Deducing a function

Via @MathsRH on twitter:

Not required for what this question was asking but does anyone want to play guess the function? pic.twitter.com/xohiIjj2gV

— RHMaths (@MathsRh) October 15, 2022

In case you can’t see that, it’s a challenge to come up with a function that matches a given picture, which I’ll describe in a moment; if you can see it, working out the key points is part of the problem-solving, so the description might constitute a spoiler or an accessibility aid depending on where you are in the process. I’ll put a line before it in any event.

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The picture shows a continuous curve that drops down from a large $y$ value to a minimum at $(-2,0)$. If then rises to a maximum at $(0,6)$, before dropping towards a horizontal asymptote at $y=3$.

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This is a point where I would bring out the DATAS framework.

• Domain. We’re not told anything about the domain, but there’s no vertical asymptote marked and I’ll assume the function is defined for all $x$.
• Asymptotes. As $x\to +\mathrm{\infty }$, $y\to 3$, and I’m assuming that as $x\to -\mathrm{\infty }$, $y\to +\mathrm{\infty }$.
• Turning points. Clearly $(-2,0)$, minimum, and $(0,6)$, maximum.
• Axes. Touches the $x$-axis at $(-2,0)$, crosses the $y$-axis at $(0,6)$.
• Shape. I think we’re looking at some variant on an $e^{-x}$ function.

But what variant? There are infinitely many possibilities, but I’ve zoomed in on one: $y=e^{-x}p(x)+3$, for some polynomial.

Why this? I want something that’s dominated by the 3 when $x$ is large, and by the $e^{-x}$ when $x$ is large-negative, but also has the possibility to have some turning points in between. The question is, what polynomial?

Let’s look at its derivative. This rises from steep negative, hits 0 at $x=-2$, peaks and drops to 0 again at $x=0$, then approaches the $x$ axis from below. Something like $e^{-x}q(x)$, where $q(x)$ is a quadratic, might work.

The quadratic would need to be of the form $q(x)=kx(x+2)$ to have the zeros in the right place, so we have $y^{\prime }=k{e}^{-x}x(x+2)$.

If we integrate that (tediously, by parts twice), we get $y=k{e}^{-x}(x^2+4x+4)+C$.

At first glance, that looks great – it even goes through $(-2,0)$! But that’s not great. The horizontal asymptote is at the same height as the turning point, and there’s no way around that in this form.

At this point, dear readers, I sighed, and wondered if I should break out the quartics. But then my brain suggested that there’s no particular reason to pick $e^{-x}$ over, say, $e^{-2x}$, other than laziness.

And indeed, if we make $y^{\prime }=k{e}^{-2x}x(x+2)$, we get $y=K{e}^{-2x}(2x^2+6x+3)$, which is better – although still not quite right. The turning points need to be equally spaced about the $x$-axis. So we have another parameter to throw in.

Deep breath

If we let $y^{\prime }=k{e}^{-ax}x(x+2)$, we get $y=k{e}^{-ax}(\frac{1}{a}x(x+2)+\frac{2}{a^2}(x+1)+\frac{2}{a^3})$, an answer only Wolfram|Alpha could love.

Where are the turning points? When $x=0$, we get $y=2k(\frac{1}{a^2}+\frac{1}{a^3})$.

When $x=-2$, we get $y=2k{e}^{2a}(-\frac{1}{a^2}+\frac{1}{a^3})$.

These need to sum to zero, so $2k((\frac{1}{a^2}+\frac{1}{a^3})+e^{2a}(\frac{1}{a^3}-\frac{1}{a^2}))=0$

Clearly $k$ can’t be 0, so let’s multiply by $\frac{a^3}{2k}$ to get $(a+1)+e^{2a}(1-a)=0$.

I suspect this isn’t going to come out neatly. Let’s try: $1+e^{2a}=a(e^{2a}-1)$, so we have $a=\coth (a)$. That is not a pleasant number – it’s a bit short of 1.2, and I’ll call this $A$.

If $k=1$, then this crosses the $y$-axis at $\frac{2}{a^3}(a+1)$, which is about 2.548 – but we want it to be 3. So $k=3÷(\frac{2}{A^3}(A+1))$, or $\frac{3A^3}{2(A+1)}$.

And, after all of that, we need to add 3 to get the asymptote and turning points in the right place.

So, our final curve is $y=\frac{3}{2(A+1)}{e}^{-Ax}(A^{2}x(x+2)+2A(x+1)+2)$, where $A$ satisfies $A=\coth (A)$.

Check it out on Desmos!

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That came out significantly less nicely than I expected it to! There are, presumably, any number of functions that fit the bill. Can you find a nicer one?