Let us understand what we need to do for #(fcircg)(x)# and #(gcircf)(x).# Before moving ahead let us understand evaluating a function.
Example: Evaluating a function
#f(x) = x^2 + 1#
Let us evaluate this function at #x=2#
#f(2) = (2)^2+1##color(red)" Note here " x # #color(red)" is replaced by " 2#
#f(2) =4 + 1#
#f(2) = 5#
If we have to evaluate this function at #x=a# then
#f(a) = (a)^2+1 # #color(red)" Note here " x # #color(red)" is replaced by " a#
If we have to replace it with a function #g(x)# then
#f(g(x)) = (g(x))^2 + 1#
We can see for evaluating we just plug in place of #x#
#f(g(x))# is same as #(fcircg)(x)#
#(fcircg)(x) = f(g(x))#
#(gcircf)(x) = g(f(x))#
Let us use the same thing on our problem
#f(x) =(-x+1)^(1/2)#
#g(x) = x^2 - 8#
#(fcircg)(x)#
# = (-g(x)+1)^(1/2)#
#=(-x^2+8+1)^(1/2)#
#=(-x^2+9)^(1/2)#
#(fcircg)(x)=(-x^2+9)^(1/2)#
Now the other composition
#(gcircf)(x)#
#(gcircf)(x) = (f(x))^2 -8#
# = ((-x+1)^(1/2))^2-8#
#= -x+1 - 8#
# = -x-7#
#(gcircf)(x) = -x-7#
Final answer
#(fcircg)(x)=(-x^2+9)^(1/2)#
#(gcircf)(x) = -x-7#
