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

Step by step

The question

Question: find the linear approximation of f(x) = x² at x = 3.

1 The point

Put 3 into f.

We get 9.

The point is (3, 9).

2 The slope there

The slope is the derivative: f′(x) = 2x.

At x = 3, the slope is 6.

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) = 9 + 6(x − 3).

Careful!

The line goes through the point at x = 3, so we need x minus 3, not just x.

4 Check

Let's check.

At x = 3, L gives 9, the same as f.

Good.

And a bit further, at 3.1: L gives 9.6, and f gives about 9.61.

Very close.

Right!

Let's look back.

First: the point.

Then: the slope there.

Then: the tangent line.

Last: check.