Review question

# When does a cubic curve have two turning points? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7861

## Solution

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.

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

Now $$$f'(x) = 3x^2 +2ax + b,\label{eq:1}$$$

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

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 $x_2 - x_1 = \frac{2}{3} \sqrt{a^2 - 3b}.$

The coordinates of the turning points satisfy the quadratic $\eqref{eq:1}$, 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 $x_2 - x_1 = \frac{1}{3} \left(-a + \sqrt{a^2 - 3b} + a + \sqrt{a^2 - 3b} \right) = \frac{2}{3} \sqrt{a^2 - 3b}.$

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 $g(x) = x^2 \left(x - \sqrt{a^2 - 3b}\right).$

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.

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.
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.

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

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.