Answer
$$K = \frac{8}{{5\sqrt 5 }},{\text{ }}r = \frac{{5\sqrt 5 }}{8}$$
Work Step by Step
$$\eqalign{
& y = 2x + \frac{4}{x},{\text{ }}x = 1 \cr
& {\text{Calculate the curvature, use }}K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Find }}y'\left( x \right){\text{ and }}y''\left( x \right),{\text{ }} \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {2x + \frac{4}{x}} \right] \cr
& y'\left( x \right) = 2 - \frac{4}{{{x^2}}} \cr
& y''\left( x \right) = \frac{d}{{dx}}\left[ {2 - \frac{4}{{{x^2}}}} \right] \cr
& y''\left( x \right) = \frac{d}{{dx}}\left[ {2 - 4{x^{ - 2}}} \right] \cr
& y''\left( x \right) = \frac{8}{{{x^3}}} \cr
& \underbrace {K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}}}_ \Downarrow \cr
& K = \frac{{\left| {\frac{8}{{{x^3}}}} \right|}}{{{{\left( {1 + {{\left[ {2 - \frac{4}{{{x^2}}}} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Evaluate at }}x = 1 \cr
& K = \frac{{\left| {\frac{8}{{{{\left( 1 \right)}^3}}}} \right|}}{{{{\left( {1 + {{\left[ {2 - \frac{4}{{{{\left( 1 \right)}^2}}}} \right]}^2}} \right)}^{3/2}}}} \cr
& K = \frac{8}{{{{\left( 5 \right)}^{3/2}}}} \cr
& K = \frac{8}{{5\sqrt 5 }} \cr
& {\text{The radius of curvature is }}r = \frac{1}{K} \cr
& r = \frac{1}{{\frac{8}{{5\sqrt 5 }}}} \cr
& r = \frac{{5\sqrt 5 }}{8} \cr} $$
