This is the first part of a problem in the high-school exit exam of this year, in Italy.

The differentiable function $y=f(x)$ has, for $x\in[-3,3]$, the graph $\Gamma$ below:

enter image description here

$\Gamma$ exhibits horizontal tangents at $x=-1,1,2$. The areas of the regions $A,B,C$ and $D$ are respectively $2,3,3$ and $1$. (Also, note that we are supposed to deduce $f(-2)=f(0)=f(2)=0$ from the graph)

If $f(x)$ is a polynomial, what is its minimum degree?

Let me explain the issue with this. In fact, the question in bold is a reformulation of mine, while the original was

In case $f(x)$ were expressible with a polynomial, what could be its minimum degree?

The use of "could" has been criticized because in fact it does not exclude incorrect answers such as $0$. Then again, it is argued that such a lexical choice was due to the high difficulty (for a high-school student) of an answer to the more precise question "what is its minimum degree?", therefore "necessary, not sufficient, accessible and not trivial conditions have been provided." (there are several articles on the subject, in Italian) Nonetheless, there is an answer generally regarded as correct: $4$. Most students came up with that, and important websites provided it as well. Their reasoning is simple: since $f'(x)$ has $3$ zeros, its degree is at least $3$, and thus $f(x)$ is at least a quartic. However, it is also relatively simple, solving a system with enough of the information we are given, that the assumption $f(x)=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0$ yields null coefficients. I personally didn't go further, but according to some articles I would have stopped at degree $9$, thus the answer to the question in bold; though this polynomial "in any case doesn't abide by $\Gamma$".

Here's my objection. It is clearly specified that $\Gamma$ is the plot of $f(x)$ in the considered interval, hence the minimum degree cannot be that of a polynomial which does not abide by it. The polynomial $P(x)$ must satisfy

\begin{cases} \int_{-3}^{-2}P(x)\,dx+2=\int_{-2}^0 P(x)\,dx-3=\int_0^2 P(x) \, dx + 3 = \int_2^3 P(x)\,dx+1=0 \\[6pt] P(-2)=P(0)=P(2)=0 \\[6pt] P'(-1)=P'(1)=P'(2)=0\\[6pt] P''(x)=0 \ \text{twice in $[-3,3]$, at the same points where $\Gamma$ changes concavity} \end{cases}

Of course not knowing the exact coordinates of the inflection points is problematic, but in such an exam a strong resemblance would be enough.

With these constraints, is there really no hope?

  • $\begingroup$ Perhaps it's an Italian thing, but in English "could" excludes incorrect answers. $\endgroup$ – Matt Samuel Jun 27 '15 at 16:47
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    $\begingroup$ If it were a polynomial of degree 4, by the position of the roots and local extrema this would force it to be $Cx(x+2)(x-2)^2$ for some $C$. $\endgroup$ – Matt Samuel Jun 27 '15 at 16:52
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    $\begingroup$ To solve this problem with all the constraints, i.e. $f(-2)=f(0)=f(2)=0,f'(-1)=f(1)=f(2)=0,(A,B,C,D)=(2,3,3,1)$ requires a polynomial of degree nine. The correct solution is computationally expensive, the intended solution is just wrong, shame on the problem poser. $\endgroup$ – Jack D'Aurizio Jun 27 '15 at 16:52
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    $\begingroup$ Also notice that you do not need to request your very last condition since it follows from $f'(-1)=f'(1)=f'(2)=0$ that $f''$ vanishes for some point in $(-1,1)$ and some point in $(1,2)$. $\endgroup$ – Jack D'Aurizio Jun 27 '15 at 16:54
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    $\begingroup$ @JackD'Aurizio The fact that $f''$ vanishes in the intervals $(-1,1)$ and $(1,2)$ does not imply strong resemblance to $\Gamma$, according to the articles I've read. // Of course, shame on the problem poser. $\endgroup$ – Vincenzo Oliva Jun 27 '15 at 17:02

Here we have the unique nine-degree polynomial that fulfills the ten constraints $f(-2)=f(0)=f(2)=f'(-1)=f'(1)=f'(2)=0, (A,B,C,D)=(2,3,3,1)$

$$ p(x) = -\frac{13960909 x}{3829050}-\frac{224 x^2}{9525}+\frac{26462017 x^3}{38290500}+\frac{8 x^4}{1905}+\frac{17935383 x^5}{34036000}+\frac{14 x^6}{1905}-\frac{6421193 x^7}{38290500}-\frac{11 x^8}{6350}+\frac{761753 x^9}{61264800}$$ together with its graph:

$\hspace1in$enter image description here

So the "minimal" solution has two unexpected stationary points in $(-3,-2)$ and $(2,3)$. To remove them both in order to have "strong resemblance", we need at least degree $\color{red}{11}$.

What an embarassing moment for the Italian mathematical instruction.

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  • $\begingroup$ Embarassing indeed. How did you figure out $11$? $\endgroup$ – Vincenzo Oliva Jun 27 '15 at 17:29
  • $\begingroup$ @VincenzoOliva: two extra constraints are enough to ensure "strong resemblance", for instance $f'(-2)=-1$ and $f(3)=-10$, so degree $11$ works. Degree $10$ has the same issue near one of the endpoints of $[-3,3]$. $\endgroup$ – Jack D'Aurizio Jun 27 '15 at 17:31
  • $\begingroup$ Oh, extra conditions. I see, thanks. One last question: is it possible to know (and/or prove) if there is a number of extra conditions (including the two you came up with) which is sufficient in order to have a plot identical to the one given? $\endgroup$ – Vincenzo Oliva Jun 27 '15 at 18:12
  • $\begingroup$ @VincenzoOliva: what do you mean by "identical"? It is trivial that there is no polynomial that is "identical" to the function $e^{x}$ over $[0,1]$, for instance. $\endgroup$ – Jack D'Aurizio Jun 27 '15 at 20:29
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    $\begingroup$ Well, I'm assuming that $f(x)$ is indeed a polynomial. But I guess infinitely many constraints would be needed, never mind. $\endgroup$ – Vincenzo Oliva Jun 27 '15 at 22:27

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