The third-degree Taylor polynomial for f ( x ) {\displaystyle f(x)} , expanded about a = 3 {\displaystyle a=3} is
T 3 ( x ) = f ( 3 ) + f ′ ( 3 ) ( x − 3 ) + f ″ ( 3 ) 2 ! ( x − 3 ) 2 + f ‴ ( 3 ) 3 ! ( x − 3 ) 3 , {\displaystyle T_{3}(x)=f(3)+f'(3)(x-3)+{\frac {\color {RedOrange}f''(3)}{2!}}(x-3)^{2}+{\frac {f'''(3)}{3!}}(x-3)^{3},}
Comparing this with the formula given in the question we get
f ( 3 ) + f ′ ( 3 ) ( x − 3 ) + f ″ ( 3 ) 2 ! ( x − 3 ) 2 + f ‴ ( 3 ) 3 ! ( x − 3 ) 3 = 24 + 6 ( x − 3 ) + 12 ( x − 3 ) 2 + 4 ( x − 3 ) 3 . {\displaystyle f(3)+f'(3)(x-3)+{\color {OliveGreen}{\frac {f''(3)}{2!}}}(x-3)^{2}+{\frac {f'''(3)}{3!}}(x-3)^{3}=24+6(x-3)+{\color {OliveGreen}12}(x-3)^{2}+4(x-3)^{3}.}
Comparing the coefficient of ( x − 3 ) 2 {\displaystyle (x-3)^{2}} , we get
f ″ ( 3 ) 2 ! = 12 , {\displaystyle {\frac {\color {RedOrange}f''(3)}{2!}}=12,}
and hence
f ″ ( 3 ) = 12 × ( 2 ! ) = 12 × 2 = 24 . {\displaystyle {\color {RedOrange}f''(3)}=12\times (2!)=12\times 2={\color {RedOrange}24}.}
Therefore, the answer is F.
, expanded about 
is
, we get
