Answer
$$4.3258667 \times 10^{-6}$$
Work Step by Step
Since
\begin{array}{ll}
{f(x)=x^{-\frac{1}{2}},} & {f(4)=\frac{1}{2}} \\
{f^{\prime}(x)=-\frac{1}{2} x^{-\frac{3}{2}},} & {f^{\prime}(4)=-\frac{1}{16}} \\
{f^{\prime \prime}(x)=\frac{3}{4} x^{-\frac{5}{2}},} & {f^{\prime \prime}(4)=\frac{3}{128}} \\
{f^{\prime \prime \prime}(x)=-\frac{15}{8} x^{-\frac{7}{2}},} & {f^{\prime \prime \prime}(4)=-\frac{15}{1024}}
\end{array}
Then $$T_{3}(x)=\frac{1}{2}-\frac{1}{16}(x-4)+\frac{3}{256}(x-4)^{2}-\frac{5}{2048}(x-4)^{3}$$
Since
$$ \left|f(x)-T_{n}(x)\right| \leq K \frac{|x-a|^{n+1}}{(n+1) !}$$
Then
\begin{align*}
\left|f(4.3)-T_{3}(4.3)\right| &\leq \frac{105}{8192} \frac{|4.3-4|^{3+1}}{(3+1) !} \\
\left|f(4.3)-T_{3}(4.3)\right| &\leq \frac{105}{8192} \frac{(0.3)^{4}}{(4) !}\\
&=\frac{105}{8192} \frac{(0.3)^{4}}{(24)} \\
&=\frac{105}{8192} \frac{(0.3)^{4}}{(24)} \\
&\approx\left[4.3258667 \times 10^{-6}\right]
\end{align*}
To check
\begin{aligned}
\left|f(4.3)-T_{3}(4.3)\right| & \approx|0.4822428222-0.4822387695| \\
& \approx 4.0527 \times 10^{-6}<4.3258667 \times 10^{-6}
\end{aligned}
