设函数f(x)=x(x-1)(x-2)(x-3),则f''(x)=0有几个实数根是f(x)的二阶导有几个实数根,

设函数f(x)=x(x-1)(x-2)(x-3),则f''(x)=0有几个实数根是f(x)的二阶导有几个实数根,
设函数f(x)=x(x-1)(x-2)(x-3),则f''(x)=0有几个实数根
是f(x)的二阶导有几个实数根,

设函数f(x)=x(x-1)(x-2)(x-3),则f''(x)=0有几个实数根是f(x)的二阶导有几个实数根,
有两个根.f（0）=f（1）=f（2）=f（3）,则存在0-1,1-2,2-3之间各存在一点a,b,c使得f‘（x）=0,即
f’（a）=f’（b）=f’（c）=0,同理则在a-b即b-c之间各存一点使得f‘’（x）=0,
所以二阶导有两个实数根

拆开，求两次导，变成二次的式子然后就是解二次方程的问题了

先求f(x)',再求f"(x),另f"(x)=0,就可以算到了啊

f(x)=x(x-1)(x-2)(x-3)
f(x)=x^4-6x³+11x²-6x
f'(x)=4x³-18x²+22x-6
f''(x)=12x²-36x+22=0
6x²-18x+11=0
Δ=18²-4·6·11=60>0
所以f''(x)=0有两个不等实根

f(x)=x(x-1)(x-2)(x-3)
f'(x) =(x-1)(x-2)(x-3)+ x(x-2)(x-3) + x(x-1)(x-3)+x(x-1)(x-2)
f''(x) =(x-2)(x-3)+(x-1)(x-3)+(x-1)(x-2)+(x-2)(x-3)+x(x-3)+x(x-2)
+(x-1)(x-3)+x(x-3)+x(x-1)+(x-1...

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f(x)=x(x-1)(x-2)(x-3)
f'(x) =(x-1)(x-2)(x-3)+ x(x-2)(x-3) + x(x-1)(x-3)+x(x-1)(x-2)
f''(x) =(x-2)(x-3)+(x-1)(x-3)+(x-1)(x-2)+(x-2)(x-3)+x(x-3)+x(x-2)
+(x-1)(x-3)+x(x-3)+x(x-1)+(x-1)(x-2)+x(x-2)+x(x-1)
= 2(x(x-1)+x(x-2)+x(x-3)+(x-1)(x-2)+(x-1)(x-3)+(x-2)(x-3) )
= 2(6x^2-18x+11)
判别式>0
f''(x)有2个实数根

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