Linear approximation of f(x) = x^3 at x = 2

Step by step

The question

Question: find the linear approximation of f(x) = x³ at x = 2.

1 The point

Put 2 into f.

We get 8.

The point is (2, 8).

2 The slope there

The slope is the derivative: f′(x) = 3x².

At x = 2, the slope is 12.

3 The tangent line

The tangent line: start at the height f of a, and go up by the slope times the distance from a.

So L(x) = 8 + 12(x − 2).

Let's see it.

The curve.

And the tangent line at x = 2.

Near 2, they are almost the same.

Careful!

The line goes through the point at x = 2, so we need x minus 2, not just x.

4 Check

Let's check.

At x = 2, L gives 8, the same as f.

Good.

And a bit further, at 2.1: L gives 9.2, and f gives about 9.261.

Very close.

Right!

Let's look back.

First: the point.

Then: the slope there.

Then: the tangent line.

Last: check.