I came across this weird function $$F(x) = \sqrt{1-\cosh(x)}$$ When you study the range of his composite function, you will find that the domain is $\mathrm{D}(F): x = \{0\}$ and the image is $\mathrm{R}(F): y = \{0\}$.
I find this very odd.
What is the reason for this, and how can this be useful in anyway?
Start from the inside of the radical sign and work your way out.
cosh(x) is defined on all of $\mathbb{R}$, and its image is $\{x\in{}\mathbb{R}|1\leq{}x\}$, which we could also write as $[1,\infty{})$.
If $x=0$, then $\cosh{x}=1$ and $F(x)=\sqrt{1-1}=0$. Therefore $0\in{}\text{domain}(F)$ and $0\in{}\text{image}(F)$.
However, notice what happens when cosh(x) outputs something greater than 1. Then the expression $(1-\cosh(x))$ is negative. The square root of a negative number is undefined, so the domain of $F$ cannot include any $x$ which makes $\cosh{x}>1$. As 1 is the absolute minimum of $\cosh{x}$ on $\mathbb{R}$ and it occurs only when $x=0$, any other value of $x$ besides $0$ fails to be in $\text{domain }F$. Thus $\text{domain }F=\{0\}$ and $\text{image }F=\{0\}$.
As for how this function could be useful in an applied situation: I don't believe it could be. It just maps $0$ to $0$ and does nothing else.