How many roots of $x(1-x)^{2}=s$ are there in $(0,1)$?

Question

This is a self-answered question, which is part of answering this related question. Alternative solutions are welcomed.

Let $0<s < \frac{4}{27}$. Prove that the cubic equation $x(1-x)^{2}=s$ has exactly two solutions in $(0,1)$.

Moreover, the third solution is a real number greater than $1$.

The limitation on the range of $s$ is due to $\frac{4}{27}=\max_{x \in [0,1]}x(1-x)^{2}$.

Answer 1

Define $F(x)= x(1-x)^2.$ Then, $F'(x)=(1-x)(1-3x)$ hence:

• for all $x<0,$ $F(x)<F(0)=0;$
• as $x$ increases from $0$ to $1/3,$ $F(x)$ continuously increases from $0$ to $F(1/3)=4/27;$
• as $x$ increases from $1/3$ to $1,$ $F(x)$ continuously decreases from $4/27$ to $F(1)=0;$
• as $x$ increases from $1$ to $+\infty,$ $F(x)$ continuously increases from $0$ to $\lim_{+\infty}F=+\infty.$

This ends the proof.

Answer 2

Define $F(x)= x(1-x)^2$. Then $F(0)=F(1)=0$, and $$F(\frac{1}{3})=\frac{4}{27}=\max_{x \in [0,1]}F(x).$$

Since $0<s < \frac{4}{27}$, by the Intermediate Value Theorem, the equation $F(x)=s$ admits at least one solution in each of the intervals $(0,\frac{1}{3})$ $(\frac{1}{3},1)$.

It remains to show that each interval contains at most one solution. This follows from monotonicity:

$$ F'(x)=3x^2 - 4x + 1>0 \iff x < \frac{1}{3} \,\,\text{ or }\,\, x>1. $$ Thus $F|_{(0,\frac{1}{3})}$ is decreasing, while $F|_{(\frac{1}{3},1)}$ is increasing.

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For any $s>0$, since $F(1)=0$, and $\lim_{x \to \infty} F(x)=\infty$, it follows that there exists $y>1$ such that $F(y)=s$.

Answer 3

The equation in question may be treated as $ \ x(1-x)^{2} \ = \ s \ \rightarrow \ x^3 - 2x^2 + x - s \ = \ 0 \ \ . \ $ One way we can investigate the number of real zeroes as $ \ s \ $ varies is to examine the properties of $ \ f(x) \ = \ x^3 - 2x^2 + x \ \ . $

If we "horizontally shift" this function $ \ \frac23 \ $ unit "to the left", we obtain $ \ g(x) \ = \ f \left( x + \frac23 \right) $ $ = \ x^3 - \frac13 x + \frac{2}{27} \ \ . \ $ In turn, we can shift (translate) this "vertically downward" to produce the odd-symmetry function $ \ h(x) \ = \ g(x) - \frac{2}{27} \ = \ x^3 - \frac13 x \ = \ x·\left(x^2 - \frac13 \right) \ = \ \left(x + \frac{1}{\sqrt3} \right)·x·\left(x - \frac{1}{\sqrt3} \right) \ \ ; \ $ this effectively translates the inflection point of $ \ f(x) \ $ to the origin. The relative extrema of $ \ h(x) \ $ lie at $ \ h'(x) \ = \ 3x^2 - \frac13 \ = \ 0 \ \Rightarrow \ x \ = \ \pm \frac13 \ \ , \ $ with $$ \ h \left( \pm \frac13 \right) \ \ = \ \ \left( \pm \frac13 \right)^3 \ - \ \frac13·\left( \pm \frac13 \right) \ \ = \ \ \mp \frac{2}{27} $$ [this can also be found through the use of inequalities: see, for instance, Korovkin, Inequalities, Theorem 5].

With the inflection of the relative function curves so described, we can translate this result back to find that $ \ f(x) \ $ has its relative maximum at $ \ \left(\frac13 \ , \ \frac{4}{27} \right) \ $ and its relative minimum at $ \ (1 \ , \ 0) \ \ . \ $ So we see that $ \ x^3 - 2x^2 + x - s \ $ has a "double zero" ("touching intercept") $ \ x \ = \ 1 \ $ for $ \ s \ = \ 0 \ \ ; \ $ the third real zero is evidently at $ \ x \ = \ 0 \ \ . \ $ (We could of course have found this by factoring $ \ x^3 - 2x^2 + x \ = \ x·(x - 1)^2 \ \ . \ ) $

Where our work with the inflection is more useful is in observing that

• for $ \ s \ = \ \frac{2}{27} \ \ , \ $ our equation is $$ x^3 - 2x^2 + x - \frac{2}{27} \ \ = \ \ h\left( x - \frac23 \right) $$ $$ = \ \left(x - \frac23 + \frac{1}{\sqrt3} \right)·\left(x - \frac23 \right)·\left(x - \frac23 - \frac{1}{\sqrt3} \right) \ \ = \ \ 0 \ \ , $$ with two zeroes in the interval $ \ (0 \ , \ 1) \ $ and the third is $ \ \frac23 + \frac{1}{\sqrt3} \ > \ 1 \ \ ; $

