Answer
$A_{0}(x)=2\\A_{1}(x)=2+\dfrac{(x-4)}{4} \\A_{2}(x)=2+\dfrac{(x-4)}{4}-\dfrac{1}{64} (x-4)^2\\A_{3}(x)=2+\dfrac{1}{4}(x-4)-\dfrac{1}{64} (x-4)^2+\dfrac{(x-4)^3}{512}$
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 $f(4)=2; f'(4)=\dfrac{1}{4};f''(4)=-\dfrac{1}{32}; f'''(4)=\dfrac{3}{256}$
This implies that $A_{0}(x)=2\\A_{1}(x)=2+\dfrac{(x-4)}{4} \\A_{2}(x)= 2+\dfrac{(x-4)}{4}+(\dfrac{-1}{32})\dfrac{(x-4)^2}{2}=2+\dfrac{(x-4)}{4}-\dfrac{1}{64} (x-4)^2\\A_{3}(x)=2+\dfrac{(x-4)}{4}-\dfrac{(x-4)^2}{64}+\dfrac{(\dfrac{3}{256})}{6}(x-4)^3=2+\dfrac{1}{4}(x-4)-\dfrac{1}{64} (x-4)^2+\dfrac{(x-4)^3}{512}$
Hence, $A_{0}(x)=2\\A_{1}(x)=2+\dfrac{(x-4)}{4} \\A_{2}(x)=2+\dfrac{(x-4)}{4}-\dfrac{1}{64} (x-4)^2\\A_{3}(x)=2+\dfrac{1}{4}(x-4)-\dfrac{1}{64} (x-4)^2+\dfrac{(x-4)^3}{512}$
