# Prove, without using Rolle's theorem, that a polynomial $f$ with $f'(a) = 0 = f'(b)$ for some $a < b$, has at most one root

Prove the following without using Rolle's Theorem:

If $$f$$ is a polynomial, $$f'(a) = 0 = f'(b)$$ for some $$a < b$$, and there is no $$c \in (a,b)$$ such that $$f'(c) = 0$$, then there is at most one root of $$f$$ in $$(a,b)$$.

I've already proven this by contraction by assuming that there is more than two roots and showing that it contradicts Rolle's Theorem. Now i'm wondering how I could prove this without using Rolle's Theorem.

• $f'(x)$ must have the same sign for $x \in (a,b)$. – copper.hat Nov 28 '18 at 5:57
• Why's that? Like I said I already proved this using Rolle's theorem and in no part did I assume f'(x) had the same sign. – Nykis Nov 28 '18 at 6:01
• In-fact if f'(x) has the same sign on (a,b) there would be no critical points correct? – Nykis Nov 28 '18 at 6:02
• It would mean that $f$ is strictly monotonic on $[a,b]$. – copper.hat Nov 28 '18 at 6:03
• @Micah: The proof uses the intermediate value theorem which does not rely on Rolle's theorem. – copper.hat Nov 28 '18 at 17:48

Well, this is not an exact answer, but we can use the following thoughts to prove the statement without using Rolle's Theorem.

Since it is given that $$f' \left( a \right) = f' \left( b \right) = 0$$ and assuming that $$f$$ is a non - constant polynomial (otherwise, if it is a constant non - zero poynomial, then it has no roots and hence the statement is true), we can see that either $$f \left( a \right) \geq 0$$ or $$f \left( a \right) \leq 0$$. Without loss of generality, I will give the arguments for $$f \left( a \right) \geq 0$$.

If $$f \left( a \right) = 0$$, clearly, $$a$$ is a root. Now, since $$f$$ is a polynomial which is non constant, there is a neighbourhood of $$a$$ where $$f$$ is either increasing or decreasing. We shall deal with the case where $$f$$ is increasing (similar argument can be given for decreasing).

Now, suppose that the neighbourhood is smaller than $$\left( a, b \right)$$ and after that neighbourhood, the function again starts decreasing (or becomes constant). Then, $$\exists c \in \left( a, b \right)$$ such that $$f' \left( c \right) = 0$$, contradicting the hypothesis. Therefore, the neighbourhood is atleast as much as $$\left( a, b \right)$$. Since $$f$$ is increasing in this interval, and $$f \left( a \right) = 0$$, there cannot be any other root in this interval. Hence, $$f$$ has at most one root.

Now, the other case where $$f \left( a \right) > 0$$. Again, from the above arguments, $$f$$ is either increasing or decreasing in $$f \left( a, b \right)$$. Hence, if it is increasing, $$f \left( b \right) > f \left( a \right) > 0$$ and it has no root in $$\left( a, b \right)$$. Similarly, if it is completely decreasing, it can cross the $$x -$$ axis atmost once.

Therefore, in all the cases, we can say that $$f$$ has atmost one root.

Suppose for the sake of contradiction that $$f$$ has two roots $$c$$ and $$d$$ in $$(a,b)$$. Since $$f$$ is a polynomial, it has finitely many roots. So we can suppose that $$c$$ and $$d$$ are consecutive roots of $$f$$: that is, $$f$$ is nonzero on $$(c,d)$$.

Since $$f$$ is a polynomial, we can write $$f(x)=(x-c)^p(x-d)^qr(x)$$ for some polynomial $$r$$ which is nonzero on $$[c,d]$$. Since $$r$$ is a polynomial, it is continuous, so by IVT it must have the same sign on all of $$[c,d]$$.

