Solution

While working on part (c) of Can you find… curvy cubics edition, which asked: “Can you find a cubic curve that has a local minimum when $x=-1$?”, two students had the following conversation:

A: I can find a cubic with a stationary point at $x=-1$ by having two of the $x$-axis intersection points at $x=-2$ and $x=0$, because the stationary point is half-way between the two intersection points. I need a third intersection point, too, not between these, so I may as well choose $x=4$, so my cubic is $y=-x(x+2)(x-4)$ (with a minus sign so the cubic is the right way up).

B: I’m not sure that’s right; is the stationary point really half-way between the $x$-axis intersection points?

Can you resolve their debate?

A good start could be to check if student A or student B is right in this case.

If student A is right in this case, does this approach always work for finding a cubic with a given stationary (turning) point, or only sometimes?

If student B is right in this case, does student A’s approach ever work?

What exactly is student A’s approach? Can you describe it precisely?

Answering this first will ensure that we agree about what we are trying to show!

It seems that student A’s approach can be described as:

• To have a stationary point at $x=p$, I will put two of my $x$-axis intercepts at $x=a$ and $x=b$, where $p$ is midway between $a$ and $b$, and my third intercept will be at $x=c$, where $c$ is not between $a$ and $b$.

This is what we will work with.

To prove that student A’s approach doesn’t always work, we only need to find one counterexample, and we have now done this.

To find out whether student A’s approach ever works, we either need to find an example where it does or to prove that it never works.

We could test several examples, and that might give us some insight (if they all fail) towards a reason why it never works, or it will give us an example of where it does. But if they all fail, that alone does not constitute a proof that student A’s approach never works.

An alternative is to try to prove that student A’s approach never works. If our proof is successful, we will be done, and if it is not, we may discover an example which works instead.

We will use this second route, and try to prove that student A’s approach never works.

How do these arguments compare? What advantages does each one offer?