Give an example (i.e. equation) of a non-constant polynomial function of the smallest degree $f(x),$ such that all of the following hold: (a) it has at least one inflexion point, (b) all inflexion points have $y$-coordinate equal to $0,$ (c) all its roots are real, and (d) it is symmetric with respect to the $y$-axis.
1. Evaluate $\frac{d}{dx}{(x^2-3)^3}.$
2. Find the roots of the polynomial $x^2-7x+12.$
3. Find the inflexion point of the curve $y=x^3-3x^2+5x-1.$

What does $f$ being symmetrical about the $y$-axis imply about its degree $n?$

Does the fact that $f$ has a point of inflexion restrict the values that $n$ can take?

How can one express a polynomial in terms of its roots?

If you arbitrarily choose one root of the polynomial, what conditions does that set on other roots?

Once you have arbitrarily selected a pair of roots, how can you select the other pair to satisfy the inflexion point condition?

Since $f$ is symmetrical about the $y$-axis, it must be of even order $n.$ At an inflexion point, the second derivative changes sign. For $n \leq 2,\,$ $f''(x)=0$ and so we deduce $n > 2.$ Hence, the smallest possible degree is $n=4.$ If the $y$-coordinate equals $0$ at all the inflexion points, then every inflexion point is a root. If all roots are real, then the polynomial may be expressed as a product of real monomials, i.e. $f(x)=(x-a)(x-b)\cdots.$ The symmetry of the function also means that the roots must be symmetrical, so $f(x)=(x-a)(x+a)(x-b)(x+b)$ $=(x^2-a^2)(x^2-b^2).$

We can arbitrarily choose a pair of symmetric roots, so let $b=1.$ We then find the second derivative of $f,$ which is $f''(x)=12x^2-2a^2-2.$ The roots of this derivative, $\pm\sqrt{(a^2+1)/6},$ are points of inflexion. In order to satisfy the condition that all inflexion points are roots of $f,$ we equate these to $\pm a$ and solve to get $a=\frac{1}{\sqrt5}.$ Therefore $f(x)=(x^2-1)(x^2-1/5)$ in this case.