Question

设二次函数f(x)=mx^2+nx+t的图像过原点,g(x)=alnx+bx,f(x),g(x)的导函数为f'(x),g

Answer 1

（1）二次函数f(x)=mx^2+nx+t的图像过原点,即f(0)=0,则 t=0,

由f'(x)=2mx + n,f'(0)=0,f'(-1)=-2,知 n = 0 ；-2 = -2m + n,即 m = 1

由f(1)=g(1),知 m+n+t = b,即 b = m=1

又g'(x) = a/x + b,f'(1)=g'(1),则 2m + n = a+b,即 a = 1

函数 f(x)= x^2,g(x)= lnx + x

（2）求F(x)=f(x)-g(x)的极小值

F(x)= x^2 - lnx - x,F'(x)= 2x - (1/x) -1 = (x-1)(2x+1)/x

令F'(x)=0 ,得 x = 1,x = - 1/2(舍去),

当00,所以F(1)= 0是极小值