Question

设函数f（x）=ax^3-2bx^2+cx+4d的图像关于原点对称，且x=1时f（x）取最小值-1/3.

当x∈[-1,1]时，图像上是否存在两点使得过此两点处的切线互相垂直？证明结论.

Answer 1

函数f（x）=ax^3-2bx^2+cx+4d的图像关于原点对称, f(x)为奇函数:
f(-x) = -ax^3 -2bx² -cx +4d = -(ax^3-2bx^2+cx+4d) = -ax^3 + 2bx^2 - cx- 4d
比较系数, b = 0, d = 0
f(x) = ax^3 +cx
f'(x) = 3ax² + c在 x=1时为0: 3a + c = 0, c = -3a
f(x) = ax^3 -3ax
f(1) = a -3a = -2a = -1/3
a = 1/6
f(x) = x^3/6 - x/2
f'前橘(x) = (x² -1)/2
如当x∈[-1,1]时, 存慧州团在两点使得过此两点处的切线互相垂直, 设其横坐标分别为p, q, 则f'(p)*f'(q) = -1
[(p² -1)/2] * [(q² -1)/2] = -1
(p² -1)(q² -1) = -4
p∈[-1,1], q∈[-1,1], 0 ≤p² ≤ 1, 0 ≤ q² ≤1
-1 ≤p² - 1 ≤ 0, -1 ≤q² - 1 ≤ 0
-1 ≤ (p² -1)(q²迹塌 -1) ≤ 1
(p² -1)(q² -1) ≠ -4
不存在此两点