奇函数则有常数项为0,因此d=0
f(x)=ax^3+cx
f'(x)=3ax^2+c
f'(1)=3a+c=0
f(1)=a+c=-2
解得：a=1, c=-3
f(x)=x^3-3x
1) f'(x)=3x^2-3=3(x+1)(x-1)=0, 极值点为-1, 1
x1, 单调增
-1