How do I find linearization for this problem

For the function f(x) = xcos(x)
a. Find the linearization L(x) of f(x) at x = 0 and use it to approximate f(0.05)
b. use differentials to approximate f(0.05)

Thank you so much

You need to use the Taylor series expansion.

cos x = 1 - x^2/2! + x^4/4! - x^6/6! + ... (! means factorial)
so
xcos x = x - x^3/2! + x^5/4! - x^7/6! + ...

a. Therefore L(x) ~ x and f(0.05) = 0.05
b. So df/dx = cos(x) - xsin(x)
at x = 0, df/dx = 1
dy = f'(x) dx = 1 * 0.05
So f(0.05) = f(0) + f'(x) dx = 0.05

f(x) = x cos(x)
f(0) = 0
f ' (x) = - x sin(x) + cos(x)
f ' (0) = 0 + 1 = 1

L(x) = f(0) + f ' (0)(x - 0)

=> L(x) = 0 + 1(x)

=> L(x) = x

f(0.05) = L(0.05) = 0.05

dy/dx = - x sin(x) + cos(x)

∆y = [ - x sin(x) + cos(x) ]∆x
use x = 0 and ∆x = 0.05
∆y = [ 0 + 1 ]0.05

= 0.05