高数导数极限。求极值点的个数。

Question

如题所述

Answer 1

f'(-x)=x[f'(x)-1]即f'(-x)=xf'(x)-x所以，
f'(x)=-xf'(-x)+x
∴f'(x)=-x²f'(x)+x²+x
f'(x)(1+x²)=x²+x
f'(x)=(x²+x)/(1+x²)
f''(x)=[(2x+1)(1+x²)-(x²+x)•2x]/(1+x²)²
=(2x-x²+1)/(1+x²)²
f''(x)=0
X=(2±√5）/2
令f'(x)=0X=0或x=-1极值点个数为2个。选B

Answer 2

f'(-x) = x[f'(x) -1] (1)

y=-x
f'(y) = -y[f'(-y) -1]

xf'(-x) = f'(x) +x (2)
x*(1) -(2)
x^2.[f'(x) -1] - [f'(x) +x ] =0
(x^2 -1)f'(x) = x
f'(x) = x/(x^2-1)

f'(x) =0
=> x=0
f''(x) = [ (x^2-1) - 2x^2]/(x^2-1)^2
= (-x^2-1)/(x^2-1)^2
f''(0) = -1 < 0 (max)
ans :A追问

错了。答案是2