Linear Approximation in One Variable Example 1


Uploaded by TheIntegralCALC on 13.04.2011

Transcript:
Linear Approximation in One Variable Example 1
Hi, everyone.
Welcome back to integralcalc.com.
Today we're going to be talking about how to find the linear approximation of a function.
The function that we're dealing with is f of x equals the quantity one plus x squared
and we're asked to find the linear approximation of this function close to the point a equals zero.
We’re going to be using this linear approximatuion formula here
where L of x is the linear approximation function.
and notice that we're evaluating at the point a equals zero
or finding the linear approximation at the point a equals zero.
And we've got a appearing on our linear approximation formula three times.
So what this tells us is that we're going to do a couple things.
First, we're going to find f of a so we'll plug a equals zero into our original function, f of x,
and that will give us this part of our formula.
Then we'll take the derivative of f of x to get f prime of x
and plug in a equals zero to the derivative to get this piece.
And then we'll plug in the actual value of a, which is zero,
for this value of a and simplify that entire formula to give us an equation for the linear approximation of this function.
So let's go ahead and get started.
Like I said, let's go ahead and take the derivative first.
So we're looking for f prime of x.
so the derivative of this function, we'll use chain rule.
And in doing so,
what we're going to do is call the inside function one plus x
and the outside function, this entire thing, the one plus x squared.
If you're having trouble with chain rule, please go check out that section on my website.
But for now we'll go through it a little bit quickly.
So rememeber that with chain rule,
we first take the derivative of the outside function,
leaving the inside function completely alone as it is,
then multiply by the derivative of the inside.
So the derivative of the outside function…
We treat the inside as just a variable. We ignore it.
So we're just looking at basically x squared here
and we'll pull this squared out in front. We’ll multiply it.
So what we'll get is we will get two times…
and then we leave the inside completely alone.
One plus x to the two minus one, right?
So that's the derivative of the outside function.
But then we have to multiply by the derivative of the inside fucntion.
The inside function is one plus x and the derivative of that is just one
so we would multiply this whole thing by one.
Now if we simplify that, obviously the one's going to go away.
We’ll get two minus one is just one so we get two times one plus x.
And we could leave it as is but I think it would be a little bit cleaner if we distribute the two.
So I'm going to multiply out the two and I'll get two plus two x.
that is our derivative function f prime of x.
So now that we have our derivative function,
we'll go ahead and plug a equals zero first into our original function
and then into our derivative function to get first this number and then this number.
So, first we'll take f of zero, plugging in a equals zero to the original function f of x.
So, first we'll take f of zero, plugging in a equals zero to the original function f of x.
So f of zero will give us one plus zero squared
which is going to be one squared, which is equal to one.
So we'll keep that in mind and now we'll plug in a equals zero to f prime of x, the derivative.
So f prime of zero is...
let's see..
oh, two plus two x.
so ignore that. Two plus two times zero.
Obviously, the two times zero goes away.
And we're just left with two.
So that's also going to get plugged into our formula.
Now we have all the information that we need to plug into our linear approximation formula.
So we'll get L of x equals...
first we're looking at...
let's do this in orange here. f of a.
So whatever we got when we plugged in a to our original functiuon f of x which was here.
so we get one plus f prime of a.
so whatever we got when we plugged in a equals zero to our derivative fiunction of f prime of x which was two.
So two times x minus a
and we already said in our original problem that a was equal to zero.
So x minus zero.
Now what we need to do is simplify this formula,
Now what we need to do is simplify this formula,
and we'll have our linear approximation formula.
So this is going to be one plus, two x minus zero is just x so one plus two x.
So that's it.
Our final answer for the linear approximation formula is L of x equals one plus two x.
So I hope that video helped you, guys
and I will see you in the next one.
Bye!