Question

How do you find the linearization of `f(x)=lnx` at x=8?

Answer 1

From simple geometry, plus some complicated things to do with general smoothness of the curve about `x = 8`, we can say for "small" `epsilon` that:

`f(8 + epsilon) approx f(8) + epsilon f'(8)`

And because `f(x) = ln x` then `f'(x) = 1/x`; and we can say that:

`f(8 + epsilon) approx ln 8 + epsilon 1/8`

We can test this in a calculator for `f(8.1)`

Actual Value: `ln 8.1 = 2.0919`

From Linearisation: `ln 8 + 0.1 * 1/8 = 2.0919`

!!

For `f(8.9)`

Actual Value: `ln 8.9 = 2.186`

From Linearisation: `ln 8 + 0.9 * 1/8 = 2.192`

:(

For `f(20)`

Actual Value: `ln 20 approx 3`

From Linearisation: `ln 8 + 12 * 1/8 approx 3.6`

:((