## Calculus 8th Edition

The inverse function is given by $$v(m)=c\sqrt{1-\frac{m_0^2}{m^2}}.$$ It says at what speed $v$ is its mass equal to $m$.
First, we have to note that the mass $m$ of the particle is a positive quantity and the speed $v$ is nonnegative. Squaring the given expression we get $$m^2=\frac{m_0^2}{1-\frac{v^2}{c^2}}.$$ By further transformations $$\frac{m_0^2}{m^2}=1-\frac{v^2}{c^2}\Rightarrow \frac{v^2}{c^2}=1-\frac{m_0^2}{m^2}\Rightarrow v^2=c^2\left(1-\frac{m_0^2}{m^2}\right).$$ Now remembering that $v\geq0$ and $c>0$ we have $$v=c\sqrt{1-\frac{m_0^2}{m^2}}.$$ This function says, given the rest mass of the particle $m_0$, at what speed $v$ will its mass be equal to $m$.