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Linear Approximations Using Differentials: Definition & Examples

Instructor: Laura Pennington

Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.

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 are 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 is 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!


linapp3


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 are 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!

Linear Approximation and Differentials

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)


linapp4


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!

To find the derivative of f(x) = √(x), let's rewrite it as f(x) = x(1/2). Now we can use the following rule:

  • The derivative of xn is nx(n-1)

Thus, we have the following:

  • f ' (x) = (1/2)x(-1/2) and f ' (9) = (1/2)(9)(-1/2) = (1/2)(1/3) = 1/6

All together, we have the following:

L(x) = f(9) + f ' (9)(x - 9)

= 3 + (1/6)(x - 9)

= 3 + (1/6)x - (9/6)

= 1.5 + (1/6)x

  • L(x) = 1.5 + (1/6)x.

All we have to do is plug 9.24 into this formula for x and we have our approximation!

  • L(9.24) = 1.5 + 1/6(9.24) = 3.04

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