Technically speaking, #f(x) = x^2 - 4# does not have an inverse, because any attempt to invert it yields two equations as follows:
Substitute #f^-1(x)# for every x in #f(x)#
#f(f^-1(x)) = (f^-1(x))^2 - 4#
The property #f(f^-1(x)) = x# makes the left side become x:
#x = (f^-1(x))^2 - 4#
#(f^-1(x))^2 = x + 4#
#f^-1(x) = {(sqrt(x + 4)),(-sqrt(x + 4)):}#
At this point, one should stop and declare the problem unsolvable but let's proceed.
Use the property #f^-1(f(g(x))) =g(x)#
#f^-1(f(g(x))) ={(sqrt(4x^2 - 8x + 4)),(-sqrt(4x^2 - 8x + 4)):}#
Simplify the left side and remove a factor of 2 from the right:
#g(x) ={(2sqrt(x^2 - 2x + 1)),(-2sqrt(x^2 - 2x + 1)):}#
Please observe the what is under the radical is a perfect square, #x^2 - 2x + 1 = (x - 1)^2#:
#g(x) ={(2(x - 1)),(-2(x -1)):}#
#g(x) ={(2x - 2),(-2x + 2):}#
check:
#f(g(x)) = {((2x-2)^2 - 4),((-2x + 2)^2 - 4):}#
#f(g(x)) = {(4x^2- 8x + 4 - 4),(4x^2 - 8x + 4 - 4):}#
#f(g(x)) = {(4x^2- 8x),(4x^2 - 8x):}#
#f(g(x)) = 4x^2- 8x#
Both cases reduce to a single equation that checks.
I think that the desired answer is #g(x) = 2x - 2# but I have no good reason to discard the negative case, #g(x) = -2x + 2#