• and for $ \ s \ = \ \frac{4}{27} \ \ , \ $ the relative maximum in the inflection is $ \ \left(\frac13 \ , \ 0 \right) \ $ and the relative minimum lies at $ \ \left(1 \ , \ -\frac{4}{27} \right) \ \ ; \ $ hence, we have a double zero at $ \ x \ = \ \frac13 \ \ $ and polynomial or synthetic division leads to $$ x^3 - 2x^2 + x - \frac{4}{27} \ \ = \ \ \left(x - \frac13 \right)^2·\left(x - \frac43 \right) \ \ = \ \ 0 \ \ , $$

So there are two real zeroes ( $ x \ = \ \frac13 \ $ with multiplicity $ \ 2 $ ) in $ \ (0 \ , \ 1) \ $ for $ \ s \ = \ \frac{4}{27} \ $ and no real zeroes in that open interval for $ \ s \ = \ 0 \ \ , \ $ but three at its "endpoints" ( $ x \ = \ 0 \ $ with multiplicity $ \ 1 \ $ and $ \ x \ = \ 1 \ $ with multiplicity $ \ 2 ) \ . $ These observations are consistent with the Viete relations that the sum of the three zeroes is $ \ 2 \ \ , \ $ their product is $ \ s \ \ , \ $ and the sum of their pairwise-products is $ \ 1 \ \ . \ $ We also note that the relative extrema cannot contact the $ \ x-$axis for $ \ s \ < \ 0 \ $ or for $ \ s \ > \ \frac{4}{27} \ \ . $

We now have some guidance as to how the zeroes depend upon the value of $ \ s \ \ $ without needing to solve for them. Consider a small "perturbation" $ \ x^3 - 2x^2 + x - \varepsilon \ = \ 0 \ \ , \ $ corresponding to finding the intersections of the $ \ x-$axis with the curve for $ \ x^3 - 2x^2 + x \ \ $ "shifted vertically downward" by a small amount $ \ \varepsilon \ . \ $ The double zero at $ \ x \ = \ 1 \ $ "bifurcates" in such a way that for the three real zeroes, $ \ r \ = \ 0 \ $ increases toward $ \ \frac23 - \frac{1}{\sqrt3} \ \ , \ \ p \ $ decreases from $ \ 1 \ $ toward $ \ \frac23 \ \ $ and $ \ q \ $ increases from $ \ 1 \ \ $ toward $ \ \frac23 + \frac{1}{\sqrt3} \ \ . \ $ This is consistent with $ \ f'(x) \ = \ 3x^2 - 4x + 1 \ $ for which $ \ f'(0) \ = \ +1 \ > \ 0 \ $ and $ \ f'(1 \pm |\delta|) \ = \ (3 \delta^2 \pm 6 |\delta| + 3 ) \ - \ (4 \pm 4 |\delta| ) \ + \ 1 $ $ = \ \pm 2| \delta | + 3 \delta^2 \ \ . $

By a similar argument, for $ \ x^3 - 2x^2 + x - \left( \frac{4}{27} - \varepsilon \right) \ = \ 0 \ \ , \ $ increasing $ \ \varepsilon \ $ causes the double zero at $ \ \frac13 \ $ to bifurcate into $ \ r \ $ decreasing toward $ \ \frac23 - \frac{1}{\sqrt3} \ $ and $ \ p \ $ increasing toward $ \ \frac23 \ \ , \ $ while $ \ q \ $ decreases from $ \ \frac43 \ $ toward $ \ \frac23 + \frac{1}{\sqrt3} \ \ . \ $ We see that $ \ f' \left(\frac43 \right) \ = \ +1 \ > \ 0 \ \ , \ $ so decreasing $ \ s \ $ will decrease $ \ q \ \ , \ $ while $ \ f' \left(\frac13 \pm |\delta| \right) $ $ = \ \left(3 \delta^2 \pm 2 |\delta| + \frac39 \right) \ - \ \left(\frac43 \pm 4 |\delta| \right) \ + \ 1 \ \ $ $ = \ \mp 2| \delta | + 3 \delta^2 \ \ , \ $ thus increasing $ \ \varepsilon \ $ (or decreasing $ \ s \ = \ \frac{4}{27} - \varepsilon \ $ ) causes $ \ r \ < \ \frac13 \ $ to decrease and $ \ p \ > \ \frac13 \ $ to increase.

Thus, for $ \ 0 \ < \ s \ < \ \frac{4}{27} \ \ , \ $ the smaller two zeroes remain in the intervals $ \ 0 \ < \ r \ < \ \frac13 \ $ and $ \ \frac13 \ < \ p \ < \ 1 \ \ , \ $ while the third zero lies in $ \ 1 \ < \ q \ < \ \frac43 \ \ ; \ $ the equation $ \ x·(1-x)^{2} \ = \ s \ \ $ always has two roots in $ \ (0 \ , \ 1) \ \ . $