Finding a closed form to a minimum of a function

Partial results since I am still working the problem.

Using the Inverse Symbolic Calculator, the point $x_*$ where the derivative cancels seems to be very close to $$x_*=10 ,(gamma , K_1(1)){}^{Gamma left(frac{1}{4}right)}$$ which numerically is $$x_*=0.216453828qquad implies g(x_*)=0.99066450008687554$$ while the exact results are $$x_*=0.216453839qquad implies g(x_*)=0.99066450008687550$$

A much better approximation seems to be $$x_*=frac{97+800 e-307 e^2}{-263-205 e+113 e^2}$$ which is in an absolute error of $2.87 times 10^{-19}$; for this value, $g(x_*)$ is in an absolute error of $2.76 times 10^{-28}$.

Because $x_0$ the abscissa of the minimum we know that $g'(x_0)=0$ then you can calculate the function $g'(x)$: $$g'(x) = bigr(x^{2(1−x)} bigl)'+bigr((1−x)^{2x}bigl)' = bigr(e^{ln{(x^{2(1−x)})}} bigl)'+bigr(e^{ln{((1−x)^{2x})}}bigl)' = bigr(e^{{2(1−x)} ln{(x)}} bigl)'+bigr(e^{{2x} ln{(1−x)}}bigl)' = bigl[ bigr(e^{{2(1−x)} ln{(x)}} bigl)bigr({{2(1−x)} ln{(x)}} bigl)' bigl]+bigl[ bigr(e^{{2x} ln{(1−x)}}bigl)({2x} ln{(1−x)})' bigr] = $$ $$ = bigl[x^{2(1−x)}bigl(2(1-x){1over x}-2ln(x) bigr) bigr]+bigl[(1-x)^{2x}bigl(2ln(1-x)+2{xover (1-x)}bigr)bigr] = 2bigl[x^{2(1−x)}({1over x}-1-ln(x))+(1-x)^{2x}(ln(1-x)+{xover (1-x)})bigr]$$ Now because $g'(x_0)=0$: $$ 2bigl[x_0^{2(1−x_0)}({1over x_0}-1-ln(x_0))+(1-x_0)^{2x_0}(ln(1-x_0)+{x_0over (1-x_0)})bigr]=0$$ so we can get: $$x_0^{2(1−x_0)}({1over x_0}-1-ln(x_0))=-(1-x_0)^{2x_0}(ln(1-x_0)+{x_0over (1-x_0)})$$ $$$$ The solutions are: $x_0 = 0.216... , 0.783... and 0.5$, but you can easily find $x$ such as $g(x)<g(0.5)$ so $0.5$ is the max point and $x_0 = 0.216... , 0.783...$ are the minimum points. You can sea the solution here: https://www.wolframalpha.com/input/?i=-2+x%5E%281+-+2+x%29+%28x+%2B+x+log%28x%29+-+1%29+-+2+%281+-+x%29%5E%282+x+-+1%29+%28x+%2B+%28x+-+1%29+log%281+-+x%29%29+%3D0.