Find the Taylor series for f (x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that Rn (x) → 0.] f (x) = 4 x, a = - 2

4x

Step-by-step explanation:

The Taylor series of a function f (x) about a value x = a is given by f (x) = f (a) + f' (a) (x - a) / 1! + f'' (a) (x - a) ²/2! + f''' (a) (x - a) ³/3! + ... where the terms in f prime f' (a) represent the derivatives of x valued at a.

For the given function, f (x) = 4x and a = - 2

So, f (a) = f (-2) = 4 (-2) = - 8

f' (a) = f' (-2) = 4

All the higher derivatives of f (x) evaluated at a are equal to zero. That is f'' (a) = f'" (a) = ... = 0

Substituting the values of a = - 2, f (a) = f (-2) = - 8 and f' (-2) = 4 into the Taylor series, we have

f (x) = f (-2) + f' (-2) (x - (-2)) / 1! + f'' (-2) (x - (-2)) ²/2! + f''' (-2) (x - (-2)) ³/3! + ...

= - 8 + 4 (x + 2) / 1! + (0) (x + 2) ²/2! + (0) (x + 2) ³/3! + ...

= - 8 + 4 (x + 2) + 0 + 0

= - 8 + 4x + 8

= 4x