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Edit in hopes of clarifying the question:

I think that for every polynomial $f(x)$ having at least one real zero there is an $x$ for which $f(x) = f'(x)$. Am I right?

Original question:

Let $f(x)$ be a polynomial having at least one real zero. Is it possible for $f(x)$ to be equal to its derivative function anywhere for some value(s) of $x$? Well, as far as I am concerned it does.

Ethan Bolker
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Arijit Sinha
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1 Answers1

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For any polynomial $f(x)$ with a real root, there is always a $y$ such that $f'(y) = f(y)$.

Method I

Let's say $f(a) = 0$. Consider the function $g(x) = e^{-x}f(x)$. We have $$f(a) = 0 \quad\implies\quad g(a) = 0$$ Pick any arbitrary $b > a$, either $g(b) = 0$ or $g(b) \ne 0$.
For the second case,

  • By IVT, there is a $p \in (a,b)$ such that $g(p) = \frac12 g(b)$.
  • Since $\lim_{x\to\infty} g(x) = 0$, there is a $q > b$ such that $|g(q)| < \frac12 |g(b)|$.
  • By IVT, there is a $r \in (b,q)$ such that $g(r) = \frac12 g(b)$.

In both cases, we get a pair of numbers $(u,v) = \begin{cases} (a,b),& 1^{st}\text{ case}\\ (p,r) & 2^{nd}\text{ case} \end{cases}$ which satisfies $g(u) = g(v)$.
By Rolle's theorem, there is a $w \in (u,v)$ such that $$g'(w) = 0 \quad\iff\quad f(w) = f'(w)$$

Method II

Following is another argument that is hopefully easier to follow.

WOLOG, we will assume the leading coefficient of $f(x)$ is positive. This means $$\lim_{x\to+\infty} f(x) = \lim_{x\to+\infty} (f(x) - f'(x)) = +\infty$$

Let's say $f(x)$ has one or more real roots. Since $f(x)$ is a polynomial, it has finitely many real roots. Let $a$ be the largest one.

Notice $f(x) > 0$ for all $x > a$ implies $\frac{f(a+h) - f(a)}{h} > 0$ for all $h > 0$. This leads to $$f'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = \lim_{h\to 0+}\frac{f(a+h)-f(a)}{h} \ge 0$$

If $a$ is a double root, then $f'(a) = 0$ and hence $f(a) = f'(a)$ and we are done.

Otherwise, $f'(a) > 0$ and $f(a) - f'(a)$ is negative. Since $f(x) - f'(x)$ is positive for sufficiently large $x$, IVT tells us there is a $b > a$ such that $f(b)-f'(b) = 0$.

achille hui
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  • I could not quite catch your solution. Would you please explain it? – Arijit Sinha Sep 10 '17 at 20:43
  • @ArijitSinha do you know IVT (intermediate value theorem) and Rolle's theorem? $g(x) = f(x)e^{-x}\implies g'(x) = (f'(x) - f(x))e^{-x}$. So finding a root for $f(x) = f'(x)$ is equivalent to finding a root of $g'(x)$ which in turn equivalent to finding an extremum of $g(x)$. The most common tool to show the existence of extremum is Rolle's theorem. I believe the intention of this question is test your understanding of Rolle's theorem. – achille hui Sep 10 '17 at 20:59
  • You can generalize this argument by making explicit just what part of "polynomial" you actually need. – Ethan Bolker Sep 10 '17 at 23:09
  • Thanks for your answer. Now it makes sense. But I argued differently. If the polynomial is of even degree then its derivative function is of odd degree. So from the nature of even and odd degree curves and Rolle's theorem the proof follows(graphically). Again when the polynomial is of odd degree it will immediately create at least one point of inflection on the x axis where the primary and derivative curves intersect. So proof again holds. My main objective was to get my method verified. Is my method correct? – Arijit Sinha Sep 10 '17 at 23:11
  • @ArijitSinha I don't think so. No where in your argument use the info $f(x)$ has a real root. If you drop that requirement, let's say $f(x) = x^2+2$, you will find $f(x) = f'(x)$ doesn't have a real solution. If you want to fix your argument, you need to elaborate what the phrase "from the nature of even and odd degree" mean explicitly. – achille hui Sep 11 '17 at 03:24
  • I mean that , since the polynomial has at least one real zero so it will have at least 2 real zeros ( complex roots occur in pairs ). So f(x) will cut the x axis at least twice. This is limiting case. From Rolle's theorem, in between 2 real zeros the derivative function will be zero somewhere. As the derivative curve is of odd degree so the ends of that curve will proceed on opposite directions and so both curves have to cut each other.@Achillehui, it is mentioned in the question that f(x) has at least one real zero. So your counterexample violates that and so my method is not valid. – Arijit Sinha Sep 11 '17 at 06:41
  • The aforementioned method is valid for the even degree polynomials only. When it is of odd degree it will immediately create at least one point of inflection on x axis (where real zero occurs) and there the derivative value will also be zero. So that is the intersection point of both curves again. – Arijit Sinha Sep 11 '17 at 06:48
  • @ArijitSinha your argument for even case may be valid. However, I fail to see why an odd polynomial vanish at $a$ will have a point of inflection at $a$. e.g $f(x)=x^3+x$ vanish at $x=0$ but $f'(x)\ne 0$ there. – achille hui Sep 11 '17 at 07:32
  • @achille hui, yes, I have missed another possible case, i.e., the case like f(x)=x^3+x. Let us take the limiting case of f(x) (f is of odd degree) that has only one real zero c. WLOG take c +ve. The curve of f cuts x axis at (c,0). Now it goes on increasingly upto (b,d), where it has a point of inflection. Then it comes down decreasingly upto (k,l). All b,d,k,l are +ve. After that it goes increasingly again upto infinity. Here (b,d) and (k,l) may coincide. Denote g(x) as the derivative function. g will come decreasingly through (b,0), then it gets minima and goes on increasingly through (k,0). – Arijit Sinha Sep 11 '17 at 09:46
  • Since g(x) is of an even degree , its ends will approach in the same direction and hence, it cuts f(x) somewhere. This completes the proof. – Arijit Sinha Sep 11 '17 at 09:48
  • $f(x)$ need not have any point of inflection, eg. $f(x) = x^3 + x$. 2) even if it do, since $g(x)$ is glowing slower than $f(x)$, it may not be able to catch up $f(x)$ as $x \to \infty$. If you can find a point $x$ such that $g(x) > f(x)$, then I will agree $g$ and $f$ need to meet at some point.
    • – achille hui Sep 11 '17 at 14:35