Given
#color(white)("XXX")f(x)=2x-7#
and since
#color(white)("XXX")fcircg(x)=f(g(x))=2 * g(x) -7#
But we are also told that
#color(white)("XXX")fcircg(x)=4x+3#
So
#color(white)("XXX")2 * g(x)-7 = 4x+3#
#color(white)("XXX")rarr 2 * g(x)=4x+10#
#color(white)("XXX")rarr g(x)=2x+5#
Therefore
#color(white)("XXX")gcircf(x)=(g(f(x))#
#color(white)("XXXXXXxX")=2 * f(x) +5#
#color(white)("XXXXXXxX")=2(2x-7)+5#
#color(white)("XXXXXXxX")=4x-14+5#
#color(white)("XXXXXXxX")=4x-9#
For simplicity, let #y=gcircf(x)#
and we are asked to find #(gcircf)^(-1)(x)=y^(-1)#
Given an equation with #y# defined in terms of #x#
the easiest way to find the inverse, #y^(-1)# is to solve the equations
to define #x# in terms of #y#
then replace #x# with #y^(-1)# and #y# with #x#:
We have
#color(white)("XXX")y=4x-9#
#color(white)("XXX")4x=y+9#
#color(white)("XXX")x=(y+9)/4#
then doing the replacement
#color(white)("XXX")y^(-1)=(x+9)/4#
that is
#color(white)("XXX")(gcircf)^(-1)(x)=(x+9)/4#
