Why are the domain and image of $F(x) = \sqrt{1-\cosh(x)}$ only $\{0\}$? [closed]

Solution 1:

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.