Step by step
The question
Question: find the linear approximation of f(x) = eˣ at x = 0.
f(x) = ex
1 The point
Put 0 into f.
We get 1.
The point is (0, 1).
f(0) = e0 = 1
2 The slope there
The slope is the derivative: f′(x) = eˣ.
f′(x) = ex
At x = 0, the slope is 1.
f′(0) = e0 = 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) = 1 + x.
L(x) = 1 + x
Let's see it.
The curve.
And the tangent line at x = 0.
Near 0, they are almost the same.
Careful!
The line goes through the point at x = 0, so we need x minus 0, not just x.
Careful: not L(x) = 1 + 1x ✗
4 Check
Let's check.
At x = 0, L gives 1, the same as f.
Good.
L(0) = 1 = f(0)
And a bit further, at 0.1: L gives 1.1, and f gives about 1.1052.
Very close.
Right!
x = 0.1: L = 1.1, f ≈ 1.1052
Let's look back.
First: the point.
Then: the slope there.
Then: the tangent line.
Last: check.
Answer
Careful: not L(x) = 1 + 1x ✗