f(x) = ax^3 + bx^2 + cx
f'(x) = 3ax^2 + 2bx + c,
0 = f'(1) = 3a + 2b + c,...(1)
0 = f'(2) = 12a + 4b + c....(2)
因方程有2个不同的根,所以 a 不等于0.
0 = 9a + 2b,
2b = -9a.
f''(x) = 6ax + 2b
f''(1) = 6a + 2b = 6a - 9a = -3a,
f''(2) = 12a + 2b = 12a - 9a = 3a.
若a < 0.
则 f''(2) < 0.f'(2) = 0.f(x) 在 x = 2处达到极大值5.
5 = f(2) = 8a + 4b + 2c....(3)
由（1）,（2）,（3）解得
a = 5/2与a < 0矛盾.
所以,
a > 0.
则 f''(1) < 0.f'(1) = 0.f(x) 在 x = 1处达到极大值5.
5 = f(1) = a + b + c....(4)
由（1）,（2）,（4）解得
a = 2,b = -9,c = 12.
因此,
x0 = 1.