Let k and K be the minimum and the maximum values of the function $\mathrm{f}\left(\mathrm{x}\right)=\frac{(1+\mathrm{x}{)}^{0.6}}{1+{\mathrm{x}}^{0.6}}$ in (0, 1] respectively, then the ordered pair (k, K) is equal to

$\begin{array}{l}f\left(x\right)=\frac{(1+x{)}^{0.6}}{1+x^{0.6}}\\ f^{\mathrm{\prime }}\left(x\right)=\frac{0.6\left(1+x^{0.6}\right)(1+x{)}^{-0.4}-0.6(1+x{)}^{0.6}{x}^{-0.4}}{{\left(1+x^{0.6}\right)}^2}\end{array}$
$\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{0.6\left({\mathrm{x}}^{0.4}-1\right)}{{\left(1+{\mathrm{x}}^{0.6}\right)}^2(1+\mathrm{x}{)}^{0.4}{\mathrm{x}}^{0.4}}<0\Rightarrow \mathrm{f}\left(0\right)>\mathrm{f}\left(1\right)$
$$\begin{gathered} \text{ Minimum occurs at }\mathrm{x}=1,\text{ value is }{2}^{-0.4} \\ \text{ and Maximum occurs at }\mathrm{x}=0,\text{ value is }1\text{. } \end{gathered}$$