Find the degree of Taylor Polynomial

Question

Find the degree 3 Taylor polynomial T_3(x) of function f(x) = (−3x+26)^(4/3) at a=6.

T_3(x) = 

Answer 1

Here to obtained the Taylor series of order 3 we have Taylor polynomial

T_n(x)=∑_(n=0)^∞(f^(n)(a))((x-a)^n)/(n!).

Here given a=6 and the function f(x)=(26-3x)^(4/3).

The function value is f(6)=(26-18)^(4/3)=16

The derivatives gives

f'(x)=4/3(26-3x)^(4/3-1)(-3)=-4(26-3x)^(1/3)

=>f'(6)=(-4)(8)^(1/3)=-8

f''(x)=(-4/3)(26-3x)^(1/3-1)(-3)=4(26-x)^(-2/3)

=>f"(6)=4(8^(-2/3))=1

f"'(x)=(-8/3)(26-x)^(-5/3)(-3)

=>f"'(6)=(8)(8^(-5/3))=(8)(1/32)=1/4

So now the series is

T_3(x)=16+(-8)(x-6)+(1)(1/2)(x-6)^2+(1/4)(1/6)(x-6)^3

=16-8(x-6)+(1/2)(x-6)^2+(1/24)(x-6)^3

Answer: 16-8(x-6)+(1/2)(x-6)^2+(1/24)(x-6)^3