Linear Approximations Using Differentials: Definition & Examples

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  • 0:03 Linear Approximation
  • 1:58 Differentials
  • 3:56 Lesson Summary
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Lesson Transcript
Instructor: Laura Pennington

Laura received her Master's degree in Pure Mathematics from Michigan State University. She has 15 years of experience teaching collegiate mathematics at various institutions.

This lesson will demonstrate how to use linear approximation with differentials to approximate function values near a given point. We will go over the steps and formulas involved in linear approximation.

Linear Approximation

Suppose you're at a carnival, and the announcer says that they are having a contest to see who can guess closest to the value of √(9.24). You know that √(9) = 3, so it must be that √(9.24) is a little over 3, but you want to be as close as possible to win the contest.

Good news! We've got a way of approximating this value without a calculator, and it's called using linear approximation. Linear approximation involves finding the equation of a line tangent to the function at a given value of x, and using it to approximate the function value for points nearby.

To put this into perspective, let's consider the function f(x) = √(x). If we could find a line that's close to the function at x = 9.24, we can use it to approximate f(9.24), or √(9.24), which is exactly what we want to do.


Notice that the tangent line of f(x) = √(x) at x = 9 has values that are very close to the function values of f(x) when x is near 9, so if we could find the equation of this line, we could use it to estimate the value of f(x) at 9.24, and find an approximate value for √(9.24).

In general, if we're finding a line to use in linear approximation that passes through a point (a, f(a)), we use the following formula:

L(x) = f(a) + m(x - a), where m is the slope of the line at x = a.

In this case, our line is tangent to f(x) = √(x), and passes through the point (9, f(9)) = (9, 3). Plugging these values in gives:

L(x) = f(9) + m(x - 9) = 3 + m(x - 9)

Wait a second, what about m? How do we find the slope at x = 9? This is where differentials come into play. Let's explore!


When finding the linear approximation for a function f(x) at the point (a, f(a)), we just saw that the general formula is L(x) = f(a) + m(x - a), where m is the slope of the line at x = a. To find the m, we use the fact that the derivative of a function at a point is equal to the slope of the tangent line at that point. Therefore, this general formula for the linear approximation can actually be rewritten as follows:

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


Great! So, all we have to do is find the derivative of f(x) and plug 9 into it, and we'll have all of the parts of our formula!

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