Posted by **Web10** on Friday, December 2, 2011 at 5:53am.

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

- Calculus -
**Steve**, Friday, December 2, 2011 at 11:00am
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! (x-2)^1 + f''(2)/2! (x-2)^2 + f^{(3)}(2)/3! (x-2)^3 + ...

= 81 - 12(x-2) + 2/9 (x-2)^2 + 4/729 (x-2)^3 + ...

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