$f(x)=x^3-3x^2+1, \forall x\in\mathbb R$,

$g(x)=1-\frac{1}{x} ,\forall x\in\mathbb R, x \neq 0$.

i) Show that $f(x)$ has $3$ distinct and real roots.

ii) It is given $\gamma < \beta < \alpha$, where $\gamma, \alpha, \beta$ are the roots of $f(x)$. Show that $g(\alpha)=\beta, \ g(\beta)=\gamma, \ g(\gamma)=\alpha$.

iii) Given $h(x)$ is a quadratic function such that $h(\alpha)=\beta, \ h(\beta)=\gamma, \ h(\gamma)=\alpha$.

Part (i) and (ii) are quite easy to show.

(i): \begin{align} D(f)&=−27𝐴^2𝐷^2+18𝐴𝐵𝐶𝐷−4𝐴𝐶^3−4𝐵^3𝐷+𝐵^2𝐶^2 \\ &=-27(1)(1)-4(-3^3)(1)>0. \end{align}

(ii): $g'(x)=\frac{1}{x^2} \implies g(x)$ is strictly increasing from the interval $(-\infty, 0)$ & $(0, \infty)$. We also know that $f(g(\alpha))=f(\beta)=0$. Similarly for $g(\beta)$ & $g(\gamma)$. So $g(\beta), g(\gamma), g(\alpha)$ are roots to $f(x)$.

Consider $g(\gamma)$ which now can either equal $\alpha, \beta$ or $\gamma$. Since the function is strictly increasing and not monotonically increasing, we conclude $g(\gamma) \neq \gamma$.

So $g(\alpha)=\gamma, \beta$ |$g(\beta)=\alpha, \gamma$ | $g(\gamma)=\alpha, \beta$. We use the fact that $g(x)$ is one to one to conclude that $g(\gamma) \neq g(\alpha)$. Thus only one of these 2 possible solutions are true. Suppose $g(\alpha)=\gamma$, $g(\beta)=\alpha$, $g(\gamma)=\beta$. Since $g(x)$ is strictly increasing, then it implies that one of the roots must lie within the negative interval and that root is $\gamma$. We can show then that $\beta>1$ which would make $\alpha<1$, which is a contradiction, so the other possibility must be true, proving $g(\alpha)=\beta, \ g(\beta)=\gamma, \ g(\gamma)=\alpha$.

(iii): This part is the part I'm stuck at. This is what I've tried.

$f(g(x))= -\frac{1}{x^3}+3(\frac{1}{x})-1$. So if $g(\alpha)$ is a root to $f(x)$, it implies $\frac{1}{\alpha}$ is a root for $x^3-3x+1=0$.

$\gamma=\frac{1}{1-\alpha}, \beta=\frac{1}{1-\gamma}, \alpha=\frac{1}{1-\beta}$.

$g^2(x)=\frac{1}{1-x}$ which potentially could be easier to work with.

That the distance between the roots can be modelled by the distance between the $g(x)$ and $g^2(x)$ graphs. The coloured segments are lines of the same length:

enter image description here

That the following are true:

$$(\alpha - \beta)(\gamma) = \gamma - \beta$$ $$(\alpha - \gamma)(\beta) = \alpha - \beta$$ $$(\beta - \gamma)(\alpha) = \alpha - \gamma$$

But after this I'm stuck. How can I proceed?


The traditional thing is this: take three distinct real numbers $u,v,w$ so the pairwise differences are nonzero. To get a quadratic $q(x)$ that gives $q_u(u) =1$ while $q_u(v)=q_u(w) = 0.$

$$ q_u(x) = \frac{(x-v)(x-w)}{(u-v)(u-w)} $$

Back to your $\alpha, \beta, \gamma$ do the same and make $$ \beta q_\alpha + \gamma q_\beta + \alpha q_\gamma $$

I'm not sure what these are called, but one can always arrange such "indicator" functions: given distinct numbers $x_1, x_2, ..., x_n$ we can make a polynomial function $f_1(x_1) = 1,$ $f_1(x_2) = 0,$ $f_1(x_3) = 0,$ and so on, where the degree of each $f_i$ is $n-1.$

Alright, Matthew points out that these are called Lagrange Polynomials, https://en.wikipedia.org/wiki/Lagrange_polynomial

  • $\begingroup$ Lagrange polynomials $\endgroup$ – Matthew Daly Aug 18 at 2:09
  • $\begingroup$ @MatthewDaly thanks. $\endgroup$ – Will Jagy Aug 18 at 2:10

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