Answer
$A_{0}(x)=\dfrac{1}{2} \\A_{1}(x)=\dfrac{1}{2}-\dfrac{x}{4} \\A_{2}(x)=(\dfrac{1}{2})-(\dfrac{1}{4})x+(\dfrac{1}{8})x^2\\A_{3}(x)=(\dfrac{1}{2})-(\dfrac{1}{4})x+(\dfrac{1}{8})x^2-(\dfrac{1}{16})x^3$
Work Step by Step
The Taylor polynomial is defined for the function $f(x)$ at the point $a$ of order $n$ as:
$A_n(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+....+\dfrac{f^{n}(a)}{n!}(x-a)^n$
Now, we get $f(0)=\dfrac{1}{2} \\ f'(x)=-\dfrac{1}{(x+2)^2} \implies f'(0)=\dfrac{-1}{4} \\f''(x)=\dfrac{2}{(x+2)^3} \implies f''(0)=\dfrac{1}{4} \\ f'''(x)=-\dfrac{6}{(x+2)^4} \implies f'(0)=\dfrac{-3}{8}$
This implies that $A_{0}(x)=\dfrac{1}{2} \\A_{1}(x)=\dfrac{1}{2}-\dfrac{x}{4} \\A_{2}(x)=(\dfrac{1}{2})-(\dfrac{1}{4})x+(\dfrac{1}{8})x^2\\A_{3}(x)=(\dfrac{1}{2})+(\dfrac{-1}{4})(x-0)+(\dfrac{1}{4})\dfrac{(x-0)^2}{2!}+(\dfrac{-3}{8})\dfrac{(x-0)^3}{3!}=(\dfrac{1}{2})-(\dfrac{1}{4})x+(\dfrac{1}{8})x^2-(\dfrac{1}{16})x^3$
