Step by step
The question
Question: estimate ∛26.9 with a linear approximation.
The function is f(x) = ∛x.
Near 26.9 there is a friendly number: a = 27.
27 is a perfect cube.
1 The point
Put 27 into f.
We get 3.
The point is (27, 3).
2 The slope there
The slope is the derivative: f′(x) = 1/(3 ∛(x²)).
At x = 27, the slope is 1/27.
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) = 3 + 1/27 × (x − 27).
4 Estimate
Put 26.9 into L.
Careful!
Use the distance from 27: 26.9 minus 27.
Not 26.9 itself.
26.9 minus 27 is −0.1.
1/27 times 0.1 is about 0.00370.
Add it up: 2.99629.
So ∛26.9 ≈ 2.99629.
5 Check
Let's check with a calculator: ∛26.9 is 2.99629 and a bit.
Our estimate 2.99629 is very close.
Right!
Let's look back.
First: the point.
Then: the slope there.
Then: the tangent line.
Then: estimate.
Last: check.