Step by step
The question
Question: estimate √15.8 with a linear approximation.
√15.8 ≈ ?
The function is f(x) = √x.
Near 15.8 there is a friendly number: a = 16.
16 is a perfect square.
f(x) = √x
a = 16
1 The point
Put 16 into f.
We get 4.
The point is (16, 4).
f(16) = √16 = 4
2 The slope there
The slope is the derivative: f′(x) = 1/(2√x).
f′(x) = 12√x
At x = 16, the slope is 1/8.
f′(16) = 12√16 = 12 × 4 = 18
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) = 4 + 1/8 × (x − 16).
L(x) = 4 + 18(x − 16)
4 Estimate
Put 15.8 into L.
L(15.8) = 4 + 18(15.8 − 16)
Careful!
Use the distance from 16: 15.8 minus 16.
Not 15.8 itself.
Careful: not 4 + 18(15.8) ✗
15.8 minus 16 is −0.2.
= 4 − 18 × 0.2
1/8 times 0.2 is 0.025.
= 4 − 0.025
Add it up: 3.975.
= 3.975
So √15.8 ≈ 3.975.
√15.8 ≈ 3.975
5 Check
Let's check with a calculator: √15.8 is 3.97492 and a bit.
Our estimate 3.975 is very close.
Right!
calculator: √15.8 = 3.97492…
Let's look back.
First: the point.
Then: the slope there.
Then: the tangent line.
Then: estimate.
Last: check.
Answer
√15.8 ≈ 3.975