Please help this question many thanks According to special r

Please help this question many thanks

According to special relativity, the mass of a particle moving with speed v greaterthanorequalto 0 is, = f (v) = m_0/Squareroot 1 - v^2/c^2 where m_0 > 0 is the mass of the particle at rest, and c is the speed of light. Find the domain of f. Show that f is one-to-one. Find the inverse function of f, and explain its meaning.

Solution

f(v) = mo/sqrt( 1- v^2/c^2)

1 - v^2/c^2 >0

v^2/c^2 < 1

v^2 < c^2

(v^2- c^2) <0

-c < v < 2 ; ( - c , c)

To show f is one to one :

find inverse of f(v) :

y = mo/sqrt( 1- v^2/c^2) ; v = mo/sqrt(1 - y^2/c^2)

v^2 = mo^2/( 1- y^2/c^2)

v^2(1 - y^2/c^2) = mo^2

(v^2 - mo^2)= v^2y^2/c^2

y^2 = (v^2 - mo^2)c^2/v^2

y = sqrt( v^2 - mo^2)/vc

f^-1(v) = sqrt( v^2 - mo^2)/v*c