Find the Taylor polynomials of orders 0, 1, 2, and 3 generated by ƒ at...

Consider the provided function,
${\displaystyle f\left(x\right)=\sqrt{x},a=4}$
Find the Taylor polynomials of orders 0, 1, 2, and 3 generated by ƒ at a.
The Taylor polynomial is shown below,
${\displaystyle P_n\left(x\right)=f\left(a\right)+\frac{f{}^{″}\left(a\right)}{2!}{(x-a)}^2+\dots +\frac{f^k\left(a\right)}{k!}{(x-a)}^k+\dots }$
First find the derivative of the function,
${\displaystyle f\left(x\right)=\sqrt{x}}$
${\displaystyle f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}}$
${\displaystyle f{}^{″}\left(x\right)=-\frac{1}{4x^{\frac{3}{2}}}}$
${\displaystyle f{}^{‴}\left(x\right)=\frac{3}{8x^{\frac{5}{2}}}}$
Substitute a=4 in the above functions we get,
${\displaystyle f\left(4\right)=\sqrt{4}\Rightarrow 2}$
${\displaystyle f^{\prime }\left(4\right)=\frac{1}{2\sqrt{4}}\Rightarrow \frac{1}{4}}$
${\displaystyle f{}^{″}\left(4\right)=\frac{1}{4{\left(4\right)}^{\frac{3}{2}}}\Rightarrow \frac{1}{32}}$
${\displaystyle f{}^{‴}\left(4\right)=\frac{3}{8{\left(4\right)}^{\frac{5}{2}}}\Rightarrow \frac{3}{256}}$
So, the Taylor polynomials of orders 0, 1, 2, and 3 is shown below.${\displaystyle P_0=2}$
${\displaystyle P_1=2+\frac{1}{4}(x-4)}$
${\displaystyle P_2=2+\frac{1}{4}(x-4)+\frac{1}{64}{(x-4)}^2}$
${\displaystyle P_3=2+\frac{1}{4}(x-4)+\frac{1}{64}{(x-4)}^2+\frac{3}{1536}{(x-4)}^3}$
Hence.