设二次函数f=mx^2+nx+t的图线过原点,g=ax^3+bx-3(x>0) f,g的导函数为f'和g',且f'=0f

f(x)=mx^2+nx+t,f(0)=t=0
f(x)=mx^2+nx
f'(x)=2mx+n
g(x)=ax^3+bx-3 (x>0)
g'(x)=3ax^2+b
f'(-1)=n-2m=-2 ①
f(1)=g(1)==> m+n=a+b-3 ②
f'(1)=g'(1)==> 2m+n=3a+b ③
①②③==>
n=2m-2 ,a=(m-3)/2 ,b=5(m+1)/2
f(x)=mx^2+(2m-2)x
g(x)=(m-3)/2x^3+5(m+1)/2x-3
当m=3时,
g(x)=10x-3 ,f(x)=3x^2+4x
g(x)=f(x)解得(x-1)^2=0
∴g(x)是f(x)在(1,7)点处的切线
f(x)≥10x-3恒成立,g(x)=10x-3
此时,符合题意,k=10,m=3
f'(x)=6x+4,f'(x)=10==>x=1
m≠3时,
f(x)为二次函数,若f(x)≥kx+m恒成立
f(x)图像开口必需朝上,因此需m>0
g(x)为三次函数,若g(x)≤kx+m恒成立,
需x^3系数(m-3)/2