Question

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

Answer 1

Answer:

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