设f(x)=x^3 +ax^2 +bx+c在x=1处取极值-2,a=3c,b=-2c-3 ,求f(x)单调区间.算出来f'(x)=3x^2+6cx-2c-3f(x)单调区间怎么求?

设函数f(x)=ax^2+bx+c (a 设f(x)=ax^2+bx且-1 设f(x)=x^5+ax^3+bx-8,且f(-2)=10,则f(x)= 设函数f(x)=1/3ax^3+bx^2+cx(a 设三次函数f(x)=ax^3+bx^2+cx+d(a 设f(x)=ax^3+bx^2+cx+d,(a 设三次函数f(x)=ax^3+bx^2+cx+d(a 设f(x)=ax^3+bx^2+cx+d,(a 设f(x)=1/ax^2-bx+c,不等式不等式f(x) 设函数f(x)=InX-1/2ax^2-bx令F（X）=f(x)+1/2ax^2+bx+a/x(0 1、设二次函数f(x)=ax(平方)+bx+c满足f(x+1)-f(x)=2x 设f(x)=ax的平方+bx,且1 设函数f(x)=ax²+bx+c(a 设f(x)=ax^3+bx+1,并且f（-3）=2,则f(3)= 设f(x)=ax^1/3+bx^3+1,且f(2)=0,求f(-2) 设函数f(x)=ax五次方+bx三次方,且f(2)=3,则f(-2)= 设f(x)=ax^3+bx+1,且f(2)=0,求f(—2)的值 设f(x)=3ax+2bx+c,若a+b+c=0,f(0)f(1)>0