Step by step
The question
Question: estimate √25.3 with a linear approximation.
√25.3 ≈ ?
The function is f(x) = √x.
Near 25.3 there is a friendly number: a = 25.
25 is a perfect square.
f(x) = √x
a = 25
1 The point
Put 25 into f.
We get 5.
The point is (25, 5).
f(25) = √25 = 5
2 The slope there
The slope is the derivative: f′(x) = 1/(2√x).
f′(x) = 12√x
At x = 25, the slope is 1/10.
f′(25) = 12√25 = 12 × 5 = 110
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) = 5 + 1/10 × (x − 25).
L(x) = 5 + 110(x − 25)
4 Estimate
Put 25.3 into L.
L(25.3) = 5 + 110(25.3 − 25)
Careful!
Use the distance from 25: 25.3 minus 25.
Not 25.3 itself.
Careful: not 5 + 110(25.3) ✗
25.3 minus 25 is 0.3.
= 5 + 110 × 0.3
1/10 times 0.3 is 0.03.
= 5 + 0.03
Add it up: 5.03.
= 5.03
So √25.3 ≈ 5.03.
√25.3 ≈ 5.03
5 Check
Let's check with a calculator: √25.3 is 5.02991 and a bit.
Our estimate 5.03 is very close.
Right!
calculator: √25.3 = 5.02991…
Let's look back.
First: the point.
Then: the slope there.
Then: the tangent line.
Then: estimate.
Last: check.
Answer
√25.3 ≈ 5.03