f(x)=x³/6-x+sinxf'(x)=x²/2-1+cosx考察函数g(x)=x²/2-1+cosxg'(x)=x-sinx再次求导:g"(x)=1-cos(x)≥0从而g'(x)是增函数,所以g'(x)≥g'(0)=0从而g(x)是增函数,所以g(x)≥g(0)=0所以f'(x)=g(x)≥0所以...