let #x = f^-1(x)# and then substitute everywhere there is an x:
#f(f^-1(x)) = (f^-1(x) + 3)/(f^-1(x) - 2)#
Let left side becomes x by definition:
#x = (f^-1(x) + 3)/(f^-1(x) - 2)#
Add zero to the numerator in the form -2 + 2:
#x = (f^-1(x) -2 + 2 + 3)/(f^-1(x) - 2)#
Regroup:
#x = ((f^-1(x) -2) + (2 + 3))/(f^-1(x) - 2)#
Break into two fractions:
#x = (f^-1(x) -2)/(f^-1(x) - 2) + (2 + 3)/(f^-1(x) - 2)#
Substitute 1 for #(f^-1(x) -2)/(f^-1(x) - 2)# and 5 for #3 + 2#
#x = 1 + 5/(f^-1(x) - 2)#
Subtract 1 from both sides:
#x - 1 = 5/(f^-1(x) - 2)#
Multiply both sides by #(f^-1(x) - 2)/(x - 1)#
#f^-1(x) - 2 = 5/(x - 1)#
Add 2 to both sides:
#f^-1(x) = 5/(x - 1) + 2#
To prove that it is an inverse, one must show that #f(f^-1(x)) = x and f^-1(f(x)) = x#.
#f(f^-1(x)) = ((5/(x - 1) + 2) + 3)/((5/(x - 1) + 2) - 2)#
#f(f^-1(x)) = (5/(x - 1) + 5)/(5/(x - 1)#
#f(f^-1(x)) = ((5/(x - 1) + 5)/(5/(x - 1)))((x - 1)/(x - 1))#
#f(f^-1(x)) = (5+ (x - 1)5)/5#
#f(f^-1(x)) = (5+ 5x - 5)/(5#
#f(f^-1(x)) = (5x)/5#
#f(f^-1(x)) = x#
Half of the proof is done.
Now for #f^-1(f(x))#:
#f^-1(f(x)) = 5/((x + 3)/(x - 2) - 1) + 2#
#f^-1(f(x)) = 5/((x + 3)/(x - 2) - 1)(x - 2)/(x - 2) + 2#
#f^-1(f(x)) = (5(x - 2))/(x + 3 - (x - 2)) + 2#
#f^-1(f(x)) = (5(x - 2))/5 + 2#
#f^-1(f(x)) = x - 2 + 2#
#f^-1(f(x)) = x#
Q.E.D.