Then

\begin{align} f'(x)&=(x-c)^p(x-d)^qr'(x)+(x-c)^pq(x-d)^{q-1}r'(x)+p(x-c)^{p-1}(x-d)^qr'(x)\\ &=(x-c)^{p-1}(x-d)^{q-1}[r'(x)(x-c)(x-d)+r(x)(p(x-c)+q(x-d))]\\ &=(x-c)^{p-1}(x-d)^{q-1}s(x) \end{align} for some polynomial $$s$$.

By direct computation, $$s(c)=r(c)(c-d)$$ and $$s(d)=r(d)(d-c)$$. But $$r(c)$$ and $$r(d)$$ have the same sign. Therefore $$s(c)$$ and $$s(d)$$ have different signs. So again by IVT there is some $$x_0 \in (c,d)$$ (and thus also in $$(a,b)$$) with $$s(x_0)=0$$, which also means $$f'(x_0)=0$$.

Note that $$f$$ is smooth, and since $$f'(x) \neq 0$$ for $$x \in (a,b)$$ the intermediate value theorem shows that $$f'(x)$$ has the same sign for $$x \in (a,b)$$, so we can assume that $$f'(x) > 0$$ for $$x \in (a,b)$$.

Suppose $$f(x^*) = 0$$ for $$x^* \in (a,b)$$. Since $$f(x) = \int_{x^*}^x f'(t)dt$$ we see that $$f(x) > 0$$ for $$x \in (x^*,b)$$ and $$f(x) < 0$$ for $$x \in (a,x^*)$$. Hence $$f$$ has at most one zero in $$(a,b)$$.

Addendum: Note that for $$p(x) = x^k$$ we have $$p(x) = p(x^*)+ \int_{x^*}^x p'(t)dt$$ without using the fundamental theorem of calculus. Linearity shows that it holds for any polynomial.

• This direction of the fundamental theorem of calculus needs the fact that two functions with identical derivatives differ by a constant, which in turn uses the mean value theorem. In general the problem statement is equivalent to Rolle's theorem, except that it has more restrictive hypotheses. So if you give a proof that works in as much generality as Rolle's theorem, you must be either proving or using Rolle's theorem itself. – Micah Nov 28 '18 at 18:25
• @Micah: Integration is algebraic process for polynomials, it does not need the fundamental theorem of calculus to demonstrate that $p(x) = p(x^*)+\int_{x^*}^x p'(t)dt$, it is a purely algebraic computation (given that you accept that $\int_{x^*}^x t^k dt = {1 \over 1+k} (x^{k+1}-(x^*)^{k+1})$ can be computed without using Rolle's theorem). – copper.hat Nov 28 '18 at 19:47
• You can make that work, but you need some proof that the integral of a positive polynomial is positive which doesn't ultimately go through Rolle's theorem. I think this probably means you're going to be explicitly computing a bunch of Riemann sums, à la Fermat. – Micah Nov 28 '18 at 21:28
• @Micah: I'm not sure I follow your persistence here. It works without much additional effort. Proving that the integral is positive does not require an explicit computation. If $[c,d] \subset (a,b)$ we have $f'(x) \ge \delta >0$ for some $\delta >0$ and so, trivially, $\int_c^d f'(t) dt \ge (d-c)\delta > 0$. Rolle's theorem doesn't appear on the horizon. At all. – copper.hat Nov 28 '18 at 21:38
• I think the problem that I have is that you're switching between two different notions of integration. On the one hand, there's the linear map $I$ which is defined to linearly extend $I[x^k]=\frac{x^k}{k+1}$. This satisfies the identity $I[p']=p$ (i.e., the fundamental theorem of calculus) by definition, but it is not clear that it satisfies positivity (that is, that $I[p](b)-I[p](a)$ is positive whenever $p$ is positive on $(a,b)$). On the other hand, there's the actual Riemann integral on polynomials. This clearly satisfies positivity, but we need Rolle's theorem to get FTC. – Micah Nov 28 '18 at 22:11