解:f(x)=(sinx+cos)^2+2cos^2x =1+2sinxcosx+cos2x+1 =2+sin2x+cos2x =2+√2sin(2x+π/4) ∵x∈【-π/2,π/4】 ∴2x+π/4∈【-3π/4,3π/4】 则sin(2x+π/4)∈【-1,1】 所以fmax=f(π/8)=2+√2 fmin=f(-3π/8)=2-√2