How do you determine the third and fourth Taylor polynomials of #x^3 + 9x - 1# at x = -1?

Firstly note that as we have a polynomial of degree #3# then its TS will be exact for #O(x^3)#.

Let use define the function, #f(x)#, by:

# f(x) = x^3+9x-1 #

Let us find all the derivatives:

# f^((1))(x) = 3x^2+9 #
# f^((2))(x) = 6x #
# f^((3))(x) = 6 #
# f^((4))(x) = 0 #, along with all higher derivatives:

Now let us find the values of the above at #a=-1#

# f^((0))(-1) = 11 #
# f^((1))(-1) = 12 #
# f^((2))(-1) = -6 #
# f^((3))(-1) = 6 #
# f^((4))(-1) = 0 #, along with all higher derivatives:

The the TS about #x=a# is given by:

# f(x) = f(a) + f^((1))(x)(x-a) + (f^((2))(x))/(2!)(x-a)^2 + (f^((3))(x))/(3!)(x-a)^3 + (f^((4))(x))/(4!)(x-a)^4 + ... #

So, if we want to write the truncated TS, we can just truncate the series as required. Thus the #3^(rd)# order TS about #a=-1# is:

# T_3(x) = (-11) + (12)(x+1) + (-6)/(2!)(x+1)^2 #
# " " = -11 + 12(x+1) -6/(2)(x+1)^2 #
# " " = -11 + 12(x+1) -3(x+1)^2 #

And similarly, truncating at the next term we have:

# T_4(x) = T_3(x) + (6)/(3!)(x+1)^3 #
# " " = T_3(x) + 6/(6)(x+1)^3 #
# " " = -11 + 12(x+1) -3(x+1)^2 + (x+1)^3 #

And as initially, as the starting function is a polynomial of degree #3# then this truncated polynomial #T_4(x)# is in fact exact as all higher derivatives (and therefore terms) are zero.