The vertex of the sum of the functions $f(x-a)$ and $f(x-b)$ with vertices $a$ and $b$ is $\frac{a+b}{2}$

Question

Suppose I have the functions $f(x-a)$ and $f(x-b)$ such that these have global maxima at $a$ and $b$ respectively. It seems plausible that $f(x-a)+f(x-b)$ will have a global maximum at $\frac{a+b}{2}$. Why is this?

Answer 1

Here's a counter-example.

Let $f(x)=(1-3x)(x+1)^3$

It can be shown that $f(x)$ has a global maximum at $x=0$

Then $f(x-1)$ has a global maximum at $x=1$ and $f(x-2)$ has a global maximum at $x=2$ because they are just translations of $f(x)$.

$f(x-1)=\left (1-3(x-1)\right)\left ((x-1)+1\right)^3=\left (4-3x\right)\left (x\right)^3=4x^3-3x^4$

$f(x-2)=\left (1-3(x-2)\right)\left ((x-2)+1\right)^3=\left (7-3x\right)\left (x-1\right)^3=\left (7-3x\right)\left (x^3-3x^2+3x-1\right)$

$f(x-2)=-3x^4+16x^3-30x^2+24x-7$

Adding gives $f(x-1)+f(x-2)=-6x^4+13x^3-30x^2+24x-7$

This has a global maximum at $x=1$: not at $\frac {1+2}2$

Answer 2

For $a$ and $b$ unequal the conjecture is, in general, not true.

Let $f(0)=1$ but $f(x)=0$ elsewhere. Then $f(x-a)+f(x-b)=0$ at $\frac{a+b}{2}$ whereas it is $1$ at both $a$ and $b$.