y = f(x)满足f(x + y) = f(x)f(y),令x = y = 1可得f(2) = f(1)*f(1) = 1/9 => f 2 (1) = 1/9,而f(x)是定义域在R上的减函数,所以f(1) = 1/3（负值舍去）； 再令x = 2,y = 1代入可得f(3) = f(2)*f(1) = (1/9)*(1/3) = 1/27； 所以f(x)*f(3x 2 –1) = f(x + 3x 2 –1) < 1/27 = f(3),因为f(x)是定义域在R上的减函数,所以3x 2 + x – 1 > 3 => 3x 2 + x – 4 > 0 => (3x + 4)(x – 1) > 0 => x < -4/3 或者 x > 1 .