Answer
$f(x)=8+10(x-2)+6(x-2)^2+(x-2)^3$
Work Step by Step
We have $f(x)=x^3+2x+4 \implies f(2)=8$; $f'(x)=3x^2-2 \implies f'(2)=10$;
$f''(x)=6x \implies f''(2)=12\\ f'''(x)=6 \implies f'''(2)=6$
Therefore, the Taylor's series at $x=2$ is as follows:
$f(x)=f(2)+f'(2) (x-2)+\dfrac{f''(2)(x-2)^2 }{2!}+\dfrac{f'''(2)(x-2)^3 }{3!}$
or, $=8+10(x-2)+\dfrac{(12) (x-2)^2 }{2 \cdot 1}+\dfrac{(6)(x-2)^3 }{3 cdot 2 \cdot 1}$
or, $f(x)=8+10(x-2)+6(x-2)^2+(x-2)^3$
