f is uni-variate polynomial of degree d. I am interested in lower bounding modulus first derivative of f (i.e. |f'|)in interval [a,b] given the promise that in interval [a,b] f' don't have any root.

Like in Markoff theorem..to upper bound first derivative of polynomial in open interval (a,b) is
|f'(x)|<=(2Md*d)/(b-a) where M is upper bound of f in (a,b),but I don't know proof of this theorem.

My question is analogues to this inequality can we lower bound first derivative of f.

Thanks Ram

f is uni-variate polynomial of degree d. I am interested in lower bounding modulus first derivative of f (i.e. |f'|)in interval [a,b] given the promise that in interval [a,b] f' don't have any root.

Like in Markoff theorem..to upper bound first derivative of polynomial in open interval (a,b) is
|f'(x)|<=(2Md*d)/(b-a) where M is upper bound of f in (a,b),but I don't know proof of this theorem.

I know polynomial f i.e. its coefficient,degree and upper bound as well as lower bound of f in interval [a,b], Now I want to lower bound f' in [a,b] in terms of these terms..say like Markoff theorem.

Pls let me know whether this time I able to put my question clearly :)

Thanks Ram

f is uni-variate polynomial of degree d. I am interested in lower bounding modulus first derivative of f (i.e. |f'|)in interval [a,b] given the promise that in interval [a,b] f' don't have any root.

Thanks Ram

f is uni-variate polynomial of degree d. I am interested in lower bounding modulus first derivative of f (i.e. |f'|)in interval [a,b] given the promise that in interval [a,b] f' don't have any root.

Like in Markoff theorem..to upper bound first derivative of polynomial in open interval (a,b) is
|f'(x)|<=(2Md*d)/(b-a) where M is upper bound of f in (a,b),but I don't know proof of this theorem.

My question is analogues to this inequality can we lower bound first derivative of f.

Thanks Ram