Question

函数y=4x²+8x+13\6(1+x)的最小值为多少?

Answer 1

y=(4x^2+8x+13)/6(x+1)

6y(x+1)=4x^2+8x+13

4x^2+(8-6y)x+13-6y=0

方程有解,则判别式>=0

即:(8-6y)^2-4*4(13-6y)>=0

64-96y+36y^2-208+96y>=0

36y^2>=144

y^2>=4

y>=2或y