Question

What is the linear approximation of a function?

Answer 1

The linear approximation of a function f(x) is the linear function L(x) that looks the most like f(x) at a particular point on the graph y = f(x). This depends on what point (a, f(a)) you want to focus in on. Spoiler Alert: It's the tangent line at that point! The tangent line matches the value of f(x) at x=a, and also the direction at that point.

Let's find the equation of the tangent line, then. The limit definition of the derivative at a point tells us that the difference quotient `(f(x)-f(a))/(x-a)` is about equal to `f'(a).`

So `f(x)-f(a) = f'(a)(x-a)`,

meaning that the equation of the tangent line at x=a is

`L(x) = f(a) + f'(a)(x-a).`

That's your linear approximation, brought to you by calculus \and dansmath!/

P.S. This is the first two terms of the Taylor Series; with more terms you can approximate a function at any point with higher degree polynomials!

One of the reasons to use linear approximations is that we can estimate an answer without the use of a calculator because it is usually a simple multiplication/division and addition. The other thing to note is the approximation is better the closer you are to `a`. A typical textbook example is estimating square roots.

Let's look at estimating `sqrt(4.01)`:

`f(x)=sqrt(x)`
`f'(x)=1/(2sqrt(x))`

We want to choose an `f(a)` that is easy to compute, that would be choosing a perfect square; in this case `a=4`:

`L(x)=f(4)+f'(4)(x-4)`
`=2+1/4(x-4)`

So, we evaluate for the desired value:

`L(4.01)=2+1/4(4.01-4)`
`=2+.01/4`
`=2.0025` done without a calculator (mathamagics)!

This is a very good estimate because `sqrt(4.01)=2.002498439`