What is the minimum of the univariate function below?

Question

I have the following function:

$f(x)=\max[(x+a)^2,(cx+b)^2]$,

where $a,b,c \in \mathbb{R}$ are fixed parameters and $x \in \mathbb{R}$ is a real variable. What would be the minimum of this function, $x^*$? Is there an analytical form for $x^*$?

Answer 1

Both $(x+a)^2$ and $(cx+b)^2$ are parabolas that open upwards; they are equal to each other at $x_1=\frac{a-b}{c-1}$ and $x_2=\frac{-a-b}{c+1}$. Lets just assume for convention that $x_1>x_2$ but in reality it doesn't matter since the argument that is about to be made is symmetric.

On the interval $(-\infty, x_1]$ the function $f(x)$ is decreasing, and on the interval $[x_2, \infty)$ the function $f(x)$ is increasing. So the minimum value of $f(x)$ must be on the interval $[x_1, x_2]$.

Notice how it is the endpoints of this interval in which $(x+a)^2=(cx+b)^2$, so within that interval of course our f(x) is just a parabola. Now between $(x+a)^2$ and $(cx+b)^2$, the one who is greater in the range $[x_1, x_2]$ simply cannot have its global min within the range $[x_1, x_2]$. That is because both parabolas have a global min at 0. If a parabolas global min was contained in f(x), then it could not have been outputted by the max[] function. Therefore on the range $[x_1, x_2]$ the function $f(x)$ is either increasing or decreasing. Which means the minimum value of $f(x)$ must be at one of the endpoints.

You can evaluate the function at each of the end points and derive conditional statements for when the max will occur at $x_1=\frac{a-b}{c-1}$ or at $x_2=\frac{-a-b}{c+1}$.

Note: In the case where $x_1=x_2$ the global min of the parabolas coincide, and ofcourse the max of the two global mins which are equal is just the global min (which is $0$ occuring at $x_1=x_2$)