By taylor series,

$f(x)= f(a)+ (x-a)f'(a)+ \dfrac {(x-a)^2}{2!} f''(a)+ \dfrac {(x-a)^3}{3!}f'''(a)+ \cdots \ \cdots $

$Here , \ a=3, $

$f(x)= x^4-3x^3+ 2x^2 -x+1 $

$f(3) = 3^4 -3 (3^3)+ 2(3^2)-3+1 = 18 -2 =16 $

$ f'(x)=4x^3 -9x^2 +4x-1 $

$f'(3) = 4(3^3) - 9 (3^2)+12-1=38 $

$ f''(x)=12 x^2 - 18x+4 $

$f''(3)=12(3^2)-54+4=58 $

$f''(x)=24x-18 $

$f'''(3)=24 \times 3 -18 =54 $

$f^{iv} (x)=24 $

$f^{3}=24$

$f(x)= f(3)+ (x-3)f'(3)+ \dfrac {(x-3)^2}{2!}f''(3)+ \dfrac {(x-3)^3}{3!} f'''(3)+ \dfrac {(x-3)^4}{4!}f^{iv}(3)$

$f(x)= 16 +38 (x-3) + \dfrac {(x-3)^2}{2} 58+ \dfrac {(x-3)^3}{6}54 + \dfrac {(x-3)^4}{24} 24 $

$f(x)=16+38 (x-3)+ 29 (x-3)^2+ 9 (x-3)^3 +(x-3)^4$