数学 导数题 求助 急需答案!谢谢
1个回答

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.