#### Answer

$$0.00334$$

#### Work Step by Step

Since
\begin{array}{ll}
{f(x)=x^{-\frac{2}{3}},} & {f(1)=1} \\
{f^{\prime}(x)=-\frac{2}{3} x^{-\frac{5}{3}},} & {f^{\prime}(1)=-\frac{2}{3}} \\
{f^{\prime \prime}(x)=\frac{10}{9} x^{-\frac{8}{3}},} & {f^{\prime \prime}(1)=\frac{10}{9}}
\end{array}
Then
\begin{aligned}
T_{2}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2} \\
&=1-\frac{2}{3}(x-1)+\frac{5}{9}(x-1)^{2}
\end{aligned}
and
$$\left|f(1.2)-T_{2}(1.2)\right|=|0.885549-0.888889|=0.00334$$