as always, it is
g(5)/0! + g'(5)(x-5)/1! + g"(5)(x-5)^2/2! + g"'(5)(x-5)^3/3! + ...
117 + 73(x-5) + 15(x-5)^2 + (x-5)^3 + ...
but all the higher derivatives are zero, so that's all you get.
What is the Taylor series generated by g(x) = x^3-2x+2 at a=5?
2 answers
f(x) = x³ -2x+2----> f(5) = 117
f'(x) = 3x² -2-------> f'(5) =73
f''(x) = 6x --------------> f''(5) = 30
f'''(x) = 6 --------------------> f'''(1) = 6
f(x) = 117/0! + 73/1! (x−5) + 30/2! (x−5)² + 6/3! (x−5)³
f(x) = 117+ 73(x−5) + 15(x−5)² + (x−5)³
f(x) = 117 + 73x−365 + 15x²−150x +375 + x³ -15x²+75x-125
f(x) = x³ -2x +2 which is the same as the original function!
f'(x) = 3x² -2-------> f'(5) =73
f''(x) = 6x --------------> f''(5) = 30
f'''(x) = 6 --------------------> f'''(1) = 6
f(x) = 117/0! + 73/1! (x−5) + 30/2! (x−5)² + 6/3! (x−5)³
f(x) = 117+ 73(x−5) + 15(x−5)² + (x−5)³
f(x) = 117 + 73x−365 + 15x²−150x +375 + x³ -15x²+75x-125
f(x) = x³ -2x +2 which is the same as the original function!