Finding the Linear Estimate of a Function: Help Needed

[funktion]

Hey guys I was just taking a look at a sample exam and I came across this, with no recollection of ever learning it.

I don't really want anyone to solve a problem for me per se, I just want an explanation for what is being asked.

So anyway, I am asked to find the linear estimate for a function f(x) for all small values of x (close to 0).

I'm not quite sure what a linear estimate is. Is it some sort of application of the linearization formula?

L(x) = F(a) = F'(a)(x-a)

Help would be appreciated. Thanks.

 

[cepheid]

I'm not quite sure what a linear estimate is. Is it some sort of application of the linearization formula?

L(x) = F(a) = F'(a)(x-a)

Help would be appreciated. Thanks.

Yeah, exactly. Except that you mistyped your formula. It should be:

L(x) = F(a) + F'(a)(x - a)

In this problem, a = 0

This is just a Taylor series expansion (about 0) in which you only keep the terms up to the first order term (that's called a first order Taylor expansion for short).

Think about it...some well known functions do look approximately linear for values of x close to zero. Example:

f(x) = sin(x)

You can see that this is the case if you take a first order Taylor expansion, but you can also just look at a plot of sin(x) and see that it sure looks that way. As a result, for small values of x:

sin(x) ~ x

 

[HallsofIvy]

This is just a Taylor series expansion (about 0) in which you only keep the terms up to the first order term (that's called a first order Taylor expansion for short).

Or, equivalently, use the tangent line approximation.

 

[Gib Z]

Basically you could think of this approximation as replacing a curve with its tangent. That is quite accurate close to the point of tangency, because at that point it has the same value and gradient, sort of like heading in a similar direction to the curve and therefore still somewhat accurate near the point of tangency.