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

##### 1 Answer

# T_3(x) = -11 + 12(x+1) -3(x+1)^2 #

# T_4(x) = -11 + 12(x+1) -3(x+1)^2 + (x+1)^3 #

#### Explanation:

Firstly note that as we have a polynomial of degree

Let use define the function,

# 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

# 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

# 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

# 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