Question

f(x)=ax^2+bx+c ,(a>0) ,方程f(x)-x=0的两根X1和X2,0

Answer 1

1）证明：令F(x)=f(x)－x,

∵ x1,x2是f(x)－x=0的根,

∴ F(x)=a(x－x1)(x－x2)．

当x∈(0,x1)时,由于x10,

又a>0,得F(x)=a(x－x1)(x－x2)>0,即x