Question #37dc6

1 Answer
Jan 19, 2017

Compute #f^-1(x)#
Use the property #f^-1(f(g(x))) =g(x)#

Explanation:

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#