What is the linear approximation of a function?

Aug 17, 2014

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 $\frac{f \left(x\right) - f \left(a\right)}{x - a}$ is about equal to $f ' \left(a\right) .$

So $f \left(x\right) - f \left(a\right) = f ' \left(a\right) \left(x - a\right)$,

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

$L \left(x\right) = f \left(a\right) + f ' \left(a\right) \left(x - a\right) .$

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 $\sqrt{4.01}$:

$f \left(x\right) = \sqrt{x}$
$f ' \left(x\right) = \frac{1}{2 \sqrt{x}}$

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

$L \left(x\right) = f \left(4\right) + f ' \left(4\right) \left(x - 4\right)$
$= 2 + \frac{1}{4} \left(x - 4\right)$

So, we evaluate for the desired value:

$L \left(4.01\right) = 2 + \frac{1}{4} \left(4.01 - 4\right)$
$= 2 + \frac{.01}{4}$
$= 2.0025$ done without a calculator (mathamagics)!

This is a very good estimate because $\sqrt{4.01} = 2.002498439$ 