What is linear approximation?

0:00 // What is linear approximation?
0:44 // When do you use linear approximation?
1:28 // Estimating square roots using linear approximation
5:23 // Estimating trig functions using linear approximation
6:37 // How to find the error in a linear approximation
7:48 // Summary

Linear approximation, or linearization, is a method we can use to approximate the value of a function at a particular point. The reason liner approximation is useful is because it can be difficult to find the value of a function at a particular point.

Square roots are a great example of this. We know the value of sqrt(9); it’s 3. That’s easy to figure out. But we don’t know the value of sqrt(9.2). We can guess that it’s a little bit more than 3, since we know that sqrt(9) is 3, and 9.2 is a little bit more than 9, but other than that, we don’t know how to find a better estimate of sqrt(9.2). That’s where linear approximation comes in to help us.

Since we’re dealing with square roots, if we imagine the graph of the function sqrt(x), we know one point on that function is (9,3). If we find the tangent line to the function sqrt(x) through the point (9,3), then we can see that, since the tangent line is really close to the graph of the function around the area of (9,3), that the value of the function and the value of the tangent line will be pretty close to each other at x=9.2.

So to get an estimate for sqrt(9.2), we’ll use linear approximation to find the equation of the tangent line through (9,3), and then plug x=9.2 into the equation of the tangent line, and the result will be the value of the tangent line at x=9.2, and very close to the value of the function at x=9.2.

That’s why linear approximation is so helpful to us, because it’s a quick, simple method that let’s us estimate a value that would otherwise be very difficult to find.

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Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)

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Phenomenal explanation! Clear as day. I just watched Khan Academy and was very confused. I came here and within 10 minutes I understood it completely - and realized it's actually just simple common sense - as long as you've got a value really close, which isn't many.

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[robwhitlock]

Delta y, as commonly defined, doesn't give the error in the estimation, it gives the difference between the y values of two points on a function, f(x_2)-f(x_1). The error, on the other hand, would be the difference in y coordinates of two points, one on the function and one on the tangent line, but both with the x coordinate that you are trying to estimate the value of the function at.

When you used f(x_2)-f(x_1) = f'(x_1)(x_2-x_1) at 6:47, the form of your equation requires that both (x_1, f(x_1)) and (x_2, f(x_2)) lie on the tangent line of f(x) that touches (x_1, f(x_1)). This is not true for all differentiable functions, and it's certainly not true for the square root function. Note that in your previous equation you had an approximate equality, which was fine, but at 6:47 you changed it to equality, which made it incorrect for most differentiable functions. Finding the value of f'(x_1)(x_2-x_1) doesn't tell you about the error in your estimation, it just tells you how much bigger or smaller your estimate is as compared to where you started, f(x_1).

To find the error in your estimation you want to take the difference between your estimate and the true value:

true value = f(x_2)
estimated value = f(x_1) + f'(x_1)(x_2-x_1)
error = (estimated value) - (true value) = f(x_1) + f'(x_1)(x_2-x_1) - f(x_2)

Thanks for working to make high quality, easy to understand math videos! :)

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