∴ There is 0; f(b)-Ab

∴f(b)-f(c)<； Ab<0, and B-C > 0

∴(f(b)-f(c))/(b-c)<； Ab/(b-c)

And b/(b-c) >; 1,∴ab/(b-c)<； A

That is, ξ∈(c, b) exists, so f' (ξ) = (f (b)-f (c))/(b-c).

That is, f'(x) has no lower bound on the right side of x=0.