The 4th term in this series is f"'(2)(x-2)3/3! so f"'(2)/6 = 9/32 and f"'(2) = 27/16
Jack S.
asked • 03/08/21Please Help I don't understand
The third-degree Taylor polynomial for a function f about X=2 is
1+(1/3)(x-2)+(1/2)(x-2)^2+(9/32)(x-2)^3
What is the value of f'''(2)
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Jack S.
Thank you for the help but I'm sure I understand. Where does the F'''(2)(x-2)^3/3! come from03/08/21