Step by step
The question
Question: estimate ∛8.3 with a linear approximation.
3√8.3 ≈ ?
The function is f(x) = ∛x.
Near 8.3 there is a friendly number: a = 8.
8 is a perfect cube.
f(x) = 3√x
a = 8
1 The point
Put 8 into f.
We get 2.
The point is (8, 2).
f(8) = 3√8 = 2
2 The slope there
The slope is the derivative: f′(x) = 1/(3 ∛(x²)).
f′(x) = 133√x2
At x = 8, the slope is 1/12.
f′(8) = 133√82 = 13 × 4 = 112
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.
L(x) = f(a) + f′(a)(x − a)
So L(x) = 2 + 1/12 × (x − 8).
L(x) = 2 + 112(x − 8)
4 Estimate
Put 8.3 into L.
L(8.3) = 2 + 112(8.3 − 8)
Careful!
Use the distance from 8: 8.3 minus 8.
Not 8.3 itself.
Careful: not 2 + 112(8.3) ✗
8.3 minus 8 is 0.3.
= 2 + 112 × 0.3
1/12 times 0.3 is 0.025.
= 2 + 0.025
Add it up: 2.025.
= 2.025
So ∛8.3 ≈ 2.025.
3√8.3 ≈ 2.025
5 Check
Let's check with a calculator: ∛8.3 is 2.02469 and a bit.
Our estimate 2.025 is very close.
Right!
calculator: 3√8.3 = 2.02469…
Let's look back.
First: the point.
Then: the slope there.
Then: the tangent line.
Then: estimate.
Last: check.
Answer
3√8.3 ≈ 2.025