Step by step
The question
Question: find the linear approximation of f(x) = 1/x at x = 2.
f(x) = 1x
1 The point
Put 2 into f.
We get 1/2.
The point is (2, 1/2).
f(2) = 12
2 The slope there
The slope is the derivative: f′(x) = −1/x².
f′(x) = −1x2
At x = 2, the slope is −1/4.
f′(2) = −122 = −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) = 1/2 − 1/4 × (x − 2).
L(x) = 12 − 14(x − 2)
Let's see it.
The curve.
And the tangent line at x = 2.
Near 2, they are almost the same.
Careful!
The line goes through the point at x = 2, so we need x minus 2, not just x.
Careful: not L(x) = 1 over 2 + −1 over 4x ✗
4 Check
Let's check.
At x = 2, L gives 1/2, the same as f.
Good.
L(2) = 1 over 2 = f(2)
And a bit further, at 2.1: L gives 0.475, and f gives about 0.4762.
Very close.
Right!
x = 2.1: L = 0.475, f ≈ 0.4762
Let's look back.
First: the point.
Then: the slope there.
Then: the tangent line.
Last: check.
Answer
Careful: not L(x) = 1 over 2 + −1 over 4x ✗