Find Taylor's series of the function f(x) = 3 x^3-8 x^2-5x+8?

Question

i have to find it from C0 to C5
can someone please give me a step by step soultion to this problem?/
thanks

Answer 1

The Taylor Series of an equation is:
∞
Σ (f(n)(a)/n!) * (x - a)^n
n=0

where
f(n)(a) is the nth derivative of f(x) at point a.
We'll use a = 0, making it a Maclaurin series, of the form:

n = 0:
f(0) = 3x^3 - 8x^2 - 5x + 8 = 8
8/0! * x^0 = 8

n = 1
f'(0) = 9x^2 - 16x - 5 = -5
-5/1! * x^1 = -5x

n = 2
f''(0) = 18x - 16 = -16
-16/2! * x^2 = -8x^2

n = 3
f'''(0) = 18 = 18
18/3! * x^3 = 3x^3

n = 4
f''''(0) = 0 = 0
0/4! * x^4 = 0

So, the Taylor Series for a = 0 = 3x^3 - 8x^2 - 5x + 8