Worked examples
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 ✗
▶ Watch it solved
The question
Question: estimate √4.1 with a linear approximation.
√4.1 ≈ ?
The function is f(x) = √x.
Near 4.1 there is a friendly number: a = 4.
4 is a perfect square.
f(x) = √x
a = 4
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.
4 Estimate
Put 4.1 into L.
L(4.1) = 2 + 14(4.1 − 4)
Careful!
Use the distance from 4: 4.1 minus 4.
Not 4.1 itself.
Careful: not 2 + 14(4.1) ✗
4.1 minus 4 is 0.1.
= 2 + 14 × 0.1
1/4 times 0.1 is 0.025.
= 2 + 0.025
Add it up: 2.025.
= 2.025
So √4.1 ≈ 2.025.
√4.1 ≈ 2.025
5 Check
Let's check with a calculator: √4.1 is 2.02485 and a bit.
Our estimate 2.025 is very close.
Right!
calculator: √4.1 = 2.02485…
Let's look back.
First: the point.
Then: the slope there.
Then: the tangent line.
Then: estimate.
Last: check.
▶ Watch it solved
The question
Question: estimate 1/1.02 with a linear approximation.
11.02 ≈ ?
The function is f(x) = 1 over x.
Near 1.02 there is a friendly number: a = 1.
f(x) = 1x
a = 1
1 The point
Put 1 into f.
We get 1.
The point is (1, 1).
f(1) = 11 = 1
2 The slope there
The slope is the derivative: f′(x) = −1/x².
f′(x) = −1x2
At x = 1, the slope is −1.
f′(1) = −112 = −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 − 1).
L(x) = 1 − (x − 1)
Let's see it.
The curve.
And the tangent line at x = 1.
Near 1, they are almost the same.
4 Estimate
Put 1.02 into L.
L(1.02) = 1 − (1.02 − 1)
Careful!
Use the distance from 1: 1.02 minus 1.
Not 1.02 itself.
Careful: not 1 − (1.02) ✗
1.02 minus 1 is 0.02.
= 1 − 0.02
= 1 − 0.02
Add it up: 0.98.
= 0.98
So 1/1.02 ≈ 0.98.
11.02 ≈ 0.98
5 Check
Let's check with a calculator: 1/1.02 is 0.98039 and a bit.
Our estimate 0.98 is very close.
Right!
calculator: 11.02 = 0.98039…
Let's look back.
First: the point.
Then: the slope there.
Then: the tangent line.
Then: estimate.
Last: check.
▶ Watch it solved
