f(x)=1+x-x^2-cosx-sinx g(x)=f'(x)=1-2x+sinx-cosx f'(x)=0解得x=0 g'(x)=-2+cosx+sinx<0 f'(x)单调减少 故x>0时f'(x)<0,f(x)单调减少,f(x) 即1+x-x^2 x<0时f'(x)>0,f(x)单调增加,f(x) 即1+x-x^2所以1+x-x^2
