Can you state how MVT implies p'(x) has at least n roots in detail? What you say sounds good but I'd like to understand how you make the claim formally.
Use M.V.T to show that polynomial p(x) of degree n>= 1 has at most n roots. Hint : the proof proceeds by induction on the degree of p(x)
I don't know how to proof by induction but here is my approach on this question.
1. we know that polynomial of degree 1 will have at most 1 root and we want to show that n+1 degree => at most n+1 root.
2. so i try proof by contradiction by assume p(x) of degree n+1 will have at least n+2 root.
by M.V.T (Rolle's theorem), this assumption implies that p‘(x) has at least n+1 roots.But p'(x) is a polynomial of degree (n+1) - 1 , so it can not have more than n roots. So it contradict my assumption. p(x) of degree n+1 has at most n+1 roots?
if p(x) is differentiable on the open interval (a,b) and continuous on the closed interval [a,b]. if p(a) and p(b) are both 0. (the roots), then there is a at least one number c in (a,b) such that p'(c) = 0
(I prefer to use "k" rather than "n" because the statement of the theorem uses "n" to represent any positive integer while your "n" (my "k"), in the proof, is a single (though undetermined) integer.