# Interesting question about finding a quadratic polynomial such that $h(\alpha)=\beta, \ h(\beta)=\gamma, \ h(\gamma)=\alpha$

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

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

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:

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 $$n-1.$$