如何判断函数的拐点?

Question

如题所述

Answer 1

y(x)=(x-1)*(x-2)^2* (x-3)^3*(x-4)^4
函数的二阶导数为零，且三阶导数不为零时，这点即为函数的拐点
考虑(x-3)^3
设P(x)=(x-1)*(x-2)^2*(x-4)^4
y(x)=[(x-3)^3]*P(x)
y'(x)=3(x-3)^2*P(x)+[(x-3)^3]*P'(x)
y"(x)=6(x-3)P(x)+3(x-3)^2*P'(x)+{[(x-3)^3]*P'(x)}'; y"(3)=0
y"'(x)=6P(x)+6(x-3)P'(x)+[3(x-3)^2*P'(x)]'+{[(x-3)^3]*P'(x)}",y"'(3)=6P(3)不为零
x=3即为函数的拐点
考虑(x-4)^4
w(x)=(x-4)^4
w'(x)=4(x-4)^3
w"(x)=12(x-4)^2, w"(4)=0
w"'(x)=24(x-4), w"'(4)=0, x=4 不为函数的拐点