Question

求f(x)=2^sinx +2^cosx的最值

Answer 1

f'(x)=2^sinxln2*cosx-2^cosxln2*sinx

令f'(x)=0,得：2^sinx/sinx=2^cosx/cosx (sinxcosx≠0)

令g(x)=2^x/x(x∈(-1,0)∪(0,1)),则g'(x)=(2^xln2*x-2^x)/x^2=2^x/x^2*(xln2-1)