Step by step
The question
Question: find the linear approximation of f(x) = √x at x = 4.
f(x) = √x
1 The point
Put 4 into f.
We get 2.
The point is (4, 2).
f(4) = √4 = 2
2 The slope there
The slope is the derivative: f′(x) = 1/(2√x).
f′(x) = 12√x
At x = 4, the slope is 1/4.
f′(4) = 12√4 = 12 × 2 = 14
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 + 1/4 × (x − 4).
L(x) = 2 + 14(x − 4)
Let's see it.
The curve.
And the tangent line at x = 4.
Near 4, they are almost the same.
Careful!
The line goes through the point at x = 4, so we need x minus 4, not just x.
Careful: not L(x) = 2 + 1 over 4x ✗
4 Check
Let's check.
At x = 4, L gives 2, the same as f.
Good.
L(4) = 2 = f(4)
And a bit further, at 4.1: L gives 2.025, and f gives about 2.0248.
Very close.
Right!
x = 4.1: L = 2.025, f ≈ 2.0248
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 4x ✗