Hint for proving that the extremum of $f(x)=(x-r_1)(x-r_2)$ is $\frac{r_1+r_2}{2}$ (without calculus)

Question

Could I get a hint (not a full solution) for how to prove this without calculus?

So far I've mainly been trying to prove the direct statement: for all $x_0\neq\frac{r_1+r_2}{2}$, $f(x_0)>f(\frac{r_1+r_2}{2})$.

If we assume $r_0<r_1$, then the only interesting case is $r_0<x_0<r_1$.
Now $$f(\frac{r_1+r_2}{2})=(\frac{r_1+r_2}{2}-r_1)(\frac{r_1+r_2}{2}-r_2)=(\frac{r_2-r_1}{2})(\frac{r_1-r_2}{2}),$$and I need to prove $(x-r_1)(x-r_2)>(\frac{r_2-r_1}{2})(\frac{r_1-r_2}{2})$. If $x>\frac{r_1+r_2}{2}$, then $x-r_1>\frac{r_1+r_2}{2}-r_1=\frac{r_2-r_1}{2}$ and $x-r_2>\frac{r_1+r_2}{2}-r_2=\frac{r_1-r_2}{2}$, but this doesnt mean that $(x-r_1)(x-r_2)>(\frac{r_2-r_1}{2})(\frac{r_1-r_2}{2})$ right? Since $(x-r_2)$ and $(\frac{r_1-r_2}{2})$ are negative. If they were all positive then we'd be done.

Should I try a completely different direction or is there something here that ive missed?

Answer 1

Hint Complete the square, to write $f(x)$ as $$f(x) = \left(x - \frac{r_1 + r_2}{2}\right)^2 + k$$ for some $k$. What can you say about $k$, and what does that imply about $f(x)$ for $x \neq \frac{r_1 + r_2}{2}$?

Answer 2

You can argue with symmetry.

$f(x)=(x-r_1)(x-r_2)$ expands as a quadratic function whose graph is an upward opening parabola. Solving $f(x)=0$ yields two roots, $r_1$ and $r_2$. Since the parabola is symmetric around its center axis, we know that the vertex must occur at the $x$ value between $r_1$ and $r_2$ which is the average: $$\frac{r_1+r_2}{2}$$

At that value of $x$ the function $f$ will attain its minimum.