By using Rolle's Theorem, prove that:
if the real polynomial p of degree n has all its roots real, then so does its derivative p'.
give an example of a cubic polynomial p for which the converse fails.
let
$\displaystyle x_1 < x_2 < x_3 < ...x_n$
be the n roots of a polynomial
Note: The derivative has degree n-1 so has at most n-1 roots
Consider the following by Rolle's theorem.
$\displaystyle f(x_{i+1})-f(x_i)=f'(c)(x_{i+1}-x_{i}) \iff 0-0=f'(c)(x_{i+1}-x_{i})$
since $\displaystyle x_i \ne x_{i+1}, f'(c)=0$
for i =1 to n-1
Then for each i There exists a $\displaystyle c \in [x_{i},x_{i+1}]$
such that $\displaystyle f'(c)=0$
These are each of the n-1 zeros of the derivative.
I guess this proof only works the zero's are not repeated(I think)
QED
I hope this helps
Good luck
I had an epiphany while playing some video games.
if $\displaystyle x_i=x_{i+1}$ it is a repeated root then the polynomial can be factored and written as
$\displaystyle p(x)=(x-x_i)^2(f(x))$ where f(x) is a polynomial
now we can take the derivative of p
$\displaystyle p'(x)=2(x-x_i)(f(x))+(x-x_i)^2(f-(x))=(x-x_i)(2f(x)-(x-x_i)f'(x))$ so
$\displaystyle p'(x_i)=(x_i-x_i)(2f(x)-(x_i-x_i)f'(x_i))=0$
So repeated real roots yield real roots of the derivative.
This can be generalized if a root is of multiplicty n the derivative will have a real zero of multiplicty n-1.
I hope this helps.