Question

How do you find the linearization function of `f(x) = sin(4x) + cos(x)`?

Answer 1

We have:

`f(x) = sin4x+cosx`

Here we will obtain an linear approximation (about `x=0`) using Maclaurin Series from First Principles.

The Maclaurin Series is define by the the infinite Power Series in ascending powers of `x`:

`f(x) = f(0) + f'(0)x/(1!) + f''(0)x^2/(2!) + f^((3))(0)x^3/(3!) + ...`

Differentiating wrt `x`' we get:

`f'(x) = 4cos4x-sinx`

Thus putting `x=0` we get:

`\ f(0) = 0+1 = 1`
`f'(0) = 4-0 = 4`

So the linear terms of the Maclaurin Series are:

`f(x) = 1 + 4x`

We can see this approximation compared with the function on this graph near `x=0`:
graph{(y-sin(4x)-cosx)(y-1-4x)=0 [-1, 1, -1, 2.5]}