Simple Functional Equation $\frac{f(a)-f(b)}{a-b}\cdot(-a)+f(a)=-ab$

Question

Compute all real-valued functions $f$ so that the line between any two points on the graph $f$ intersects the $x$-axis at the product of those two points' $x$-coordinates times $-1$.

(if we do some simplifing, we just need all functions so that $\frac{f(a)-f(b)}{a-b}\cdot(-a)+f(a)=-ab$ for all points $a$ and $b$)

We can immediately see that $f(x)=x^2$ works, and we can also see that $f(0)=0$. If we assume $x$ to be a polynomial and let $f(x)=x\cdot Q(x)$, then we can do some manipulation and see that $\frac{Q(a)-Q(b)}{a-b}=1$, from which we see that $Q(x)=x+c$ an we can easily find $f(x)=x^2$ is the only solution.

But what if $f$ is not a polynomial (and we cant suppose $f(x)=x\cdot Q(x)$)? Looking at the Taylor Polynomial of $f$, it would appear that this case fails too, but I cannot rigorously prove it.

Thanks

Answer 1

A common trick is to consider the difference of any two such functions. In this case, let $h(x) = f(x) - g(x)$, where $f$ and $g$ both satisfy your functional equation. Then we can determine that $\displaystyle \frac{h(a)-h(b)}{a-b}(-a)+h(a) = 0$, and hence $ah(b) = bh(a)$ for all $a$ and $b$. Substituting in $1$ for $b$, we get $h(a) = ah(1)$, and thus $h(x) = cx$ for some constant $c$. Since $f(x) = x^2$ is a solution, all solutions must therefore be polynomials of the form $f(x) = x^2 + cx$.

Edit: Originally this said to apply the OP's earlier work at this point, but all such functions $f(x) = x^2+cx$ are in fact solutions to the functional equation in question, as one can check by substituting.

Answer 2

$\frac{f(a)-f(b)}{a-b}\cdot(-a)+f(a)=-ab$ is equivalent to $bf(a)-af(b)=ab(a-b)$. Put $a=0,b\neq 0$ to get $f(0)=0$. Then choose $a$ constant such that $f(a)=c\neq 0$ (constant zero function does not work here) and get $$ bc-af(b)=a^2b-b^2a$$ divide by $a$ (which is not zero) and you have $f(b)=b^2+bc/a-ab$, but $a,c$ are constants so $f(b)=b^2+kb$. We verify and this works.

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Note: verification at the end is important. At the International Mathematical Olympiad (the top competition for students before university) you lose a point by default if you do not verify the solution of a functional equation in the end.

Answer 3

This is just a slight variant of Beni Bogosel's answer.

The given functional equation implies $bf(a)-af(b)=ab(a-b)$ for all $a$ and $b$, so in particular if $a=x$ and $b=1$, in which case it becomes $f(x)-xf(1)=x(x-1)$, which can be rewritten as $f(x)=x(x+c)$ where $c=f(1)-1$. So if there is a function satisfying the functional equation, then it must be of the form $f(x)=x(x+c)$, which, of course, has been verified to work.