$y'=(2x-2)2^{x^2-2x+\log_2\left(5\right)}\times\ln\left(2\right)$

$(2x-2)2^{x^2-2x+\log_2\left(5\right)}\times\ln\left(2\right)=0$

x=1 - критическая точка

f(1) = $2^{-1+\log_2\left(5\right)}=2^{\log_2\left(\frac12\right)+\log_2\left(5\right)}=2^{\log_2\left(2,5\right)}=2,5$