Calculus
posted by Web10 on .
Find the degree 3 Taylor polynomial T3(x) of the function f(x)=(−3x+33)^4/3 at a=2.

f(x) = (3x+33)^4/3
at x=2, 3x+33 = 27
f(2) = 27^4/3 = 81
f'(x) = 4(3x+33)^1/3
f'(2) = 4*27^1/3 = 4*3 = 12
f''(x) = 4(3x+33)^2/3
f''(2) = 4*27^2/3 = 4/9
f^{(3)}(x) = 8(3x+33)^5/3
f^{(3)}(2) = 8/243
p(x) = f(2) + f'(2)/1! (x2)^1 + f''(2)/2! (x2)^2 + f^{(3)}(2)/3! (x2)^3 + ...
= 81  12(x2) + 2/9 (x2)^2 + 4/729 (x2)^3 + ...