f(x) = ax³ - 2bx² + cx + 4d
f'(x) = 3ax² - 4bx + c
(1)
f(-x) = - f(x) ：a(-x)³ - 2b(-x)² + c(-x) + 4d = -(ax³ - 2bx² + cx + 4d)
f(0) = 0：4d = 0
f'(-1) = 0：3a + 4b + c = 0
f(-1) = -2/3：-a - 2b - c + 4d = -2/3
a = -1/3 ,b = 0 ,c = 1 ,d = 0
f(x) = -x³/3 + x
(2)
f'(x) = -x² + 1
x ∈ [-1,1]：f'(x)∈[ 0,+∞)
f'(x1)f'(x2) ≠ -1
当x∈[-1,1]时,图像上不存在两点,使得过此两点处的切线互相垂直还有第三小题啊(3)|f(x1) - f(x2)| = |(-x1³/3 + x1) - (-x2³/3 + x2)|= |x1 -x2||x1² + x1x2 + x2² - 3|/3-2 ≤ x1 -x2 ≤ 21 ≤ x1² + x1x2 + x2² ≤ 3|f(x1) - f(x2)| ≤ 4/3