1 The Question

In a post on BlueSky, Dave Jones asks:

Hi #Alevelmaths, tangents are currently being discussed with first years. I have a question: for which functions are the two statements logically equivalent, "$f(x)$ has a tangent that crosses at more than one point" & "the second derivative takes +ve and -ve values"? Thanks in advance.

I had a go at answering, and decided that I thought that the two statements were equivalent. However, short message systems, like BlueSky, are not a great medium for developing a mathematical argument and checking that it holds tight. So I figured that as I've done the thinking, I may as well get it written down somewhere.

We have two statements for a function $f:[a,b]\to ℝ$:

1. There exist distinct $c,d\in [a,b]$ such that:

   $$f\text{'}(c)=\frac{f(d)-f(c)}{d-c}$$

2. There exist $c,d\in [a,b]$ such that $f"(c)\ge 0$ and $f"(d)\le 0$.

The reason to believe that these two statements are related is that to get a re-intersecting tangent then the function must curve away from and then back towards the tangent line. This should mean that it has a region where it is convex and one where it is concave, and so places where the second derivative is positive and negative.

There is a special case that is a bit irritating, so let's deal with it and remove it from consideration. If $f$ has a section on which it is linear then the conditions are trivially satisfied: the function is its own tangent, and its second derivative is zero along that stretch.

To remove that situation then we assume that there is no subinterval on which the derivative is constant. This allows us to replace the weak inequalities in the second statement by strict inequalities.

I'll also assume that the function has sufficient regularity that I don't need to bother with it. There will be a minimal regularity condition, but I'm not bothered about that. So let's assume that the function is at least $C^2$ on the interval. This means that the second derivative exists and is continuous.

2 From Intersection to Derivatives

Let's assume the first statement: that the function has a tangent that re-intersects the graph.

That is, there are distinct $c,d\in [a,b]$ such that:

The argument here depends on the Mean Value Theorem. This states that if $g:[p,q]\to ℝ$ is continuous and is differentiable on $(p,q)$ then there is some $r\in (p,q)$ for which:

Apply this to $f$ on $[c,d]$, this produces some $r\in (c,d)$ such that:

and this is equal to $f\text{'}(c)$.

So there are distinct points, $r$ and $c$, where the gradients are equal. Since $r\in (c,d)$, it is also the case that $r>c$.

The assumption that $f$ has no linear sections means that $f\text{'}$ is not constant on $(c,r)$ so there is $s\in (c,r)$ such that $f\text{'}(s)\ne f\text{'}(r)$.

Now apply the Mean Value Theorem again to $f\text{'}$ on the interval $(c,s)$. This gives a $u\in (c,s)$ such that:

Likewise, apply it to $f\text{'}$ on the interval $(s,r)$. This gives a $v\in (s,r)$ such that:

Now $r>s>c$ so the denominators in both of these are positive. Since $f\text{'}(s)\ne f\text{'}(r)$ then the numerators (and hence the second derivatives) are also non-zero.

Hence $f"(u)$ and $f"(v)$ are both non-zero and differ in sign. So the second derivative of $f$ takes both strictly positive and negative values.

3 From Derivatives to Intersections

For the reverse direction, suppose that $f"$ takes both positive and negative values. We've assumed that $f"$ is continuous which means that if it is positive at a point then it is positive in an interval about that point. Therefore, there is an interval on which it is strictly positive and one on which it is negative. These intervals obviously don't overlap.

Integrating a function over a region where it is, say, strictly positive results in a strictly positive quantity. So there are $p,q,r,s$ with $p<q$ and $r<s$ such that:

Assume wlog that $q<r$. Consider the integral over the full interval $[p,s]$. This could be positive or negative, assume wlog that it is negative (including zero).

Consider what happens to the integral:

as $q$ moves to $s$. It starts strictly positive. When $q$ reaches $s$ then, by assumption, it is negative (including zero). This integral varies continuously as $q$ moves towards $s$, so by the intermediate value theorem then there is some point where this integral is exactly zero. There also must be a first such point where this happens (being zero is what is known as a closed condition, meaning that the set of zeros has a least value).

Unpacking the "wlog"s, we have an interval, say $(u,v)$, over which the integral of $f"$ is zero and such that there is no $w\in (u,v)$ such that the integral of $f"$ over $(u,w)$ is zero.

The Fundamental Theorem of Calculus then applies to show that:

So we have distinct points where the gradients of the graph are equal. Our assumption that this is minimal means that there is no point in between with the same gradient.

This means that the tangent lines at $x=u$ and $x=v$ don't re-intersect the graph in the interval $(u,v)$: if either did then the Mean Value Theorem would provide a point in between with the same gradient (as in the forward argument). The only way this can happen is if the graph lies wholly between the tangent lines.

Now consider the chord joining the graph above $x=u$ and $x=v$. At both ends, the graph is travelling in the same direction along the tangent line, so at both ends the graph travels from one side of the chord to the other in the same direction. But this means that it must cross back again in between. Hence the chord from $x=u$ to $x=v$ intersects the graph at some point strictly in between $u$ and $v$, let the $x$ value at this point be $w$.

Now consider what happens to the chord as the end point slides back along the graph from $x=v$ to $x=w$. The angle of this chord (measured at $u$) changes and then returns to its original place. At some point in between, that angle will have reached an extreme value and at that point the chord will be tangential.

Since it achieves this before it reaches $w$, which is itself distinct from $u$, we have a tangent that re-intersects the graph.