Step by step
The question
Question: find the linear approximation of f(x) = ∛x at x = 8.
f(x) = 3√x
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)
Careful!
The line goes through the point at x = 8, so we need x minus 8, not just x.
Careful: not L(x) = 2 + 1 over 12x ✗
4 Check
Let's check.
At x = 8, L gives 2, the same as f.
Good.
L(8) = 2 = f(8)
And a bit further, at 8.1: L gives 2.00833, and f gives about 2.0083.
Very close.
Right!
x = 8.1: L = 2.00833, f ≈ 2.0083
Let's look back.
First: the point.
Then: the slope there.
Then: the tangent line.
Last: check.
Answer
Careful: not L(x) = 2 + 1 over 12x ✗