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

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Answer 1 · 2017-07-26T10:37:48

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]}

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