let me use a better color, so it's right over there
and what we wanna do is approximate it with a line around that. And what we're essentially going to do is we're gonna 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 one, the approximation gets worse and worse but if we stay around
x equals negative one, what'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 the linear approximation of f around x equals negative one or they say, what is the following
is the best approximation and all of your choices are, are lines, well essentially, they're
asking you to find the equation of the tangent
line at x equals negative one. 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 a 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 x one,
so x one comma y one sits on that line some place. Actually, I like to write
this point slope form like this sometimes, y minus
y one over x minus x one is equal to b because this comes as straight out of the idea. Look, if x one and y one 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, actually let me
just write f of x again, so I'm gonna write it as x minus one to the negative one power, that makes it a little bit clearer that
we can use the power rule and a little bit of a chain rule, so the derivative of f with
respect to x is equal to, so the derivative of x minus
one of the negative one with respect to x minus one,
well that's just going to be, I'm just gonna use the power rule here, it's gonna be negative
one times x minus one to the negative two and then
we're gonna multiply that times the derivative of x
minus one with respect to x, well that's just going to be one, right. The derivative of x with
respect of x is one, the derivative of negative
one with respect to x is zero, so we could say times one here if we like or we could just not write that 'cause it don't change the value. And so let's evaluate that when
x is equal to negative one. So f prime of negative one is equal to, I could just write this is, negative, all right like this way,
negative one over negative one minus one squared and so this
is going to be negative two down here, so this is equal
to negative, negative 1/4. So, the slope of our tangent line is, so I could write this way,
m is equal to negative one, negative 1/4 and so now,
we just have to write its equation down. So we already know, an x one and a y one that sits on the line,
in fact we want to use the point when x is equals negative one so we know that the
point negative one comma, we could just input it right over here, f of negative one is negative 1/2, one over negative one
minus one, negative 1/2. So we know that this negative
one 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, under right here, y minus y one, 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's
slope version of our equation, is equal to our slope times x minus x one, so x minus or x squared
that we know sits on this, so minus negative one. 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 gonna be plus
one 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 gonna get y is
equal to negative 1/4 x and then if I already am subtracting 1/4 and I subtract another half
is gonna be negative 3/4 so minus three, 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 equation, is going to be a very
good linear approximation, about as good as you can get
for a 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, we say, okay, I can
actually use the equation of the tangent line at to approximate this function around
x equals negative one.