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November 27, 2014

November 27, 2014

Posted by **COFFEE** on Saturday, July 28, 2007 at 1:02pm.

f(x) = x3, a = -1

and what i've done so far:

f (x) = x^3

f ' (x) = 3x^2

f '' (x) = 6x^1

f ''' (x) = 6x

f (-1) = -1

f ' (-1) = 3

f '' (-1) = -6

f ''' (-1) = -6

using taylor series equation.. my final answer that was wrong:

((-1(x+1)^0)/(0!))+((3(x+1)^1)/(1!))+((-6(x+1)^2)/(2!))+((-6(x+1)^3)/(3!))

.. is this what the question was asking for? if not, what is it then? thank you very much for your assistance.

f ''' (x) = 6

f'''(-1) = 6

- Calculus - Taylor #2 -
**Steve**, Tuesday, September 24, 2013 at 4:24amLooks good, assuming your arithmetic is ok.

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