Hey amazingboxz.
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
Your first statement is only that a polynomial of degree 1 has 1 zero. Your second statement requires that p', a polynomial of degee n has only n roots. You need to include "assume a polynomial of degree k has only k roots" in which case your proof is by induction".
(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.