1. The problem statement, all variables and given/known data the question asks to find the quadratic maclaurin polynomial for f(x) Given f(x) = x sin(x) 3. The attempt at a solution i know that a maclaurin series is when a=0 in a taylor series. i did the 1st-5th derivatives of f(x) and then used the formula for taylor polynomial and set a=0. f'=sin(x) + x cos(x) f''=2cos(x) -x sin(x) f'''=-3sin(x) -xcos(x) f''''=-4cos(x) +xsin(x) f'''''=5sin(x) +xcos(x) with the formula for taylor polynomial Pn(x) = n Ʃ [f^k (a) (x-a)^k] / k! k=0 by setting a=0 i got Pn(x)=0+0+x^2 -0- (x^4)/6............ however the answer seems to be P(x)=x^2 anyone able to explain to me how that is so? i think i got it... is the quadratic maclaurin polynomial just the 4th order maclaurin polynomial??