# This question relates to the number of real roots of a polynomial equation.

So I practicing questions from Differential Calculus and came across a problem which required me to find the number of real roots of a polynomial equation, say $f(x)=0$. I initially thought of applying Sturm's theorem but was in a dilemma as to how this question relates to Differential Calculus. I looked for a solution and came across a statement which stated

Let $f(x)=0$ have $n$ number of real roots. Then by Rolle's theorem, $f'(x)=0$ will have $n-1$ real roots and $f''(x)=0$ will have $n-2$ real roots.

Can I therefore conclude that if a polynomial function has $k$ number of real roots then it's $n$th differential coefficient $f^{(n)}(x)=0$ will have $k-n$ number of real roots? Also, if that's the case, why is there a theorem as lengthy as Sturm's theorem if one could simply determine number of real roots of an equation through differential calculus?

• Rolle's theorem states that $f'$ has at least one zero between two zeros of $f$. You only get a lower bound on the number of zeros of the derivatives. – Martin R Aug 18 '17 at 20:41
• Note that your highlighted statement is not an if and only if, so you can't use it to determine the number of real roots directly from the number of roots of a derivative. (For an example, look at \$x^2+1). – rogerl Aug 18 '17 at 20:43
• The derivative of a polynomial is a polynomial. The result applies to the derivative as to the original polynomial, There are some negotiable subtleties if the roots of the original polynomial are not distinct. – Mark Bennet Aug 18 '17 at 20:54