Step by step
The question
Question: find the linear approximation of f(x) = ln x at x = 1.
f(x) = ln x
1 The point
Put 1 into f.
We get 0.
The point is (1, 0).
f(1) = ln 1 = 0
2 The slope there
The slope is the derivative: f′(x) = 1 over x.
f′(x) = 1x
At x = 1, the slope is 1.
f′(1) = 11 = 1
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) = x − 1.
L(x) = x − 1
Let's see it.
The curve.
And the tangent line at x = 1.
Near 1, they are almost the same.
Careful!
The line goes through the point at x = 1, so we need x minus 1, not just x.
Careful: not L(x) = 0 + 1x ✗
4 Check
Let's check.
At x = 1, L gives 0, the same as f.
Good.
L(1) = 0 = f(1)
And a bit further, at 1.1: L gives 0.1, and f gives about 0.0953.
Very close.
Right!
x = 1.1: L = 0.1, f ≈ 0.0953
Let's look back.
First: the point.
Then: the slope there.
Then: the tangent line.
Last: check.
Answer
Careful: not L(x) = 0 + 1x ✗