Answer
$$9.489 \times 10^{-4}$$
Work Step by Step
Since
$
\begin{array}{ll}
{f(x)=x^{\frac{11}{2}},} & {f(1)=1} \\
{f^{\prime}(x)=\frac{11}{2} x^{\frac{3}{2}},} & {f^{\prime}(1)=\frac{11}{2}} \\
{f^{\prime \prime}(x)=\frac{99}{4} x^{\frac{7}{2}},} & {f^{\prime \prime}(1)=\frac{99}{4}} \\
{f^{\prime \prime \prime}(x)=\frac{693}{8} x^{\frac{5}{2}},} & {f^{\prime \prime \prime}(1)=\frac{693}{8}} \\
{f^{(4)}(x)=\frac{3465}{16} x^{\frac{3}{2}},} & {f^{(4)}(1)=\frac{346}{16}}
\end{array}
$
Then $$T_{4}(x)=1+\frac{11}{2}(x-1)+\frac{99}{8}(x-1)^{2}+\frac{231}{16}(x-1)^{3}+\frac{1155}{128}(x-1)^{4}$$
Since $$\left|f(x)-T_{n}(x)\right| \leq K \frac{|x-a|^{n+1}}{(n+1) !} $$
Then
\begin{aligned}
\left|f(1.2)-T_{4}(1.2)\right|& \leq \frac{10395}{32} \sqrt{1.2} \frac{|1.2-1|^{4+1}}{(4+1) !} \\
\left|f(1.2)-T_{4}(1.2)\right| & \leq \frac{10395}{32} \sqrt{1.2} \frac{(0.2)^{5}}{5 !} \\
&=\frac{10395}{32} \sqrt{1.2} \frac{(0.2)^{5}}{120} \\
& \approx \sqrt{9.489 \times 10^{-4}}
\end{aligned}
To check
\begin{aligned}
\left|f(1.2)-T_{4}(1.2)\right| & \approx|2.725817989-2.7249375| \\
& \approx 8.80489 \times 10^{-4}<9.489 \times 10^{-4}
\end{aligned}
