# Relation between real roots of a polynomial and real roots of its derivative

I have this question which popped in my mind while solving questions of maxima and minima. First Case:Let $f(x)$ be an $n$ degree polynomial which has $r$ real roots. Using this can we say anything about the number of real roots of $f'(x)$?

Second Case:Suppose, $f(x)$ has all $n$ real roots. Then will all of its derivatives also have all real roots?

Also, if any of its derivatives do not have all real roots, then will $f(x)$ also have not all real roots? If the above is true then what about its converse?

Comment Case:For the third case:Suppose f'(x) is a 5 degree polynomial with 3 real roots.Then f(x) will be a 6 degree polynomial.(correct me if I am wrong).What are the possible no. of roots that f(x) can have(3,4,5 etc.?).Basically I am asking for an example.Also it would be great if you follow all cases with an example like in the 4th case.

• I don't understand the beginning. If $f(x)$ is an n-degree polynomial with $r$ real roots, then it is not the case that $f'(x)$ will have $(r-1)$ real roots. Consider $f(x) = x^2 +1$. Here $r = 0$, but $f'(x) = 2x$ which has 1 root at 0. Jul 23 '15 at 19:35
• Sorry I never realized that.I just said what I was taught.Now I know it was incorrect.I have edited it. But I dont understand why the question has been downvoted? Jul 23 '15 at 19:49
• You're probably being downvoted for not providing any work/ideas for your question. Jul 23 '15 at 20:22
• This is of interest: en.wikipedia.org/wiki/Gauss%E2%80%93Lucas_theorem Jul 23 '15 at 21:48

First case: If the number of real roots $r$ of $f(x)$ is greater than one, then $f'(x)$ has at least $r-1$ real roots. (The limitation "greater than one" is not necessary but the statement is trivial if $r\le 1$.) Given any two roots $a<b$ of $f(x)$, $f$ is continuous and differentiable on $[a,b]$, so by Rolle's theorem $f'(c)=0$ for some $a<c<b$.

There may be more roots of $f'(x)$ than those between roots of $f(x)$, so the only upper bound is the obvious one of $n-1$. Ask if you need examples. It seems to me that if multiplicity is taken into account that the number of real roots of $f'(x)$ has the same parity (even/odd) as the number of real roots of $f(x)$, but I haven't proven it yet. If multiplicity is not taken into account, the parity can be anything.

Second case: If $f(x)$ has degree $n$ and has $n$ real roots, then each consecutive pair of roots of $f(x)$ defines a root of $f'(x)$, which makes $n-1$ roots of $f'(x)$. Since $f'(x)$ is a polynomial of degree $n-1$, this is all possible roots. This continues for all later derivatives, so you are correct: all its derivatives will have all real roots.

Third case: The contrapositive of the second case tells us that if any of its derivatives have any non-real roots, then $f(x)$ also has some non-real roots.

Fourth case: The converse of the third case is not true. For example, $f(x)=x^2+1$ has two non-real roots, but its derivative $f'(x)=2x$ has one real root.

Comment case: You asked, "Suppose $f'(x)$ is a $5$ degree polynomial with $3$ real roots. What are the possible no. of roots that $f(x)$ can have($3,4,5$ etc.?)." The formulas for $f(x)$ and $f'(x)$ are given in the diagram, where $C$ is a real constant, zero in the graph. You can see that $f'(x)$ is a degree $5$ polynomial with $3$ real roots.

The dashed horizontal lines show the possible number of real roots of $f(x)$ for varying values of $C$. There are $0$ real roots for $C=3$, $1$ real root for $C\approx. 2.638$, $2$ real roots for $C=1$, $3$ real roots for $C\approx -1.757$, and $4$ real roots for $C=-1.9$. My discussion for the first case shows that there cannot be more than $4$ real roots since $f'(x)$ has $3$ real roots.

• For the third case:Suppose f'(x) is a 5 degree polynomial with 3 real roots.Then f(x) will be a 6 degree polynomial.(correct me if I am wrong).What are the possible no. of roots that f(x) can have(3,4,5 etc.?).Basically I am asking for an example.Also it would be great if you follow all cases with an example like in the 4th case. Jul 23 '15 at 20:40

First, notice that if any differentiable function $f(x)$ has $r$ real (distinct) roots, then $f'(x)$ must have at least $r-1$ distinct roots. This follows from the following:

Suppose that $a_1<...<a_r$ are (distinct) roots from $f(x)$ and $i \in \{1,...,r-1\}$. Then consider every interval of the form $[a_i,a_{i+1}]$. Since this interval is compact, any function on this interval achieves a maximum/minimum. Each minimum/maximum (depending on whether the function $f$ is dipping below or above the real line) corresponds to a root of $f'(x)$. Since we have $r -1$ such intervals, we have at least $r-1$ roots.

Since all polynomials are differentiable, we notice that if $f(x)$ has $r$ (distinct) roots, then $f'(x)$ has at least $r - 1$ distinct roots.

Your question isn't the clearest, but maybe someone else can come up with a better question to answer 1.

If a polynomial $f(x)$ is degree $n$ and has $n$ (distinct) real roots, then using the argument above (as well as the fundamental theorem of algebra (that might be over-kill, but oh well)), you can conclude that $f'(x)$ has exactly $n-1$ real roots.

Given this work, you can figure out 3 with the case for (distinct) roots.