solve in detail please:

min value of 2^(sinx-1) + 2^(cosx-1) is...???

1. root(2)^[root(2)]
2. root(2)^ [-root(2)]

2^(sinx-1) +2^(cosx-1)=(1/2)(2^sinx +2^cosx)=f(x), using AM≥GM,f(x)≥g(x)=sqrt(2^(sinx+cosx)), sqrt(2^(sinx+cosx))=

sqrt(2^(1/√2)(sin((pi/4)+x))), minimum value of g(x) is root(2)^[-root(2)], when sin((pi/4)+x)=-1,therefore minimum value of

f(x) also is equal to root(2)^[-root(2)]