Use a linear approximation to estimate 1/1.02 Step by Step

Step by step

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

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.

Answer