Answer
$${f_{avg}} > {g_{avg}}$$
Work Step by Step
$$\eqalign{
& {\text{Let the function }}f\left( x \right) = \frac{2}{{{x^2} + 1}}{\text{ on the interval }}\left[ { - 1,1} \right] \cr
& {\text{The average value is:}} \cr
& {f_{avg}} = \frac{1}{{1 - \left( { - 1} \right)}}\int_{ - 1}^1 {\frac{2}{{{x^2} + 1}}} dx \cr
& {f_{avg}} = \frac{1}{2}\left[ {2{{\tan }^{ - 1}}x} \right]_{ - 1}^1 \cr
& {\text{Evaluate and simplify}} \cr
& {f_{avg}} = {\tan ^{ - 1}}\left( 1 \right) - {\tan ^{ - 1}}\left( { - 1} \right) \cr
& {f_{avg}} = \frac{\pi }{4} - \left( { - \frac{\pi }{4}} \right) \cr
& {f_{avg}} = \frac{\pi }{2} \approx 1.5707 \cr
& \cr
& {\text{Let the function }}g\left( x \right) = \frac{7}{{4\sqrt {{x^2} + 1} }}{\text{ on the interval }}\left[ { - 1,1} \right] \cr
& {\text{The average value is:}} \cr
& {g_{avg}} = \frac{1}{{1 - \left( { - 1} \right)}}\int_{ - 1}^1 {\frac{7}{{4\sqrt {{x^2} + 1} }}} dx \cr
& {g_{avg}} = \frac{7}{8}\int_{ - 1}^1 {\frac{1}{{\sqrt {{x^2} + 1} }}} dx \cr
& {\text{Integrate using }}\int {\frac{{dx}}{{\sqrt {{a^2} + {x^2}} }} = \ln \left( {x + \sqrt {{a^2} + {x^2}} } \right) + C} \cr
& {g_{avg}} = \frac{7}{8}\left[ {\ln \left( {x + \sqrt {1 + {x^2}} } \right)} \right]_{ - 1}^1 \cr
& {\text{Evaluate and simplify}} \cr
& {g_{avg}} = \frac{7}{8}\left[ {\ln \left( {1 + \sqrt 2 } \right)} \right] - \frac{7}{8}\left[ {\ln \left( { - 1 + \sqrt 2 } \right)} \right] \cr
& {g_{avg}} = \frac{7}{8}\left( {\ln \left( {1 + \sqrt 2 } \right) - \ln \left( { - 1 + \sqrt 2 } \right)} \right) \cr
& {g_{avg}} = \frac{7}{8}\ln \left( {\frac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}}} \right) \approx 1.5424 \cr
& \cr
& {\text{Therefore, the average value of the function }}f\left( x \right){\text{ is greater}} \cr
& {\text{than the average value of }}g\left( x \right) \cr
& {f_{avg}} > {g_{avg}} \cr} $$
