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#