For every $p \in \mathbb{R} $ we have an equation $x^3 - 3x - p = 0. $ If the equation has only one real solution let $f(p)$ be the square of this solution. Otherwise let $f(p)$ be the product of the smallest and largest solutions of this equation.
(a) Find the minimum value of $f(p)$.
(b) Approximately sketch $f(p)$.
My work so far:
I solved task (a) by showing that there are three real solutions if $|p| \leq 2$ (the extremes of polynomial $x^3 - 3x$) and then using Vieta's formulas to show that $f(p) = x^2 - 3$ if there are three real solutions. If there is only one real solution $f(p)$ is always positive, so the minimum value of $f(p)$ is $-3$.
I'm having issues with task (b) as I can't figure out the shape of $f(p)$ on the interval where there is only one real solution.
In the range where there is only one real solution, we have $p = x^3 - 3x$ and $f(p) = x^2$.
This gives us an equation $p^2 = x^2(x^2 - 3)^2 = f(p)(f(p) - 3)^2$, i.e. the point $(u, v) = (f(p), p)$ lies on the curve defined by $v^2 = u^3 - 6u^2 + 9u$.
The last curve is an elliptic curve. See the linked page for some pictures showing the shapes of elliptic curves.