设函数f(x)=1/3x^3+ax^2+5x+6在区间[1,3]上是单调函数,则实数a的取值范围是?用导函数的零点和1,3比大小讨论做出来的答案为什么不对

设函数f(x)=x^3+ax^2-9x-1(a 设函数f(x)=x^3+ax^2-9x-1(a 设函数f(x)=x^3+ax^2-9x-1(a 设函数f(x)=x^3+ax^2-9x-1(a 急设函数f(x)=2{x}^{3}+ax-2,已知f(x) 设函数f(x)=-1/3x^3+2ax^2+1/3a(0 设函数f(x)=ax^2-2x+3,对于满足1 设函数f(x+1)=ax+1,且f(2)=3,则a= （x-3)(2x-1)设函数f(x)=ax^³+b,已知f(1)=0,则 设函数f(x)=ax+2,不等式｜f(x)｜ 已知函数f(x)=x^3+ax^2+x+1,设函数f(x)在区间(-2/3,-1/3)内是减函数,求a的范围? 设函数f(x)=1/3ax^3+bx^2+cx(a 设函数f(x)= -1/3x^3+2ax^2-3a^2x+a/3(0 设函数f(x)=-1/3x~3+2ax~2-3a~2x+1(0 设函数f(x)=-1/3x^3+2ax^2-3a^2x+1(0 设函数f(x)=1/3x^3-1/2ax^2-2a^2x+1(a 设函数f(x)=-1/3x^3+2ax^2-3a^2x+b,0 设函数f(x)=-1/3x^3+2ax^2-3a^2x+b,0