Question

数学问题 设f(x)=(x-1)(x-2)(x-3)(x-4) 则方程f'(x)=0实根的个数是多少?怎么求啊?

Answer 1

f(x)=(x-1)(x-2)(x-3)(x-4)
=(X^2-5X+4)(X^2-5X+6)
=(X^2-5X)^2+10(X^2-5X)+24.
令,u=x^2-5x,
V=u^2+10u
f(x)'=(u^2+10u)'
=2u*u'+10*u'
=2*(x^2-5x)*(2x-5)+10*(2x-5).
∵f(x)'=0,
2*(x^2-5x)*(2x-5)+10*(2x-5)=0,
(2x-5)(x^2-5x+5)=0,
2x-5=0,x^2-5x+5=0,
x1=5/2,x2=(5+√5)/2,x3=(5-√5)/2.
则方程f'(x)=0实根的个数是3个,为:
x1=5/2,x2=(5+√5)/2,x3=(5-√5)/2.