Question
Find the degree 3 Taylor polynomial T3(x) of the function f(x)=(−3x+33)^4/3 at a=2.
Answers
Steve
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<sup>(3)</sup>(x) = 8(-3x+33)^-5/3
f<sup>(3)</sup>(2) = 8/243
p(x) = f(2) + f'(2)/1! (x-2)^1 + f''(2)/2! (x-2)^2 + f<sup>(3)</sup>(2)/3! (x-2)^3 + ...
= 81 - 12(x-2) + 2/9 (x-2)^2 + 4/729 (x-2)^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<sup>(3)</sup>(x) = 8(-3x+33)^-5/3
f<sup>(3)</sup>(2) = 8/243
p(x) = f(2) + f'(2)/1! (x-2)^1 + f''(2)/2! (x-2)^2 + f<sup>(3)</sup>(2)/3! (x-2)^3 + ...
= 81 - 12(x-2) + 2/9 (x-2)^2 + 4/729 (x-2)^3 + ...