f(x)=sinxcosx+2(sinx+cosx)+4令sinx+cosx=t X属于(0,pai/2),t=sinx+cosx∈(1,√2】1+2sinxcosx=t^2 sinxcosx=(t^2-1)/2f(t)=t^2/2+2t+7/2=(t+2)^2/2+3/2t=1 fmin=6 取不到最小值t=√2 fmax=9/2+2√2