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[Добромир Глухаров]

$y(x)=2^{-4cosx}+2^{2-3cosx}-2^{3-2cosx}$

$2^{-cosx}=t\in[0,5;2]$

$y(x)=y_1(t)=t^4+2^2.t^3-2^3.t^2=t^4+4t^3-8t^2$

$y_1(0,5)=0,0625+4.0,125-8.0,25=0,0625+0,5-2=-1,4375$

$y_1(2)=16+4.8-8.4=16$

$y_1'(t)=4t^3+12t^2-16t=4t(t^2+3t-4)=0\Rightarrow t_1=0\notin[0,5;2];t_2=-4\notin[0,5;2];t_3=1$

$y_1''(t)=12t^2+24t-16=4(3t^2+6t-4)$

$y_1''(1)=20>0\Rightarrow t=1\to y_1(1)=-3$ - локален минимум.

$\Rightarrow max_{x\in\mathbb{R}}y(x)=y_1(2)=y(\pi+2k\pi)=16;$

$min_{x\in\mathbb{R}}y(x)=y_1(1)=y(\frac{\pi}{2}+k\pi)=-3$