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
