Given: #f(x) = 2x^2+3# and #f(v) = 5v+6#
We know that #f(x)# is going to have two inverses; this may cause a problem.
Compute #f^-1(x)# by substituting #f^-1(x)# for every #x# with within #f(x)#:
#f(f^-1(x)) = 2(f^-1(x))^2+3#
The property of a function and its inverse, #f(f^-1(x)) = x#, causes the left side to become #x#:
#x = 2(f^-1(x))^2+3#
Solve for #f^-1(x)#:
#x-3 = 2(f^-1(x))^2#
#(x-3)/2 = (f^-1(x))^2#
#f^-1(x) = sqrt((x-3)/2)# and #f^-1(x) = -sqrt((x-3)/2)#
Substitute #x = f(v)# into the left sides and #x = 5v+6# into the right sides:
#f^-1(f(v)) = sqrt((5v+6-3)/2)# and #f^-1(f(v)) = -sqrt((5v+6-3)/2)#
Use the property #f^-1(f(v)) = v# to make the left sides become v:
#v = sqrt((5v+6-3)/2)# and #v = -sqrt((5v+6-3)/2)#
We are about to square both equations but this will eliminate, therefore, the two equations degenerate into a single equation:
#v^2 = (5v+6-3)/2#
#2v^2= 5v+3#
#2v^2- 5v-3=0#
Factor:
#(2v+1)(v-3) = 0#
#v = -1/2# and #v=3#
