Let #f:RR->RR, f(x)=5x^3-7# be a function from #RR# to #RR#.
We want to prove that #f# is injective over #RR# and then find its inverse. But we will do it the other way around; ie., we will find the inverse of #f# and this will be sufficient to prove that #f# is injective.
#f(x)=5x^3-7#
so #f(x)+7=5x^3#
so #(f(x)+7)/5=x^3#
so #((f(x)+7)/5)^(1/3)=x#
so #f^-1(x)=((x+7)/5)^(1/3)#
Now we check that #f(f^-1(x))=x# and #f^-1(f(x))=x#
#f(f^-1(x))=5(((x+7)/5)^(1/3))^3-7=cancel5 *(x+cancel7)/cancel5-cancel7=x#
#f^-1(f(x))=((5x^3-7+7)/5)^(1/3)=((5x^3)/5)^(1/3)=x#
So #f^-1# is the inverse of #f#. Now, since #f# has an inverse, it must be bijective, and so it must be injective.