Solutions - 0240-07

Use the formula $a_n={\displaystyle \frac{f^{(n)}(3)}{n!}}$:

$$\begin{array}{cc}{a}_0& =\frac{f^{(0)}(3)}{0!}=1\\ & \\ a_1& =\frac{f^{(1)}(3)}{1!}=2\\ & \\ a_2& =\frac{f^{(2)}(3)}{2!}=6\\ & \\ a_3& =\frac{f^{(3)}(3)}{3!}=\frac{1}{2}\end{array}$$

Therefore:

$$f(x)=1+2(x-3)+6{(x-3)}^2+\frac{1}{2}{(x-3)}^3+\cdots$$