Answer
$p_{0}(x)=\dfrac{1}{2} \\ p_{1}(x)=\dfrac{1}{2}-\dfrac{x}{4} \\p_{2}(x)=\dfrac{1}{2}-\dfrac{x}{4}+\dfrac{x^2}{8}$
$p_{3}(x)=\dfrac{1}{2}-\dfrac{x}{4}+\dfrac{x^2}{8}-\dfrac{x^3}{16}$
Work Step by Step
Taylor polynomial of order $n$ for the function $f(x)$ at the point $k$ can be defined as:
$p_n(x)=f(k)+\dfrac{f'(k)}{1!}(x-k)+\dfrac{f''(k)}{2!}(x-k)^2+....+\dfrac{f^{n}(k)}{n!}(x-k)^n$
Here, $f(0)=\dfrac{1}{2} ; f'(x)=-\dfrac{1}{(x+2)^2} \implies f'(0)=\dfrac{-1}{4}; f''(x)=\dfrac{2}{(x+2)^3} \implies f''(0)=\dfrac{1}{4}; f'''(x)=-\dfrac{6}{(x+2)^4} \implies f'(0)=\dfrac{-3}{8}$
Thus, $p_{0}(x)=\dfrac{1}{2} \\ p_{1}(x)=\dfrac{1}{2}-\dfrac{x}{4} \\p_{2}(x)=\dfrac{1}{2}-\dfrac{x}{4}+\dfrac{x^2}{8}\\p_{3}(x)=\dfrac{1}{2}+(\dfrac{-1}{4})(x-0)+(\dfrac{1}{4})\dfrac{(x-0)^2}{2!}+(\dfrac{-3}{8})\dfrac{(x-0)^3}{3!}$
Thus, $p_{3}(x)=\dfrac{1}{2}-\dfrac{x}{4}+\dfrac{x^2}{8}-\dfrac{x^3}{16}$
