When does a cubic curve have two turning points?

When does a cubic curve have two turning points?

Question

Let $f(x) = x^3 + ax^2 + bx + c$, where the coefficients $a$, $b$ and $c$ are real numbers. The figure below shows a section of the graph of $y=f(x)$. The curve has two distinct turning points; these are located at $A$ and $B$, as shown. (Note that the axes have been omitted deliberately.)

1. Find a condition on the coefficients $a$, $b$, $c$ such that the curve has two distinct turning points if, and only if, this condition is satisfied.

It may be assumed from now on that the condition on the coefficients in (i) is satisfied.

1. Let $x_1$ and $x_2$ denote the $x$ coordinates of $A$ and $B$, respectively. Show that $x_2 - x_1 = \frac{2}{3} \sqrt{a^2 - 3b}.$

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1. Suppose now that the graph of $y=f(x)$ is translated so that the turning point at $A$ now lies at the origin. Let $g(x)$ be the cubic function such that $y=g(x)$ has the translated graph. Show that $g(x) = x^2 \left(x - \sqrt{a^2 - 3b}\right).$

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1. Let $R$ be the area of the region enclosed by the $x$-axis and the graph $y=g(x)$. Show that if $a$ and $b$ are rational then $R$ is also rational.

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1. Is it possible for $R$ to be a non-zero rational number when $a$ and $b$ are both irrational? Justify your answer.

Solution

The curve has two distinct turning points if and only if the derivative, $f'(x)$, has two distinct real roots.

Now

$$\begin{equation} f'(x) = 3x^2 +2ax + b, \end{equation}$$

which has two distinct real roots when the discriminant is greater than $0$. That is, $(2a)^2 - 4(3)(b) > 0,$ and hence $a^2 - 3b > 0.$

The coordinates of the turning points satisfy the quadratic $(1)$, so $x = \frac{-2a \pm \sqrt{4a^2 - 12b}}{6} = \frac{-a \pm \sqrt{a^2 - 3b}}{3}.$ As $x_1 < x_2$, we have $x_1 = \frac{-a - \sqrt{a^2 - 3b}}{3}\quad\text{and}\quad x_2 = \frac{-a + \sqrt{a^2 - 3b}}{3}.$ (We know that both of these are real as $a^2 - 3b > 0$.)

Therefore $x_2 - x_1 = \frac{1}{3} \left(-a + \sqrt{a^2 - 3b} + a + \sqrt{a^2 - 3b} \right) = \frac{2}{3} \sqrt{a^2 - 3b}.$

The translated graph now has a repeated root at the origin, and a third root at $t$, say, where $t>0$. Therefore the graph has equation of the form $g(x) = x^2 (x-t) = x^3 - tx^2.$ Hence $g'(x) = 3x^2 - 2tx = x(3x-2t),$ and so $g(x)$ has turning points at $x=0$ and $x=\frac{2}{3}t$.

From part (ii), we have $\frac{2}{3}t - 0 = \frac{2}{3} \sqrt{a^2 - 3b},$ and so $t=\sqrt{a^2 - 3b}.$ This gives $g(x) = x^2 \left(x-\sqrt{a^2 - 3b}\right).$

Alternatively, we could use the result from part (ii) again. Our cubic $g(x)=x^3-tx^2$ is just like the original cubic in the question, with $a=-t$, $b=0$ and $c=0$.

Therefore the difference between the $x$-coordinates of the turning points for $g(x)$ is $\frac{2}{3}\sqrt{a^2-3b}=\frac{2}{3}\sqrt{t^2-0}=\frac{2}{3}t$ as $t>0$. We can immediately deduce that $t=\sqrt{a^2-3b}$, as in the first approach.

We use integration to determine

$R$

. As the region is below the

$x$

-axis, we take the negative of the integral, so we get

$$\begin{align*} R &{} = - \int_0^t g(x) \, dx \\ &{} = - \int_0^t (x^3 - tx^2) \, dx\\ &{} = - \left[ \frac{x^4}{4} - \frac{tx^3}{3} \right]_0^t \\ &{} = - t^4( \frac{1}{4}-\frac{1}{3}) \\ &{} = \frac{1}{12} t^4\\ &{} = \frac{1}{12} (a^2 - 3b)^2. \end{align*}$$

So if $a$ and $b$ are rational, then $R$ must also be rational.

Yes, it is possible for $R$ to be a non-zero rational number when $a$ and $b$ are both irrational.

For example, by choosing $a = 2\sqrt{\sqrt{2}}$ and $b = \sqrt{2}$, we have $a^2-3b=\sqrt{2}$, and hence $R=\dfrac{2}{12}= \dfrac{1}{6}$ which is rational.