Question

已知a大于等于0,函数f(x)=x^2+ax.设x1属于（负无穷,-a/2）,记曲线y=f(x)在点M(x1,f(x1)

Answer 1

1)

f'=2x+a

f'(x1)=2x1+a

f(x1)=x1^2+ax1

曲线y=f(x)在点M(x1,f(x1))处的切线为

l: y=(2x1+a)(x-x1)+x1^2+ax1

x2为直线l的根：代入上式得：

(2x1+a)(x2-x1)+x1^2+ax1=0

x2-x1=-(x1^2+ax1)/(a+2x1)

x2=x1-(x1^2+ax1)/(a+2x1)

=(ax1+2x1^2-x1^2-ax1)/(a+2x1)

=x1^2/(a+2ax1)

得证~

2)OM=(x1,f(x1))

ON=(x2,0)

OM*ON=x1x2>9a/16

x1^3/(2x1+a)>9a/16

x1