f(x) = (x 1 ) 2 –4x 1 + (x 2 ) 2 + 4 .分别对x 1 和x 2 求导数并且令导数值为0,可得2x 1 –4 = 0 和 2x 2 = 0 => x 1 = 2 和 x 2 = 0 => f(x 1 ,x 2 ) = 0 + 0 = 0 这是函数的最小值（极小值）.