Step by step
The question
Question: estimate 2.01³ with a linear approximation.
The function is f(x) = x³.
Near 2.01 there is a friendly number: a = 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.
4 Estimate
Put 2.01 into L.
Careful!
Use the distance from 2: 2.01 minus 2.
Not 2.01 itself.
2.01 minus 2 is 0.01.
12 times 0.01 is 0.12.
Add it up: 8.12.
So 2.01³ ≈ 8.12.
5 Check
Let's check with a calculator: 2.01³ is 8.1206 and a bit.
Our estimate 8.12 is very close.
Right!
Let's look back.
First: the point.
Then: the slope there.
Then: the tangent line.
Then: estimate.
Last: check.