Step by step
The question
Question: find the linear approximation of f(x) = √x at x = 9.
f(x) = √x
1 The point
Put 9 into f.
We get 3.
The point is (9, 3).
f(9) = √9 = 3
2 The slope there
The slope is the derivative: f′(x) = 1/(2√x).
f′(x) = 12√x
At x = 9, the slope is 1/6.
f′(9) = 12√9 = 12 × 3 = 16
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) = 3 + 1/6 × (x − 9).
L(x) = 3 + 16(x − 9)
Careful!
The line goes through the point at x = 9, so we need x minus 9, not just x.
Careful: not L(x) = 3 + 1 over 6x ✗
4 Check
Let's check.
At x = 9, L gives 3, the same as f.
Good.
L(9) = 3 = f(9)
And a bit further, at 9.1: L gives 3.01666, and f gives about 3.0166.
Very close.
Right!
x = 9.1: L = 3.01666, f ≈ 3.0166
Let's look back.
First: the point.
Then: the slope there.
Then: the tangent line.
Last: check.
Answer
Careful: not L(x) = 3 + 1 over 6x ✗