General form of quartic polynomial with an extremum and two inflection points

Question

Can anyone tell me of a general form of a quartic function that has one extremum and two inflection points?

I am looking for a general form, not a specific function.

Edit: It doesn't have to be as general as possible, actually; just a form which generates solutions. Ideally with one variable other than the variable of the polynomial.

Answer 1

Completely general form

First let's find the general form of all such quartics. Let $Q(x)$ be the quartic polynomial in question. Then $Q'(x)$ must have only one real root, and $Q''(x)$ must have two real roots. We are looking for the conditions under which this is true.

Let's start with $$ Q''(x) = a(x-r)(x-s) = ax^2 - a(r+s)x + a(rs) $$ for real $a,r,s$, $r \ne s$, $a \ne 0$.

Now we can let $$ Q'(x) = \frac13 ax^3 - \frac{1}{2} a (r+s)x^2 + a(rs)x + c, $$ which is a cubic whose derivative has two changes of sign, so there is both a strict local max and a strict local min, i.e. two "turning points". The cubic $Q'(x)$ thus has only one real root iff the values of $Q'(r)$ and $Q'(s)$ have the same sign (see this answer). Finally we integrate one more timme, $$ Q(x) = \frac{1}{12} ax^4 - \frac{1}{6}a(r+s)x^3 + \frac{1}{2} a(rs)x^2 + cx + d, $$ and there are no additional conitions on $d$. Therefore, the entire solution set is given as follows:

Complete solution set:

All polynomials $$ Q(x) = \frac{1}{12} ax^4 - \frac{1}{6}a(r+s)x^3 + \frac{1}{2} a(rs)x^2 + cx + d $$ where $$ a,r,s,c,d \in \mathbb{R} $$ and $$ a \ne 0, r \ne s $$ and $$ Q'(r) Q'(s) > 0 $$ i.e. $Q'(r)$ and $Q'(s)$ have the same sign.

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Response to edit

Since you've now indicated you prefer a formula (not necessarily completely general) for quartics with only one variable other than $x$, note that we can use the above general form to get many such forms:

• The easiest thing is to vary the parameters $a$ or $d$, which don't matter. For instance we can set $r = -1$, $s = 1$, $c = 36$, $a = 12$ generating the set of solutions

  $$x^4 - 6x^2 + 12x + d\quad \text{ for all } d \in \mathbb{R}$$

  And setting $r = -1$, $s = 1$, $d = 0$, $c = a$ similarly gives the set of solutions (after scaling by $12$)

  $$ax^4 - 6ax^2 + 12ax\quad \text{ for all } a \ne 0$$

• These solution sets seem a bit trivial, though, so fix $a = 12$ and $d = 0$. We're left with $$ Q(x) = x^4 - 2(r+s)x^3 + 6(rs)x^2 + cx $$

  We may now take $r = 0$, $s > 0$; $c = Q'(0)$ has to be the same sign as $Q'(s) = 4s^3 - 6s^3 + c$, which equals $c - 2s^3$. We can guarantee this e.g. if $c = -1$. So another solution set is

  $$Q(x) = x^4 - 2sx^3 - x \quad \text{ for all } s > 0.$$

• If we want to vary $c$ instead, let's take $r = 0$, $s = 1$, so that $Q'(r) = c$, $Q'(s) = c - 2$. This will work for example for any $c < 0$. So we can have (switching sign of $c$)

  $$Q(x) = x^4 - 2x^3 - cx \quad\text{ for all } c > 0.$$

There are many other such forms we could come up with, but this should give you something to work with. The completely general form can easily be used in this way to give any number of specific forms with only one varying parameter.

Answer 2

An extremum at point $x_k$ has first derivative zero: $$f'(x_k) = 0$$ An inflection point is a point $x_k$ where $$f''(x_k) = 0$$

Given you know three such points, $x_1,x_2,x_3$ you can build a system of equations:

$$\cases{f'(x_1) = 0\\f''(x_2) = 0\\f''(x_3) = 0}$$ Where the coefficients $c_l$ are what we are solving for: $$\cases{f(x) = \sum_{l=0}^4c_lx^l\\\\ f'(x) = \sum_{l=0}^4l\cdot c_lx^{l-1}\\\\f''(x) = \sum_{l=0}^4l\cdot(l-1)\cdot c_lx^{l-2}}$$

Since differentiation is linear on polynomials this will be a homogeneous linear equation system with three equations and five unknowns. Unless conditions create a degenerate system there will be a linear subspace of dimensionality $5-3=2$ (why?) of coefficients which fulfill the criteria.