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Answer 1 · 2018-01-05T11:30:15

#g(x) = x^2+x-2#

Let #x = f^-1(x)#:

#f(f^-1(x))= 3(f^-1(x))^2+5(f^-1(x))+1#

A property of a function and its inverse is #f(f^-1(x))=x#, therefore, the left side becomes x:

#x= 3(f^-1(x))^2+5(f^-1(x))+1#

Divide both sides by 3:

#1/3x= (f^-1(x))^2+5/3(f^-1(x))+1/3#

Add #a^2-1/3# to both sides:

#1/3x-1/3+a^2= (f^-1(x))^2+5/3(f^-1(x))+a^2#

Using the pattern #(x+a)^2 = x^2+2ax + a^2# we observe that

#2a = 5/3#

#a = 5/6#

#a^2 = 25/36#

#1/3x-12/36+25/36= (f^-1(x)+5/6)^2#

#(f^-1(x)+5/6)^2 = (12x+13)/36#

#f^-1(x) + 5/6 = +-sqrt(12x+13)/6#

#f^-1(x) = +-sqrt(12x+13)/6-5/6#

We know that #g(x) = f^-1(fog(x))#:

#g(x) = +-sqrt(12(3x^4+6x^3-4x^2-7x+3)+13)/6-5/6#

#g(x) = +-sqrt(36x^4+72x^3-48x^2-84x+49)/6-5/6#

#g(x) = +-sqrt((6x^2+6x-7)^2)/6-5/6#

We shall choose the positive value, because #fog(x)# would have a different form if the negative were the correct choice:

#g(x) = x^2+x-7/6-5/6#

#g(x) = x^2+x-2#