Question

设函数f(x)具有二阶导数,试证明曲线y^2=f(x)的拐点的横坐标a适合下列关系（f'(a))^2=2f(a)f''(

Answer 1

证明:y^2=f(x),所以:y=f(x)^1/2

y'=1/2f(x)^(-1/2)*f'(x)

拐点时y''=1/2*(-1/2)f(x)^(-3/2)*f'(x)*f'(x)+1/2f(x)^(-1/2)*f''(x)=0

拐点的横坐标为a所以

1/2f(x)^(-1/2)*f''(a)=1/4f(x)^(-3/2)*[f'(a)]^2

f''(a)=1/2f(x)^(-1)*[f'(a)]^2

移项得（f'(a))^2=2f(a)f''(a)